Class 9 Physical Science

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement Synopsis

1. Scale is used for measurement of length.
2. The smallest scale division is usually 1 mm or 0.1 cm.
3. While taking the readings of the two sides during measurements of length by an ordinary scale, it is essential to look perpendicularly at the point of reading other wise the reading becomes erroneous, this error, which gives different readings due to different positions of eyes is called parallax error.
4. Common balance is used for the measurement of the mass of an object.
5. In a weight box weights are in the ratio 5 : 2: 2: 1.
6. Sensitivity of a common balance is directly proportional to its capacity of measuring the slightest difference in the mass of an object.
7. Watch is used for measurement of time.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement Short And Long Answer Type Questions

Question 1. Which instrument is used for the measurement of length? Describe this instrument in brief.

Answer: A scale is used for the measurement of length is to be placed on a straight line in such a way.

The scale is generally made up of thin wood, plastic sheet, aluminium or steel. Its length could be 30 cm, 50 cm or 1 m.On the one side of the scale, the centimetre unit is graduated (marked) and on the other side, the unit inch is marked.

1 centimetre is divided in 10 equal parts, every smallest length being 1 millimetre. One inch is divided into 8 or 16 equal parts. In a metre scale, only millimetre and centimetre markings are present whereas in a foot scale, markings are in inch or its smaller parts.

Question 2. How would you measure the length of a straight line by an ordinary scale?

Answer: Suppose the length of a straight line AB is to be measured by an ordinary scale.

The scale is to be placed on the staright line im such a way that the straight line is along the length of the scale and the graduations of the scale are coincidental to the straight line.

Side A of the line is set to coincide with a fixed marking of the scale.

random errors vs systematic errors

In this case, it is prudent if side A coincides with a fixed cm. Now looking perpendicularly, reading of the side B is taken. The difference of the two readings gives the length of AB.

Length of line AB = reading of side B – reading of side A = 4.2 cm – 1 cm = 3.2 cm

Concepts Related to Accuracy and Precision in Measurements for Class 9 Solutions

Question 3. What is meant by parallax error?

Answer:

Parallax Error

While taking the readings of the two sides during measurement of length of a line by an ordinary scale, it is essential to look perpendicularly at the points of the reading.

Otherwise, the reading becomes erroneous. This error, which gives different readings due to different positions of the eyes, is called parallax error.

D is the correct position of the eye.

Question 4. How do you measure the thickness of a page of a book with the help of an ordinary scale?

Answer: It is not possible to measure directly a thickness of less than 1 mm by an ordinary scale.

As the thickness of a page is generally less than 1 mm, the thickness of a page is measured indirectly by an ordinary scale.

Suppose there are n number of pages in a book.

By compressing only the pages (excluding the covers) such that the thickness of layers of air between the pages is excluded, the thickness of the book is measured several times. The mean of the readings is, say, b.

Then, the thickness of each page is \(\frac{b}{n}\).

Question 5. How do you measure the length of a curved line with the help of a thread and an ordinary scale?

Answer: Measurement of the length of a curved line with the help of a thread and a scale:

A long thread is taken and an ink mark is made at one end (A) of the thread.

“examples of systematic errors “

Marked end of the thread is placed at one end point of the curved line and the thread is positioned along the curve till it reaches the other end point.

Another ink mark is made on the thread at position B.

Now, this thread is stretched over a scale and the length between the two marks on the thread is measured. This gives the length of the curved line by the thread and scale method.

 

Question 6. How do you measure the diameter of a wire with the help of an ordinary scale?

Answer: It is not possible to directly measure a length less than 1 mm by scale. As the diameter of a thin wire is less than 1 mm, it is measured indirectly.

The wire is wound several times on a cylinder of small radius so that there is no gap between the rounds. Now with the help of a scale, total length of the wounds on the cylinder is measured.

Several measurements are made to take an average measure. Suppose the mean length of the wire is b. If the number of coils made is n, then diameter of wire is \(\frac{b}{n}\).

Question 7. Why is an ordinary scale made up of wood instead of metal?

Answer: Metal is a good conductor of heat. If the temperature of a metal scale is increased or decreased, its length increases or decreases.

As a result, the length between the two markings changes. Thus, a correct reading is obtained only at the temperature at which the scale was marked. If temperature increases, the correct reading is greater than the reading shown on the scale and if temperature decreases, the correct reading is smaller.

Wood is a bad conductor of heat. So, the increase or decrease of the length of wood with changing temperature is ignored. As a result, the reading shown in the scale may be assumed to be correct. Hence, an ordinary scale is made of wood instead of metal.

Question 8. Which instrument is used for the measurement of time? Describe a pendulum clock.

Answer: A clock is used for the measurement of time.

In a pendulum clock, a metal bob is attached to a metallic rod at one end and the other end is tied firmly with a fixed support and is suspended.

This is a pendulum. The length from the point of suspension to the centre of gravity of the bob is known as the working length. The pendulum oscillates in a periodic motion.

“accuracy in measurement “

There are two hands in the clock, the bigger one indicates minutes and the smaller one indicates hours. The clock works with the help of a spring which stores potential energy when it is wound.

This stored potential energy is the source of energy of a pendulum clock and is converted into kinetic energy. The clock needs to be wound at a regular interval as the clock stops working when the stored potential energy is exhausted.

Sample Solutions from WBBSE Class 9 Physical Science Chapter 1

Question 9. What type of watch is used in swimming and running competitions? What is the problem of using an ordinary clock in these cases?

Answer: Stopwatch is used in swimming and running competitions.

The ordinary clock cannot be started and stopped at will. But the stopwatch may be started and stopped according to our convenience.

Further, one can measure a minimum amount of one second by an ordinary clock whereas with the help of a modern digital stopwatch, a time interval of one-tenth of a second can be measured accurately.

Question 10. Mention different types of clocks.

Answer: We measure time with the help of clocks. The oldest clock is sundial. With the progress of science and technology, different types of clocks have been invented over the time.

Example: Pendulum clock, table clock, wrist watch, electronic digital watch, chronometer, caesium atomic clock etc.

Question 11. What is the inconvenience of expressing your age in seconds?

Answer: Let us suppose that the age of a person is 14 years. If this is converted to seconds, it becomes 14 x 365 x 86400 s = 441504000s.

Hence, if age is expressed in seconds, the number becomes enormous and is inconvenient to handle. For this reason, convenient units like year, month, day are used to express the age of a person.

Question 12. What is a metronome?

Answer: A metronome (electronic metronome) is a modern watch which measures time very accurately. This watch is used during the launching of artificial satellites.

Question 13. Which instrument is used for measurement of the mass of a body? Elaborate the principle used in the measurement of mass by this instrument.

Answer: Common balance is used for measurement of the mass of a body.

While measuring the mass of a body by a common balance, the body is kept in the left pan while some known standard weights are put in the right pan.

When the balance beam comes to a horizontal position, weight of the body in the left pan becomes equal to the standard weight placed in the right pan. This is the principle of measurement of mass.

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WBBSE Solutions for Class 9 History	WBBSE Class 9 History Very Short Answer Questions

 

Question 14. What do you mean by the sensitivity of a common balance? Write down the conditions for a common balance to be sensitive.

Answer: The sensitivity of a common balance is directly proportional to its capacity of measuring the slightest difference in the mass of a body.

The conditions for a common balance to be sensitive are:

1. The arms should be long.
2. The balance beam should be light.
3. The pointer should be long.
4. The centre of gravity of the common balance should be situated very near to the fulcrum.

Question 15. Explain whether the mass of a body or Its weight is measured by a common balance.

Answer: The mass of a body and not its weight is measured by a common balance.

While measuring the mass of a body by a common balance, it is placed in the left pan of the common balance and some standard weights are placed in the right pan.

O is the fulcrum of the balance and A, B are the left and right ends of the common balance, respectively.

if the acceleration due to gravity of the place is g, then in a horizontal state of the common balance, mass of the body x g x AO

= mass of the standard weights x g x OB or, mass of the body

= mass of the standard weights [AO = OB]

“errors of measurement “

Question 16. Why are the masses of the standard weights in a weight box kept in the ratio of 5:2:2:17.

Answer: The masses of the standard weights in a weight box are kept in the ratio of 5:2:2:1 so that any mass between 10 mg and 211.11 g can be measured using them.

Understanding Types of Measurement Errors for Solutions

Question 17. What are the qualities of a good common balance?

Answer: Qualities of a good common balance are:

1. The common balance should be sensitive, that is, it should be able to measure the slightest difference in the mass of a body.
2. The common balance must be strong.
3. The common balance should be accurate, i.e., equal amount of masses put in the two pans should keep it horizontal.
4. The common balance should be stable, i.e., its oscillation should be short-lived.
5. The lengths of the arms and the masses of the two pans should be equal.

Question 18. Why is a sensitive common balance not stable?

Answer: A common balance is considerably sensitive if the centre of gravity of the common balance is situated very near to the fulcrum. Again, the balance is stable if the centre of gravity is well below the fulcrum.

Now, it is not possible to have two opposite conditions in the same common balance simultaneously. This is the reason why a very sensitive common balance is not stable.

Question 19. The lengths of the two arms of a common balance are equal, but the masses of the two scale pans are different. How do you determine the correct mass of a body?

Answer: Let the masses of the left and the right hand sides of the common balance be M₁ and M₂ respectively and the actual mass of the body is m.

If a mass m₁ is put in the right pan by keeping the body in the left pan, the balance becomes horizontal.

∴ M₁ + m = M₂ + m₁…(1)

Again, keeping the body in the right pan, putting a mass m₂ in the left pan makes the balance horizontal.

∴ M₁ + m₂ = M₂ + m……(2)

By subtracting equation (2) from (1), we get

m₁ – m₂ = m₁ – m

or, 2m = m₁ + m₂ or, \(m=\frac{m_1+m_2}{2}\)

Question 20. The lengths of the two arms of a common balance are unequal but the masses of the two scale pans are equal. How would you measure the correct mass of a body?

Answer: Suppose the lengths of the left and the right sides of the common balance are given by x and taken as m.

The lengths of the arms and the masses of the balance becomes horizontal when by keeping two pans should be equal. the body in the left pan, a mass m₁ is put in the right pan.

If the acceleration due to gravity at that place is given by g, then mxg = m₁yg…(1)

Again, the balance becomes horizontal when by keeping the body in the right pan, a mass m₂ is put in the left pan.

∴ m₂xg = m₁yg…….(2)

By adding equations (1) and (2), we get

\(\frac{m \times g}{m_2 \times g}=\frac{m_1 y g}{m y g} \text { or, } \frac{m}{m_2}=\frac{m_1}{m}\)

or, \(m^2=m_1 m_2 \quad or, m=\sqrt{m_1 m_2}\)

Question 21. What is a measuring cylinder? How do you measure the volume of a liquid with the help of a measuring cylinder? Or, How do you measure the volume of tea in a tea cup?

Answer: A measuring cylinder is a vessel made up of a strong glass closed at one end, with uniform cross section. This is used for the measurement of volume.

It has graduation marks in millilitres (cm³) on its exterior surface, along its length. Each cm³ is further divided into 5 or 10 equal parts.

To measure the volume of a liquid, it is poured inside the measuring cylinder and the reading of its upper surface is taken. While taking a reading, eyes have to be placed in a perpendicular direction to the point of observation.

For those liquids which wet the glass (like water), for which the upper surface is concave inside the cylinder, reading of the lowest portion of the concave surface is to be taken.

For those liquids which do not wet the glass (like mercury) and for which the upper surface is convex inside the cylinder, reading of the highest portion of the convex surface is to be taken.

Question 22. How do you measure the volume of an irregular solid body with the help of a measuring cylinder?

Answer: To measure the volume of an irregular solid body with the help of a measuring cylinder, a measuring cylinder is taken whose inner volume is 4 to 5 times larger than the volume of the body and the area of its cross-section is such that the body may easily be entered inside the cylinder.

A liquid, in which the body does not dissolve or float or react chemically, is taken in the cylinder up to a certain height. Suppose the volume of the liquid in this condition is V₁ cm³.

Now, a wax-coated, thin and strong thread is fastened with the body and the body is slowly immersed completely inside the liquid.

If the volume of the liquid, with the solid inside it is V₂ cm³, then the volume of the body (V₁ – V₂) cm³.

If the volume of the immersed thread is deducted from this reading, the accurate volume of the body can be obtained.

WBBSE Class 9 Measurement of Physical Quantities Solutions

Question 23. What are the precautionary measures to be taken while measuring the volume of an irregular solid body by a measuring cylinder?

Answer: Required precautionary measures are:

1. A liquid, which neither dissolves nor reacts chemically with the body, has to be taken inside the measuring cylinder.
2. The solid body has to be immersed very slowly in the liquid so that no liquid splashes out.
3. The thread by which the solid body is wound has to be coated with wax so that the thread does not soak water.
4. Reading has to be taken in such a way that there is no parallax error.
5. No bubble of the body should stick to the wax is attached to the stone and it is dipped surface of the cylinder inside the liquid.

Question 24. How do you determine the rate of fall of water from a tap with the help of a volume-measuring cylinder and a stopwatch?

Answer: Rate of fall of water from a tap can be determined with the help of a volume-measuring cylinder and a stopwatch. Suppose water is falling from a tap at a uniform rate.

A dry and empty cylinder is held below a running tap and the stopwatch is switched on immediately. After accumulation of some water, the cylinder is removed from under the tap and the stopwatch is stopped simultaneously.

Readings of the measuring cylinder and the stopwatch gives the volume of water collected during a period. Suppose, V₁ volume of water is collected during a time t₁.

∴ rate of fall of water, \(W_1=\frac{V_1}{t_1}\)

In this way, several readings are to be taken and the mean of these gives the rate of fall of water which is reasonably error-free.

Unit of rate of fall of water is cm³/s if units of volume of water and time are taken as cm³ and second.

Question 25. How do you measure the volume of piece of stone which does not fit into a measuring cylinder?

Answer: A glass cylinder with an attached side pipe and with sufficient cross-sectional area is taken so that the piece of stone may fit into it.

If water is poured continuously in the above cylinder, it flows out by the side pipe at a particular time.

Now, if the pouring of water is stopped, water level stands just beneath the level of the side pipe.

The measuring cylinder is kept just below the side pipe. Now a thin and strong thread coated with wax is attached to the stone and it is dipped slowly into the water of the glass cylinder.

The stone displaces an equal amount of water of its volume which flows out by the side pipe into the measuring cylinder. The volume of water that is stored in the measuring cylinder is the volume of the piece of stone.

If the volume of the submerged thread is deducted from the last reading, the actual volume of the stone can be obtained.

Question 26. How do you measure the density of a solid body with the help of a common balance and a measuring cylinder?

Answer: At first, the mass of the solid body is measured by the common balance. Suppose, m is the mass of the body. Now a measuring cylinder is taken whose area of cross section is sufficient for entry of the solid body inside the cylinder.

A particular liquid, that neither dissolves nor reacts chemically with the solid, is taken in the cylinder. Suppose, V₁ is the volume of the liquid in the cylinder.

Now a thin, strong and wax-coated thread is fastened tightly around the body and is slowly dipped into the liquid of the cylinder.

Let the volume of the liquid with the solid be V₂.

Hence, the volume of the body V₂ – V₁. Here, if the volume of the submerged thread is deducted from the reading V₂-V₁, actual volume of the body is obtained.

∴ Density of the body, \(d=\frac{m}{V_2-V_1}\)

Question 27. How do you measure the density of a liquid with the help of a common balance and a measuring cylinder?

Answer: At first, the mass of the measuring cylinder in a dry condition is measured by a common balance.

Suppose, the mass is m₁.

Next the cylinder is nearly half-filled with the liquid whose density is being measured here.

Volume of the liquid is determined from the scale calibrated on the cylinder. Suppose, V is the volume of the liquid. The mass of the cylinder filled with the liquid is then measured by a common balance.

Suppose, m₂ is the mass in this case.

∴ Mass of the liquid = m₂ -m₁

“types of errors in measurement “

∴ Density of the liquid = \(\frac{m_2-m_1}{V}\)

Question 28. How do you measure the volume of one drop water of a dropper with the help of a volume measuring cylinder?

Answer: 200 to 300 drops of water are dropped in a dry and empty measuring cylinder using the dropper. Volume of this quantity of water is measured.

Let the volume of water be V cm³ and let n drops of water have been dropped.

∴  Volume of 1 drop of water = \(\frac{V}{n} \mathrm{~cm}^3\)

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement Very Short Answer Type Questions Choose The Correct Answer

Question 1. Smallest distance that can be measured by an ordinary scale is

1. 0.1 cm
2. 0.01 cm
3. 0.001 cm
4. 0.2 cm

Answer: 1. 0.1 cm

Question 2. The length which cannot be measured by an ordinary scale is

1. 4.2 cm
2. 2.13 cm
3. 7.7 cm
4. 9.5 cm

Answer: 2. 2.13 cm

Question 3. Lengths of the two arms of a common balance are equal but the two pans are of different masses. A body is found to be 20 g and 21 g when put in the left pan and the right pan, respectively. The actual mass of the body is

1. 20.2 g
2. 20.4 g
3. 20.5 g
4. 20.6 g

Answer: 3. 20.5 g

Question 4. In the weight box of a common balance, the weights are taken in the ratio of

1. 5:3:2:1
2. 5:4:2:1
3. 5:2:2:1
4. 5:3:3:1

Answer: 3. 5:2:2:1

Question 5. The physical quantity which is measured by a common balance is

1. Volume
2. Mass
3. Weight
4. Force

Answer: 2. Mass

Question 6. Which of the following instruments can be used to measure the volume of a wooden block of irregular shape?

1. common balance
2. Measuring cylinder
3. Metre scale
4. Stopwatch

Answer: 2. Measuring cylinder

Question 7. Which of the following is not a prerequisite for sensitivity of a common balance?

1. Beam of the balance should be long
2. The balance should be light
3. The pointer should be small in size
4. The centre of gravity should be very close to the fulcrum

Answer: 3. The pointer should be small in size

Question 8. Which of the following instruments does not function in a place where there is no gravity?

1. Spring balance
2. Common balance
3. Ordinary scale
4. Both A and R

Answer: 4. Both A and R

Question 9. When mass of a body is measured on the earth surface by a common balance its value becomes m kg. When the measurement is done on the moon’s surface it is m’kg.If gravitational force on the surface of the moon is only 1/6 as the gravitational force on the earth, then

1. m = m’
2. m = 1/6m’
3. m = 6m’
4. m = m’ = 0

Answer: 1. m = m’

Question 10. Time taken to complete one oscillation by a second pendulum is

1. 1s
2. 2s
3. 3s
4. 1/2 s

Answer: 2. 2s

Question 11. Which of the following can be used to measure the area of a metallic strip with irregular shape?

1. Ordinary scale
2. Common balance
3. Graph paper
4. String

Answer: 3. Graph paper

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement Answer In Brief

Question 1. What is the use of a chronometer watch?

Answer: A chronometer watch gives the correct time at Greenwich, London and this time has been adopted as a global standard time.

Question 2. Name an instrument by which a length of 0.01 cm can be measured accurately.

Answer: A length of 0.01 cm can be measured accurately by using slide calipers.

Question 3. Name an instrument by which a length of 0.001 cm can be measured accurately.

Answer: A length of 0.001 cm can be measured accurately by using a screw gauge.

Question 4. What is the effective length of a pendulum clock?

Answer: The effective length of a pendulum clock is. measured from the point of suspension to the centre of mass of its bob.

Question 5. What type of energy is stored in the spring of a pendulum clock?

Answer: Potential energy is stored in the spring of a pendulum clock.

Important Concepts in Measurement of Physical Quantities for Class 9

Question 6. Which instrument is used to measure the mass of a body?

Answer: Common balance is used to measure the mass of a body.

Question 7. What is the ratio in which the weights are kept in a weight box?

Answer: Weights are kept in the ratio of 5:2:2:1 in a weighing box.

Question 8. What is least count?

Answer: The minimum measurement which can be performed by using an instrument in measuring a physical quantity, is called least count of this instrument.

Question 9. What is the minimum length that can be measured by a metre scale correctly?

Answer: The minimum length that can be measured by a metre scale is 0.1 cm or 1 mm.

Question 10. What is the maximum length that can be measured by a metre scale accurately?

Answer: 1 m or 100 cm is the maximum length that can be measured by a metre scale accurately.

Question 11. What is the maximum time that can be measured by a wall clock?

Answer: The maximum time that can be measured by a wall clock is 12 hours.

Question 12. What is the maximum mass that can be measured by a common balance using its weight box?

Answer: The maximum mass that can be measured by a common balance by using its weight box is 211.11 g.

Question 13. What is the use of rider of a common balance?

Answer: Rider is used in a common balance To measure a mass of less than 10 mg.

Question 14. What are the minimum and maximum measurement that can be measured by a measuring cylinder of volume 100 mL and 10 mL?

Answer: The minimum and maximum measurement of volume that can be measured by a measuring cylinder of 100 mL and 10 mL are 1 mL and 0.1 mL; 100 mL and 10 mL respectively.

Question 15. For measuring 72.05 g mass of a body what are the different weights in grams and milligrams to be taken from the weight box?

Answer: The weights that can be used are 50g, 20g, 2g and 50 mg.

Question 16. What can be used to measure the length of a curve line?

Answer: The length of a curve line can be measured by using a thread and a meter scale.

Question 17. What can be done to level the base of a common balance?

Answer: levelling screw can be used to balance the base of a common balance.

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement Fill In The Blanks

Question 1. ________ cylinder and stopwatch can be used to measure the rate of flow of water from a tap.

Answer: Measuring

Question 2. _______ time is the minimum possible time that can be measured by a wrist watch.

Answer: 1s

Question 3. ________ is the minimum mass that can be measured by a common balance.

Answer: 10 mg

Question 4. Very small time intervals are measured by _______

Answer: Stopwatch

Question 5. Number of hour hand in stop watch is _______

Answer: Zero

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement State Whether True Or False

Question 1. Common balance is used for the measurement of the weight of an object.

Answer: False

Question 2. Determination of area of a regular shaped sheet can not be done by using a graph paper.

Answer: False

Question 3. Digital clock is used for accurate measurement of time.

Answer: True

Question 4. Common balance can work even in the place where there is no gravity.

Answer: True

Question 5. Volume of an irregular shaped body can be measured by a measuring cylinder.

Answer: True

Question 6. Mass of a body can be measured by a spring balance even in a place where there is no gravity.

Answer: False

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement Numerical Examples

Percentage of error in the measurment of length by a metre scale is

= \(\frac{\text { value of the smallest division of the scale }}{\text { measured length }}\) x 100%

Then actual mass of the body m = √m1m2.

“types of errors in measurement “

A common balance have equal lengths of arms but to different masses of pans.

If m₁ and m₂ be the measured of a body when it is kept on right and left pan respectively

Percentage of error in the measurement of but of different masses of pans. If m1 and m2 be the measured masses of a body when it is kept on right and left pan respectively.

Then actual mass of the body \(m=\frac{m_1+m_2}{2}\)

Vernier constant (c) = length of 1 main scale division-length of 1 vernier scale division measurement of length (l)= main scale reading + vernier scale reading x vernier constant.

Question 1. The two arms of the balance beam of a common balance are unequal but the masses of the two scale pans are equal. When a body is weighed at first in the left pan and then in the right pan, 8 g and 12.5 g are obtained respectively as masses. What is the real mass of the body?

Answer: If the masses of the body are m₁ and m₂ in the two cases,

m₁ = 8g and m₂ = 12.5 g

∴ real mass of the body,

\(m =\sqrt{m_1 m_2}=\sqrt{8 \mathrm{~g} \times 12.5 \mathrm{~g}}\)

= \(\sqrt{100 \mathrm{~g}^2}=10 \mathrm{~g}\)

Question 2. 5.00 cm length is measured by a scale whose vernier constant is 0.01 cm. Find percentage of error in this measurement.

Answer: Maximum possible error in the measurement by this scale is = 0.01 cm.

∴ Percentage of error = \(\frac{0.01}{5.00} \times 100 \%=0.20 \%\)

Practice Problems for Chapter 1 Measurement Errors

Question 3. In a measuring cylinder 1 ml is divided into 10 equal divisions. Volume of liquid is measured by it and the reading is 25 ml. Find percentage of error in this measurement.

Answer: Here least count of the measuring cylinder is = \(\frac{1}{10}\) mL = 0.1 mL

Measured volume of the liquid = 25 mL.

“types of errors in measurement “

∴ Percentage of error in measurement of volume \(\frac{0.1}{25}\) = 0.4

Question 4. The two arms of a common balance are equal but masses of the two pans are different. The measured mass of a body when placed in the left pan is 10 g and that of the body when placed in the right pan is 10.2 g. Find exact mass of the body.

Answer: The measured mass of a body in the two cases are m₁ = 10 g and m₂ = 10.2 g.

∴ Exact mass of the body

\(m=\frac{m_1+m_2}{2}=\frac{10+10.2}{2}=10.1 \mathrm{~g}\)

Chapter 1 Topic C Measurement Of Different Physical Quantities And Errors In Measurement

1. Match the physical quantities in column A with their respective units in coloumn B.

Answer: 1. D, 2. A, 3. B, 4. C

2.

Answer: 1. C, 2. A, 3. D, 4. B

Chapter 1 Topic B Dimension Synopsis

Dimension of physical quantity is the power or index to which fundamental units like mass, length, time, etc are raised to express that quantity.

The dimensional formula is the relationship between a physical quantity and the dimension of its fundamental units.

Example: Dimensional formula of force = MLT^-2.

Dimensional equation is the equation by which any physical quantity is expressed in terms of dimensional formula.

Example: Dimensional equation of force is [F]= MLT^-2.

1. The value of a constant in a dimensional equation can not be determined by using dimensional analysis.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Example: The time period of a simple pendulum is \(T=k \sqrt{\frac{1}{g}}\)

Where l is length, g is acceleration due to gravity and k is a constant.

We can not determine the value of the constant k by using dimensional analysis.

2. The relation containing a constant which is not dimensionless can not be established by dimensional analysis.

Example: We can not find the law of mm2 from dimensional gravitation \(F=G \frac{m_1 m_2}{d^2}\) analysis because G is not dimensionless.

Class 9 Physics Chapter 1 Dimensions

Chapter 1 Topic B Dimension Short And Long Answer Type Questions

Question 1. What do you mean by the dimension of a physical quantity? What is dimensional formula?

Answer:

1. Dimension of a physical quantity is the power or index to which fundamental units like mass, length, time, etc are raised to express that quantity.
2. Dimensional formula is the relationship between a physical quantity and the dimension of its fundamental units.

Question 2. What is dimensional equation?

Answer:

Dimensional equation is the equation by which any physical quantity is expressed in terms of dimensional formula.

Question 3. How are the dimensional formulae of fundamental physical quantities expressed

Answer:

There are seven fundamental units in SI. These are length, mass, time, temperature, electric current, luminous intensity, and quantity of matter, units of which are considered as fundamental units.

These dimensional formulas are denoted as L, M, T, Θ, I, J, and N, respectively.

Question 4. Write down the dimensional formula and dimensional equation of velocity.

Answer:

Velocity is the rate of change of displacement with time.

∴ \(\text { unit of velocity }=\frac{\text { unit of displacement }}{\text { unit of time }}\)

= \(\text { unit of length } \times(\text { unit of time })^{-1}\)

∴ dimension of velocity is 1 in length and -1 in time.

Dimensional formula of speed is LT^-1.

If velocity is expressed by v, then dimensional equation of velocity is [v] = LT^-1.

WBBSE Class 9 Dimensions Solutions

Question 5. Write down the dimensional formula and dimensional equation of force.

Answer:

Force = mass x acceleration

∴ unit of force

= unit of mass x unit of acceleration

= \(\text { unit of mass } \times \frac{\text { unit of length }}{(\text { unit of time })^2}\)

= \(\text { unit of mass } \times \text { unit of length }\) x \(\times(\text { unit of time })^{-2}\)

∴ dimension of force is 1 in mass; 1 in length and -2 in time.

Dimensional formula of force is MLT^-2.

If force is expressed by F, then dimensional equation of force is [F] = MLT^-2.

Question 6. Write down the dimensional formula and dimensional equation of a physical quantity x whose dimension is 0 in mass, 1 in length, and -2 in time.

Answer:

Since the dimension of the quantity is 0 in mass, 1 in length, and -2 in time, the dimensional formula of the quantity is M⁰LT^-2.

Dimensional equation of the quantity is [x] = M⁰LT^-2.

Question 7. The dimensional equation of a physical quantity x is [x] = M^-2L³T^-1. Write down the dimension and the dimensional formula of the quantity.

Answer:

Here, [x] = M^-2L³T^-1.

∴ Dimension of the quantity is -2 in mass, 3 in length, and -1 in time.

Dimensional formula of the quantity is M^-2L³T^-1

Class 9 Physics Chapter 1 Dimensions Long Answer Questions

Question 8. Show that plane angle is a dimensionless physical quantity.

Answer:

Magnitude of an angle in radian

\((\theta)=\frac{\text { length of arc }(s)}{\text { radius of the circle }(r)}\)

∴ Dimentional formula of angle

= \(\frac{\text { Dimensional formula of length of arc }}{\text { Dimensional formula of radius }}\)

= \(\frac{L}{L}=L^0\)

∴ Plane angle is a dimensionless physical quantity.

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WBBSE Solutions for Class 9 History	WBBSE Class 9 History Very Short Answer Questions

 

Chapter 1 Topic B Dimension Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following quantities has a unit but no dimension?

1. Strain
2. Atomic weight
3. Angle
4. None of the above

Answer: 3. Angle

Question 2. Dimensional formula of force is

1. MLT^-3
2. M²LT^-2
3. ML^-1T^-2
4. MLT^-2

Answer: 4. MLT^-2

Question 3. Solid angle has

1. Both dimension and unit
2. Only dimension but no unit
3. Only unit but no dimension
4. Neither dimension nor unit

Answer: 3. Only unit but no dimension

Question 4. Strain has

1. Both dimension and unit
2. Only dimension but no unit
3. Only unit but no dimension
4. Neither dimension nor unit

Answer: 4. Neither dimension nor unit

Practice Questions for Chapter 1 Dimensions

Question 5. Dimension of time in the dimensional formula of power is

1. -1
2. -2
3. -3
4. -4

Answer: 3. -3

Question 6. Dimension of absolute humidity is ML^-3. It is similar to the dimension of

1. Mass
2. Density
3. Specific gravity
4. Area

Answer: 2. Density

Question 7. Dimensional formula of a dimensionless physical quantity is

1. M⁰L⁰T⁰
2. MLT
3. M²LT^-2
4. None of these

Answer: 1. M⁰L⁰T⁰

Class 9 Physics Chapter 1 Dimensions very short Answer questions

Question 8. Dimensional formula of weight is

1. MLT^-2
2. ML^-1T^-2
3. M^-1LT^-2
4. MLT^-3

Answer: 1. MLT^-2

Question 9. Which two physical quantities of the following have same dimensional formula?

1. Velocity, speed
2. Displacement, work done
3. Force, momentum
4. Velocity, acceleration

Answer: 1. Velocity, speed

Question 10. Dimensionless physical quantity is

1. Mass
2. Specific heat
3. Weight
4. Atomic weight

Answer: 4. Atomic weight

Question 11. ML^-1T^-2 is dimensional formula of

1. Acceleration
2. Density
3. Force
4. Pressure

Answer: 4. Pressure

Question 12. Dimensional formula of surface tension is

1. MLT^-2
2. MT^-2
3. MT^-1
4. LT^-1

Answer: 2. MT^-2

Question 13. In A=B+C equation

1. Dimension of A and C are same but that of B is different
2. Dimension of A, B, and C are equal
3. A, B, and C all have different dimension
4. Dimension of a and b are same but that of c is different

Answer: 2. Dimension of A, B, and C are equal

Question 14. in (P+a/V²) (V-b) = RT equation dimensional formula is

1. MLT^-2
2. ML²T^-1
3. ML⁵T^-2
4. ML³T^-2

Answer: 3. ML⁵T^-2

Chapter 1 Topic B Dimension Answer In Brief

Question 1. Give example of two physical quantities which do not have any dimension or unit.

Answer: Atomic mass and specific gravity are two physical quantities which do not have any dimension or unit.

Question 2. Give an example of a dimensionless physical quantity which has unit.

Answer: Angle is a physical quantity which is dimensionless but has unit (radian).

Question 3. Write dimension of acceleration.

Answer: Acceleration has dimension 1 in length, 0 in mass, and -2 in time.

Question 4. Give an example of a physical quantity which have unit but no dimension.

Answer: Angle is such a physical quantity which have unit (radian) but no dimension.

Question 5. Give example of two physical quantities which have no unit and dimension.

Answer: Atomic weight and specific density are two such physical quantities which have no unit and dimension.

Important Concepts in Dimensions for Class 9

Question 6. Write dimensional formula of velocity.

Answer: Dimensional formula of velocity is LT^-1.

Question 7. Which physical quantity have dimensional formula ML^-1T^-2?

Answer: Pressure is a physical quantity which have dimensional formula ML^-1T^-2. (strain also have dimensional formula ML^-1T^-2).

Question 8. A physical quantity have unit °F^-1. Write dimensional formula of the physical quantity.

Answer: Dimensional formula of the physical quantity is K^-1.

Chapter 1 Topic B Dimension Fill In The Blanks

Question 1. _____________ of two sides of a correct equation or relation is always same.

Answer: Dimension

Question 2. Angle is a ____________ physical quantity.

Answer: Dimensionless

Question 3. Nuclear density is a dimensionless quantity. Its dimensional formula is __________

Answer: M⁰L⁰T⁰

Question 4. All the terms on the two sides of a physical equation must have the _______ dimension.

Answer: Same

Question 5. Dimension of atomic weight is _______

Answer: M⁰L⁰T⁰

Question 6. _________ is such a physical quantity whose dimension is T^-1

Answer: Frequency

Question 7. Pressure and _________  both have the dimensional formula.

Answer: Stress

Chapter 1 Topic B Dimension State Whether True Or False

Question 1. Dimensional formula is the relationship between a physical quantity and the dimension of its fundamental units.

Answer: True

Question 2. Dimensional formula of kinetic energy is ML²T^-2.

Answer: True

Chapter 1 Topic B Dimension Numerical Examples

Key Information:

In any mathematical equation involving physical quantities, each term on either side of the equation must have the same dimension.

Question 1. The two arms of the balance beam of a common balance are equal but the masses of the two scale pans are different. When a body is weighed first in the left pan and then in the right pan, 10 g and 10.2 g are obtained respectively as masses. What is the real mass of the body?

Answer:

If the masses of the body are m₁ and m₂ in the two cases, m₁ = 10g and m₂ = 10g.

∴ Real mass of the body,

\(m=\frac{m_1+m_2}{2}=\frac{10 \mathrm{~g}+10.2 \mathrm{~g}}{2}=10.1 \mathrm{~g}\)

Question 2. In P = \(\frac{W}{t}\) equation p, W, and t are power, work done, and time. Find dimensional formula of P.

Answer:

Here P = \(\frac{W}{t}\)

∴ Dimensional formula of P is

\([P]=\left[\frac{W}{t}\right]=\frac{M L^2 \mathrm{~T}^{-2}}{\mathrm{~T}}=\mathrm{ML}^2 \mathrm{~T}^{-3}\)

Concepts Related to Fundamental and Derived Dimensions for Class 9 Solutions

Question 3. Buoyancy of a body is equal to the weight of the displaced liquid by the body. By using this relation find dimensional formula of buoyancy.

Answer:

Buoyancy = weight of the displaced liquid

= v.ρ.g[where v is the volume of the displaced liquid of density ρ and g is acceleration due to gravity]

∴ Dimensional formula of buoyancy = [vρg]=L³.ML^-3.LT^-2 = MLT^-2

Question 4. Velocity of a particle is \(v=a t^2+\frac{b}{t+c}\) where t is time. Find dimensional formula of a, b, and c.

Answer:

Here, the equation is \(v=a t^2+\frac{b}{t+c}\)

∴ Dimensional formula of the left hand side is [v] = LT^-1

Now [at²] = [a]T⁺² = LT^-1

∴ [a] = LT^-3

Again, \(\frac{[b]}{[t]+[c]}=\mathrm{LT}^{-1} \text { or, } \frac{[b]}{\mathrm{T}}=\mathrm{LT}^{-1}\)

∴ Dimensions of a, b, and c are LT^-3, L, and T.

Question 5. Vander Waals equation is \(\left(p+\frac{a}{v^2}\right)(v-b)\) = RT. Find dimension of a and b.

Answer:

\(\frac{a}{v^2}\) is added to pressure. So its dimension would be the same as that of pressure P.

∴ \({[P]=\left[\frac{a}{v^2}\right] \text { or, } \mathrm{ML}^{-1} \mathrm{~T}^{-2}=\frac{[a]}{\left(\mathrm{L}^3\right)^2}}\)

∴ \({[a]=M L^{6-1} \mathrm{~T}^{-2}=\mathrm{ML}^5 \mathrm{~T}^{-2}}\)

Again b is subtracted from v. So its dimension would be the same as that of volume v.

∴ [v] = [b]

or, [b] = L³

Question 6. In \(s=\frac{1}{2} a t^2\) equation, s, a and t are displacement, acceleration, and time. By using dimensional analysis check whether the equation is correct or not.

Answer:

The equation is \(s=\frac{1}{2} a t^2\)

∴ Dimension of the left hand side is = [s] = L and dimension of the right-hand side is

= \(\left[\frac{1}{2} a t^2\right]=\mathrm{LT}^{-2} \cdot \mathrm{T}^2=\mathrm{L}\)

∴ Dimension of the L.H.S= dimension of the R.H.S.

∴ The equation is dimensionally correct.

Question 7. In the K = It + mt² + nt³ equation t and K are expressed in second and m/s respectively. Find SI unit of I and n, where l, m, and n are three physical quantities.

Answer:

According to the principle of dimensional homogeneity, in any mathematical expression or equation involving physical quantities, each term on either side of the equation must have the same dimension.

Here, dimension of K is = [K] = LT^-1

∴ Dimension of It is [It] = [K] or, [I]T=LT^-1

∴ [l] = LT^-2

∴ Unit of in SI is m · s^-2.

Again, dimension of nt³ is [nt³] = [n]T³ = LT^-1

∴ \([n]=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^3}=\mathrm{LT}^{-4}\)

∴ Unit of n in Sl is m · s^-4.

Study Guide for Class 9 Measurement and Dimensions Questions

Question 8. Dimensional formula of three physical quantities A, B, and C are MLT^-2, ML²T^-3, and LT^-1 respectively. Show that the equation A = \(\frac{B}{C}\) is dimensionally correct.

Answer: For the given equation, dimensional formula of the left hand side is

[A] = MLT^-2 and dimensional formula of the right-hand side is

= \(\left[\frac{B}{C}\right]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-3}}{\mathrm{LT}^{-1}}=\mathrm{MLT}^{-2}\)

∴ \({[A]=\left[\frac{B}{C}\right] . \quad therefore[A]=\left[\frac{B}{C}\right]}\) equation is dimensionally correct.

Sample Solutions from WBBSE Class 9 Physical Science Chapter 1

Question 9. Time period of a simple pendulum is T = \(2 \pi \sqrt{\frac{1}{g}}\), where l is effective length of the pendulum and g is acceleration due to gravity. Using dimensional analysis verify whether the equation is correct or not.

Answer:

Here time period of a simple pendulum is T = \(2 \pi \sqrt{\frac{1}{g}}\)

∴ Dimension of the left hand side of the equation is [T] = T and dimension of the right hand side is

= \(2 \pi \sqrt{\frac{1}{g}}=\left[L^{\frac{1}{2}} \cdot L^{-\frac{1}{2}} T^{+\frac{2}{2}}\right] \text {, as }[g]=L T^{-2}\)

= \(L^{\frac{1}{2}}-\frac{1}{2} \cdot T^{+1}=T\)

∴ T= \(2 \pi \sqrt{\frac{1}{g}}\) equation is dimensionally correct.

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle Synopsis

The force acting normally on unit area of a surface is called pressure. It is given by

\(P=\frac{\text { normal force }}{\text { area }}=\frac{F}{A}\)

In CGS system and SI, units of pressure are dyn/cm² and N/m² (or Pa), respectively and they are related as 10 dyn/cm² = 1 N/m² = 1 Pa.

Those materials whose molecules move freely past one another (or the materials which flow) are called fluid. Liquids and gases are fluid.

Pressure Of A Liquid at a point inside the liquid is defined as the force applied by the liquid in a perpendicular direction on a surface of unit area surrounding that point.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Thrust Of A Liquid is defined as the force applied by the liquid in a perpendicular direction on a surface adjacent to the liquid. It is given by

thrust(F) = pressure(P) x area(A)

Pressure due to the liquid at a depth of h and density d is given by P = hdg

and the total pressure is given by P’ = P_a + hdg

where Pa denotes the atmospheric pressure.

Height of water barometer under standard atmospheric pressure is 10.336 m (taking, g = 980 cm/s²).

what is archimedes principle

The force per unit area applied perpendicularly against a surface by the weight of the atmosphere at a point is known as atmospheric pressure.

The pressure of a mercury column of 76 cm at a temperature of 0°C at sea level and at 45° latitude is known as the standard atmospheric pressure.

Standard atmospheric pressure = 1.01325 x 10⁶ dyn/cm² = 101325 Pa .

Barometer is the instrument by which atmospheric pressure is measured.

Siphon is an arrangement that carries a liquid from a higher level up and over a barrier and then to a lower level. Here, the flow is maintained by gravity and atmospheric pressure as long as tube in the arrangement remains full.

Density of a substance is defined as its mass per unit volume.

Specific Gravity of a material is defined as the ratio of its mass to the mass of an equal volume of water at 4°C. It is a dimensionless and unitless quantity.

The upward thrust exerted by a liquid at rest on a body partially or fully immersed in it is known as buoyant force and the phenomenon is known as buoyancy.

Buoyant force is given by F_b = vdg, where v is the volume of the immersed portion, d is the density of the liquid and g is acceleration due to gravity.

Archimedes’ Principle:

When a body is partially or fully immersed in a stationary liquid or gaseous material, there is an apparent reduction in the weight of the body. This apparent reduction is equal to the weight of the liquid or gaseous material displaced by the body.

Archimedes’ principle is not applicable for a freely falling body or for an artificial satellite.

Floatation And Submersion Of A Body:

Suppose, weight of a body in air = W and weight of the displaced water by the fully submerged body = W₁.

1. If W > W₁, the body moves downward inside the liquid.
2. If w = w₁, the body floats completely immersed inside the liquid.
3. If W < W₁, the body tends to move upward and floats in an equilibrium condition.

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle Short And Long Answer Type Questions

Question 1. What is pressure? Write its mathematlcal expression.

Answer:

The force acting normally on unit area of a surface is called pressure.

If a force(F) is applied in a perpendicular direction on an area (A) of a surface, then pressure, P = \(\frac{F}{A}\)

Question 2. What are the units of pressure in CGS system and SI? Establish a relationship between them.

Answer:

Units of pressure in CGS system and in SI are dyn/cm² and N/m² or pascal (Pa), respectively.

The relationship between them is given by

applications of archimedes principle

\(1 \mathrm{~N} / \mathrm{m}^2=\frac{10^5 \mathrm{dyn}}{10^4 \mathrm{~cm}^2}=10 \mathrm{dyn} / \mathrm{cm}^2\)

Question 3. What do you mean by a fluid?

Answer:

A material which can flow is called a fluid, The molecules of liquid and gaseous materials can move freely past one another. Hence, liquid and gas are called fluids.

Question 4. With the help of a simple experiment, show that liquid exerts pressure.

Answer:

A vessel is taken. A hole is made on the surface of thevessel and a cork is used to plug it. The vessel is filled up with water and then the cork is taken out. It is found that water is coming out (gushing out) with great speed.

Now if the hole is blocked by hand, it is found that water has stopped coming out. But to make this happen, one has to use a good amount of force. The reason is that water applies a force on the wall of the vessel and to stop that flow, equal force has to be applied in the opposite direction.

The force that water applies on an unit area of the wall in a perpendicular direction is the pressure of water. It can be inferred from this experiment that liquid exerts pressure.

Question 5. What do you mean by pressure and thrust of a liquid? Thrust of a liquid can be categorised as which type of quantity?

Answer:

Pressure:

Pressure of a liquid at a point inside the liquid is defined as the force applied by the liquid in a perpendicular direction on a surface of unit area surrounding that point.

Thrust:

Thrust of a liquid is defined as the force applied by the liquid in a perpendicular direction on a surface adjacent to the liquid. Suppose the liquid exerts a force (F) in a perpendicular direction on an area (A) around a point inside the liquid.

∴ Pressure of liquid at that point,

P = \(\frac{F}{A}\) or, F = PA

i.e., Thrust of liquid = pressure x area.

archimedes principle of buoyancy

Thrust is a kind of force. So, thrust is a vector quantity.

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WBBSE Solutions for Class 9 History	WBBSE Class 9 History Very Short Answer Questions

 

Question 6. Write down the characteristics of the pressure of a liquid.

Answer:

Characteristics Of The Pressure Of A Liquid Are As Follows:

1. Pressure of a liquid at a point inside the liquid depends on the depth and not on the shape of the vessel.
2. At any point inside a stationary liquid, a liquid exerts the same amount of pressure in every direction.
3. Pressure of a liquid is same at every point on any horizontal plane inside a stationary liquid.

Question 7. Why is bottom of the dam made thicker than the top?

Answer:

Water pressure increases with the increase of depth. Hence, the lateral pressure of water is maximum at the bottom. To tolerate such huge pressure, bottom of a dam is made thicker than the top.

Question 8. Establish the mathmatical expression  for pressure inside a liquid of depth h.

Answer:

A vessel contains a liquid of density d. We have to calculate pressure of liquid at a point M at a depth h from the upper surface of the liquid. A circular surface of area A around the point M is imagined.

From each point on the circumference of this circular surface, a perpendicular is drawn on the upper surface. As a result, a right circular cylinder is obtained.

The weight of the liquid inside this right circular cylinder of surface area A acts downward in a perpendicular direction.

So the thrust of liquid on this surface,

F = weight of liquid pillar of height h

= Ahdg

∴ pressure of liquid at point M,

P = \(\frac{F}{A}=\frac{A h d g}{A}=h d g\)

Question 9. There is some liquid in a vessel. What is the pressure at a depth h of the liquid when the vessel is lifted up with an acceleration a?

Answer:

When the vessel is lifted up with an acceleration a, effective acceleration due to gravity, g₁ = g + a

If the density of the liquid is d, then pressure at a depth h of the liquid, P = hdg₁= hd(g + a)

Question 10. There is some liquid in a vessel. What is the pressure at a depth h of the liquid when the vessel is brought down with an acceleration a?

Answer:

When the vessel is brought down with an acceleration o, effective acceleration due to gravity, g₁ = g – a

If the density of the liquid is d, then pressure at a depth h of the liquid, P = hdg₁ = hd(g – a)

Question 11. With the help of an experiment, show that the free surface of a stationary liquid always remains horizontal.

Answer:

Let us assume that the upper surface of a stationary liquid is not horizontal but wavy. A horizontal plane MN is imagined inside the liquid kept in a vessel. Let us suppose that two points A and B are taken on this plane.

If density of the liquid is d and the depths of points A and B are h₁ and h₂ respectively, then pressure of liquid at point A, P₁ = h₁dg, and pressure of liquid at point B, P₂ = h₂dg.

Now, as we know that pressure of liquid is same everywhere on a horizontal plane of a stationary liquid, we may write

P₁ = P₂ or, h₁dg = h₂dg

∴ h₁ = h₂

“archimedes principle diagram “

Hence, points A and B are located at the same depth from the free surface of the liquid. Again, points A and B are located on the same horizontal plane. Therefore, our assumption was incorrect and a stationary free surface is horizontal and not wavy.

Question 12. With the help of an experiment, explain the property of a liquid by which it attains same height in several vessels connected simultaneously.

Answer:

When a liquid remains in a stationary state in several vessels of different shapes inter-connected with each other, upper surfaces of the liquid remains in the same horizontal level.

To explain the above property, we consider a U-shaped tube. In the horizontal portion at the lower end of this U-tube, a stopcock is fitted. By closing the stopcock, a liquid is poured in the left arm (A) and the right arm (S) of the stopcock so that height of liquid in the arm A is greater than that in arm B.

Now if the stopcock is opened, it is seen that the liquid flows from arm A to arm B. This flow continues till the liquid attains the same level in both the arms.

When the stopcock was closed, pressure of the liquid at the bottom surface of arm A was greater than that at the bottom surface of arm B. As a result, when the stopcock is opened, liquid starts flowing from arm A to arm B.

When heights of the liquid column in both the arms are equal, pressure of liquid at the bottom surface of both the arms becomes equal and water stops flowing. This proves that stationary liquid surface always prefers to settle at the same level which is an inherent property of a liquid.

Question 13. Write down one practical application of the inherent property of a liquid by which it attains uniform level in several vessels connected simultaneously.

Answer:

In many places of a city, the property of a liquid to attain uniform level is utilised for water supply in multi-storied buildings, Filtered water from river or lake or from underground is purified and stored in a huge tank at a height by means of motor pumps.

This water is supplied to different buildings through pipes. This water goes up to different heights from the earth’s surface without the help of pumps. Of course, this height is kept lower than the height of the tank so that even at the highest elevation, water can come out from the tap whenever required.

Question 14. What do you mean by atmospheric pressure?

Answer:

The force per unit area applied perpendicularly against a surface by the weight of the atmosphere at a point is known as atmospheric pressure.

Question 15. Define standard atmospheric pressure. Calculate the value of standard atmospheric pressure.

Answer:

The pressure of a mercury column of 76 cm at a temperature of 0°C at sea level and at 45° latitude is known as the standard atmospheric pressure.

At the sea level and at 45° latitude, if acceleration due to gravity, g = 980.6 cm/s²; density of mercury at 0°C, d = 13.596 g/cm³ and height of mercury column h = 76 cm, then the standard atmospheric pressure is given by

P = hdg = 76 x 13.596 x 986.6

= 1.01325 x 10⁶ dyn/cm²

Question 16. Describe the construction and working principle of a Fortin barometer.

Answer:

Construction of Fortin barometer:

A glass tube (AB) of uniform diameter and of length 1 meter sealed at one end is filled with mercury and immersed upside down in a vessel (D).

The lower portion of the vessel is made up of leather and the upper portion of brass. To protect it from external injury, the entire glass tube is covered with a metal tube (C).

“buoyant force equation “

To observe the upper surface of mercury, a portion is cut. For measurement of the height of the mercury column, a primary scale (M) and a vernier scale (V) are attached,

Working principle:

Through the leather coating of vessel D, air may pass freely but not mercury. So pressure at the mercury surface of the vessel is equal to the air pressure. This barometer is hung from a hook (E).

By adjusting the screw S, surface of mercury is brought in contact with the ivory pin (l) so that the mercury surface coincides with the zero marking of the primary scale (M). Height of mercury column in the glass tube is measured with the primary scale.

Reading of vernier scale is taken by adjusting the screw P. With the help of these two scales, height of mercury column in the tube is calculated.

A thermometer is kept by the side of the barometer to measure the temperature at which barometer reading is taken because with the change of temperature, density of mercury and reading of scale also change. So by making necessary corrections in the observed reading, actual reading can be obtained.

Question 17. How can one forecast the weather with the help of a barometer?

Answer:

One can forecast the weather from a change of the reading of a barometer.

1. If the reading of a barometer decreases slowly, there is a possibility of rainfall. Water vapour is lighter compared to air. If the amount of water vapour increases in air, density of air also decreases at that place. As a result, reading of the barometer comes down slowly.
2. If the reading of a barometer drops suddenly, there is a possibility of storm, i.e., low pressure has been created at that place.
3. If the reading of a barometer increases slowly, we may conclude that the amount of water pressure in air at that place is decreasing slowly. As a result, weather of that place becomes dry and clear.
4. If the reading of a barometer increases suddenly, it may be concluded that the amount of water vapour in air at that place has decreased considerably. So, weather of that place remains dry and clear temporarily.

Question 18. What are the advantages mercury in a barometer?

Answer:

The advantages of using mercury in a barometer are

1. Pure mercury is easily available. © Mercury do not wet glass, as a result it is easier to take readings.
2. Freezing point and boiling point of mercury are -39°C and 357°C, respectively. So, within this long range of temperature, mercury remains in a liquid state.
3. Mercury is opaque and bright. So one can take the reading of mercury column easily.
4. As the pressure of mercury vapour on the top of the mercury column in a barometer tube is very less, it does not influence the actual reading.
5. As mercury is a good conductor of heat, temperature of mercury throughout the tube remains uniform.
6. As the density (13.6 g/cm³) of mercury is high, height of the mercury column does not reach so high.
7. Volume expansion, density, and coefficient of thermal expansion of mercury can be measured accurately. If there is a change of temperature in the room, it is not difficult to get the original reading after making necessary corrections.

Question 19. Why is a thermometer attached to the barometer tube?

Answer:

To know the temperature at which the reading is taken, a thermometer is placed by the side of the barometer tube. With the change of temperature, density of mercury and reading of scale also change. So, necessary correction of the primary reading gives the actual reading.

Question 20. What is a siphon? Briefly write how it works.

Answer:

Siphon is an arrangement that carries a liquid from a higher level up and over a barrier and then to a lower level. Here, the flow is maintained by gravity and atmospheric pressure as long as the tube in the arrangement remains full.

.

 

Working Principle:

Siphon means a U-tube made up of glass, rubber, or plastic with unequal arms and open ends. The vessel (A) from which liquid has to be transferred is kept at a higher level and the vessel (B) in which liquid has to be transferred at a lower level.

The tube is filled up with the liquid which is to be transferred. Now when the shorter arm of the pipe is placed in the filled-up vessel kept at a higher place and the longer arm of the pipe is placed in the other vessel, then flow of liquid through the tube starts immediately.

Question 21. Explain the siphon process.

Answer:

Two points C and D are imagined in the same horizontal level in a siphon.

If h₁ is the height from liquid surface of vessel A to point C, then pressure at point C, P_c = P – h₁dg;

where P is the atmospheric pressure, d is the density of liquid and g is the acceleration due to gravity.

Again, if h₂ is the height from the open end of the longer arm to the point D, then pressure at point D, P_D = P – h₂dg

Since, h₂ > h₁, so, P_C > P_D.

This means that the liquid flows from C towards D. After coming to point 0, the liquid flows downward due to gravity and accumulates in vessel B As a result, when liquid is displaced from point C, a vacuum is created at that place. So, liquid from vessel A goes up the pipe to reach point C due to atmospheric pressure and fill up that vacuum.

A steady flow of liquid continues through the tube. This is how the siphon process goes on.

archimedes principle experiment

Question 22. Write down the conditions of siphon process.

Answer:

Conditions Of The Siphon Process Are:

1. The tube should be filled with the liquid which has to be transferred.
2. Level of liquid in the vessel from which liquid has to be transferred is to be kept higher than the level of water in the vessel in which it has to be transferred.
3. Due to the atmospheric pressure, liquid goes up the tube in the siphon process So in order to keep the siphon process continuous, atmospheric pressure has to be maintained.
4. There should not be any hole in the tube.

Question 23. Write two applications of siphon.

Answer:

Siphon is used in automatic flush system in public toilet. It is also used to transfer liquid from one vessel to another.

Question 24. Is there a change of rate of flow of a liquid through a siphon if there is a slight change of atmospheric pressure?

Answer:

Rate of change of flow of a liquid through a siphon depends on the difference of pressures between two points situated on the tube. [p = (h₂ – h₁)dg, where h₂ and h₁ are the respective heights of the long and the short arms of the siphon from the liquid surface.)

So if there is a slight change of atmospheric pressure, there is not any change of difference of pressure between the two points situated on the tube. As a result, there is not any change of rate of flow of the liquid through the siphon.

Question 25. A drum is being filled up with water through a long pipe attached to a tap. After the drum is filled up, the pipe gets disconnected from the tap for some reason ‘ and falls on the ground. If the other end of the pipe remains immersed upto the bottom surface of the drum, what happens? Give a scientific explanation of the phenomenon that takes place.

Answer:

Total amount of water of the drum is drained out through the pipe. Raising of water through the pipe, crossing the rim of the drum, and then draining out is a phenomenon that may be explained only through the siphon process.

In this process, a liquid kept in a vessel may be transferred to a lower level crossing a comparatively higher point through a continuous pipe by the atmospheric pressure. Siphon works without the help of a pump, mainly due to pressure difference inside a stationary liquid.

Question 26. What is buoyancy? Mention is cause.

Answer:

The upward thrust exerted by a liquid at rest on a body partially or fully immersed in it is known as buoyant force and the phenomenon is known as buoyancy.

Buoyancy arises from the fact that fluid pressure increases with depth and from the fact that the increased pressure is exerted in all directions so that there is an unbalanced upward force on the bottom of a submerged body.

Question 27. What is Archimedes’ principle?

Answer:

Archimedes’ principle or the physical law of buoyancy states that if any body is submerged completely or partially in a fluid at rest, there is a .reduction in its apparent weight, whose magnitude is equal to the weight of the fluid displaced by the body.

Question 28. Write down the mathematical expression for buoyancy.

Answer:

If a body is immersed partially or fully in a liquid or gaseous material, buoyancy of the liquid or fluid becomes F_b = vdg;

where v is the volume of the immersed portion of the body, d is the density of liquid or gas and g is the acceleration due to gravity.

Question 29. Buoyancy depends on which factors?

Answer:

Buoyancy depends on three factors. They are:

volume of the immersed portion of the body (v), density of the medium in which the body is immersed (d) and acceleration due gravity at that place (g).

Question 30. What is the direction of buoyancy? When completely immersed, how does buoyancy change with the depth of a body?

Answer:

Buoyancy always acts in an upward direction which is the opposite direction of the weight of a body. When completely immersed, buoyancy does not change with the depth of a body.

Question 31. What is centre of buoyancy?

Answer:

Centre of buoyancy, is the point where the centre of gravity of the liquid or gas is located before it is displaced by the immersed body. The centre of buoyancy and centre of gravity are same for totally immersed body in a fluid whereas for partial immersion, these two are different.

Question 32. What do you mean by reaction of buoyancy?

Answer:

Reaction of buoyancy is the equal and opposite force that the body applies on the liquid or gaseous material when it is partially or fully immersed in a liquid or gaseous material.

Question 33. Why is 1kg of cotton heavier than 1kg of iron in a vacuum? Or, Real weight of 1kg cotton is greater than that of 1kg iron—explain the statement.

Answer:

The weight of a body in air is its apparent weight because the body remains immersed in air.

∴ Real weight of the body = weight of the body in air (apparent weight of the body) + weight of air displaced by the body

Apparent weights of 1kg cotton and 1kg iron in air are the same. But volume of 1kg cotton is greater than the volume of 1kg iron. So, cotton displaces comparatively greater volume of air than iron.

Hence, weight of this air displaced by cotton is more than the weight of the air displaced by iron. So, the real weight of cotton is greater than that of iron.

Therefore, 1kg of cotton is heavier than 1kg of iron in vacuum.

Question 34. Two balloons of the same volume are filled up with two different gases under the same pressure. First one is filled with hydrogen gas and the second one with helium gas. Which balloon experiences higher upward force?

Answer:

If any gas lighter than air is used to fill up the balloon, upward thrust on the balloon due to buoyancy of air is higher than the weight of the balloon. So, an upward resultant force acts on the balloon.

upward resultant force on the balloon = buoyant force – weight of the balloon

Now, volume of the two balloons are the same. So, buoyancy is the same in both cases. But helium is a heavier gas than hydrogen, so the balloon which is filled up with hydrogen gas experiences more upward force acting on it.

Question 35. What are the conditions of floatation and submersion?

Answer:

Whether a body sinks or floats in a liquid depends on whether the weight of water displaced by the body is more or less than the weight of the body. Suppose, the weight of the body = W and the weight of the liquid displaced by the body when it is fully immersed in liquid = W₁.

Now the body is immersed in the liquid and then released.

1. If W > W₁, the body remains immersed in liquid.
2. If W = W₁, the body fully floats inside the liquid.
3. If W< W₁ the immersed body comes up through the liquid from the bottom. During its upward movement, it becomes stationary at a particular movement. At this juncture, a portion of the body remains outside the liquid. In this condition, the weight of the body is equal to the weight of an equal volume of liquid displaced by the immersed portion of the body.

Question 36. Is buoyancy of water equal or unequal in case of a block of wood and a block of iron, both having the same volume?

Answer:

Value of buoyancy of liquid is the weight of the liquid displaced by a fully or partially immersed body in that liquid. The portion of the volume of the body that remains immersed in the liquid is equal to the volume of the liquid displaced by the body.

It is known that a block of wood floats partially in water but a block of iron sinks. Clearly, as the two volumes are equal, iron block displaces comparatively more amount of water. As a result, value of buoyancy of water is more in case of iron block.

Question 37. Which of the following incidents signifies an equilibrium condition?

1. A gas balloon is going up in the sky.
2. A gas balloon is floating and moving with uniform velocity at a great height parallel to the earth’s surface.
3. A piece of wood is immersed fully In water and then released.
4. An iron nail is let loose slowly on a water surface.
5. An iron cauldron is immersed in water and then released.
6. A paper boat is floated in water.

Answer:

1. As the buoyant force of air on the balloon is greater than the weight of the gas balloon, it goes up flying from the ground. So, it is not in equilibrium condition.
2. The buoyant force of air on the balloon is equal to the weight of the balloon. So, it floats in the sky and moves with uniform velocity parallel to the earth’s surface. This means total force acting on it is zero. So, it is in equilibrium condition.
3. If a piece of wood is immersed fully in water and then released, it comes up with an acceleration. In this case, buoyant force is greater than the weight of the wood. Thus, this wood is not in equilibrium condition.
4. If an iron nail is let loose slowly on a water surface, it sinks fast in water with an uniform acceleration. In this case, weight of the nail is greater than the buoyant force. So the nail is not in an equilibrium condition.
5. If an iron cauldron is immersed in water and then released, it sinks in water. In this case, the cauldron is not in an equilibrium condition.
6. A paper boat floats in water. In this case, buoyant force = weight of the boat. This means that the boat is in an equilibrium condition.

Question 38. A piece of wood is floating in water in a dosed container. If some amount,of air is pumped out of the container, what is the change of state of floatation of the piece of wood?

Answer:

According to the condition of floatation, weight of the piece of wood is equal to the summation of weight of water displaced by the piece of wood and weight of air displaced by the piece of wood.

If some amount of air is pumped out of the container, density of air inside the container is reduced. So, weight of air displaced by the piece of wood also reduces but weight of the piece of wood remains unchanged. As a result, the piece of wood displaces more water than earlier. So, the piece of wood sinks a bit more in water.

Question 39. A piece of wood is floating in water in a closed container. If some amount of air is pumped into the container, what is the change of state of floatation of the piece of wood?

Answer:

According to the condition of floatation, weight of the piece of wood is equal to the summation of weight of water displaced by the piece of wood and weight of air displaced by the piece of wood. If some more air is pumped into the container, density of air inside the container increases.

So, weight of air displaced by the piece of wood also increases but weight of the piece of wood remains unchanged. As a result, the piece of wood displaces less amount of water than earlier. So, the piece floats up a bit more.

Question 40. There are two blocks of the same mass, one made up of wood and the other of iron. Which of the two has a higher value of buoyancy of water?

Answer:

Wooden block floats on water but iron block sinks in it.

Buoyancy in case of wooden block (B₁) is equal to the weight of the wooden block (W₁) but in case of iron block, buoyancy (B₂) is less than the weight of the iron block (W₂).

Now according to the question, mass of the wooden block is equal to the mass of the iron block. So, weight of the wooden block (W₁) is equal to the weight of the iron block (W₂).

∴ B₂ < B₁ i.e., value of buoyancy of water in case of wooden block is higher.

Question 41. A body is fully immersed inside a liquid or a body floating in a liquid in an equilibrium condition. What do you mean by this ‘equilibrium condition’?

Answer:

A body is shown which is fully immersed in a liquid and floating in a stationary state. The same body is shown partially immersed in another liquid, floating in stationary conditions.

In both the cases, the weight [W] of the body is working downward. But buoyant force (B₁ in the first case and B₂ in the second case) works on the body in an upward direction. In these two cases, weight of the body and buoyant force are equal but working in opposite directions, so the body is floating in a stationary state.

According to mechanics, if the resultant of total forces acting on a body is zero, then the body floats in equilibrium, i.e., it remains stationary or moves with uniform velocity. In this case, the body is stationary and hence, it is in an equilibrium condition.

Considering the magnitude of force and its direction, W + B₁ = 0 in the first case and W + B₂ = 0 in the second case.

Question 42. What is density? What is the mathematical relationship between mass, volume, and density?

Answer:

1. The density of a substance is its mass per unit volume.
2. If the volume of a mass m is V, then its density, \(\frac{m}{V}\).

Question 43. What are the units of density in CGS system and Si? Establish a relationship between these two.

Answer:

Units of density is CGS system and SI are g/cm³ and kg/m³, respectively.

The relationship between them is given by

\(1 \mathrm{~kg} / \mathrm{m}^3=\frac{1000 \mathrm{~g}}{\left(10^2\right)^3 \mathrm{~cm}^3}=\frac{1}{1000} \mathrm{~g} / \mathrm{cm}^3\)

or, \(1 \mathrm{~g} / \mathrm{cm}^3=1000 \mathrm{~kg} / \mathrm{m}^3\)

Question 44. Discuss the influence of pressure on density.

Answer:

Solid and liquid materials are generally considered as incompressible. So, there is not much change in densities of solid and liquid materials due to application of pressure. But in the case of gaseous materials, the temperature remaining constant, density is directly proportional to pressure or d ∝ P (when temperature remains constant).

Question 45. What do you mean by relative density? Does a body float or sink in water, if the value of its relative density in comparison to water is less than 1?

Answer:

1. Relative density is the ratio of the density of a body to the density of a given reference material.
2. The relative density of a body compared to water is less than 1 means that the density of the body is less than the density of water.

So, it floats on water.

Question 46. Define specific gravity. What is its unit?

Answer:

1. The ratio of the mass of a body to the mass of an equal volume of distilled water at 4°C is called the specific gravity of the material of that body.
2. Specific gravity has no unit since it is a ratio of similar quantities.

Question 47. What is the necessity of temperature correction during determination of specific gravity of a material?

Answer:

While determining the specific gravity of a material in a laboratory, if water at 4°C is not available, then necessary temperature correction has to be made. Suppose, at the time of measurement of specific gravity in the laboratory, temperature of water is t°C.

∴ Specific gravity of the material,

s = \(\frac{\text { mass of the body }}{\text { mass of equal volume of water at } 4^{\circ} \mathrm{C}}\)

= \(\frac{\text { mass of the body }}{\text { mass of equal volume of water at } t^{\circ} \mathrm{C}}\) x \(\frac{\text { mass of equal volume of water at } t^{\circ} \mathrm{C}}{\text { mass of equal volume of water at } 4^{\circ} \mathrm{C}}\)

= calculated specific gravity x specific gravity of water at t°C

Question 48. What is the relationship between the density of a material and its specific gravity?

Answer:

Specific gravity = \(\frac{\text { density of a material }}{\text { density of pure water at } 4^{\circ} \mathrm{C}}\)

In the CGS system, density of pure water at 4°C is 1 g/cm³. So in this system, numerical values of specific gravity and density of a material are the same.

On the other hand, density of pure water in SI at 4°C is 1000 kg/m³.

So, specific gravity of a material in SI = \(\frac{\text { density of the material }}{1000}\)

∴ numerical value of the density of a material = 1000 x specific gravity.

Question 49. Difference between specific gravity and density.

Answer:

Differences between specific gravity and density

Question 50. With the help of Archimedes’ principle, how do you measure the density of a solid material insoluble in water and whose density is less than that of water?

Answer:

Suppose, the weight of a solid insoluble material having density less than that of water is mgf. As the density of the body is less than the density of water, so the body does not sink in water. So, a sinker (a piece of iron or brass) is taken.

When the body is in air and the sinker is in water, the weight is m₁gf and when both are submerged in water, weight is m₂gf.

∴ if the volume of the body is V and the density of water is d,

\(V d g=\left(m_1-m_2\right) g \quad \text { or, } \quad V=\frac{m_1-m_2}{d}\)

Density of water (d) = 1g/cm³

So, V = (m₁ – m₂) cm³

∴ density of the body,

\(d_1=\frac{m}{V}=\frac{m}{m_1-m_2} \mathrm{~g} / \mathrm{cm}^3\)

Question 51. With the help of Archimedes’ principle, how do you measure the density of a solid material insoluble in water and whose density is greater than that of water?

Answer:

Suppose the weight of a solid insoluble material in water having density greater than that of water is m₁gf in air and its weight is m₂gf when fully immersed in water.

∴ Apparent reduction of its weight = (m₁ – m₂) gf

Density of water (d) = 1 g/cm³

∴ Volume of the body, \(v=\frac{m_1-m_2}{d} \mathrm{~cm}^3=\left(m_1-m_2\right) \mathrm{cm}^3\)

and density of the body, \(d_1=\frac{m_1}{V}=\frac{m_1}{m_1-m_2} \mathrm{~g} / \mathrm{cm}^3\).

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle Very Short Answer Type Questions Choose The Correct Answer

Question 1. Dimensional formula of pressure is

1. MLT^-2
2. ML^-1T^-3
3. ML^-1T²
4. ML^-2T^-2

Answer: 3. ML^-1T²

Question  2. 1N/m² = how many dyn/cm²?

1. 10
2. 100
3. 0.1
4. 0.01

Answer: 1. 10

Question 3. Which of the following does not express a general property of a liquid?

1. Elasticity
2. Buoyancy
3. Surface tension
4. Viscosity

Answer: 2. Buoyancy

Question 4. Which of the following are fluid?

1. Solid, liquid and gas
2. Solid and liquid
3. Liquid and gas
4. Solid and gas

Answer: 3. Liquid and gas

Question 5. The relationship between pressure and thrust is

1. pressure = thrust x area
2. pressure = \(\frac{\text { thrust }}{\text { area }}\)
3. pressure = thrust x area²
4. pressure = \(\frac{\text { thrust }}{\text { area }^2}\)

Answer: 2. pressure = \(\frac{\text { thrust }}{\text { area }}\)

Question 6. A liquid is kept in a vessel. If the vessel falls freely, pressure at a depth h of the liquid is

1. hdg
2. 0
3. 1/2hdg
4. 3/2hdg

Answer: 2. 0

Question 7. Buoyant force acting on a body does not depend on the

1. Depth of a body completely immersed inside a liquid
2. Volume of the immersed portion of the body
3. Value of acceleration due to gravity
4. Density of displaced liquid

Answer: 1. Depth of a body completely immersed inside a liquid

Question 8. If density of a material is 8000 kg/m³, then the value of its specific gravity is

1. 6
2. 16
3. 4
4. 8

Answer: 4. 8

Question 9. The weight of a piece of metal is 200 gf in air and when fully immersed in water, it iS 150 gf. The specific gravity of the metal is

1. 3
2. 5
3. 6
4. 4

Answer: 4. 4

Question 10. A body is completely submerged in a liquid and then released. It starts rising upward. In this condition

1. Buoyancy = weight of the body
2. Buoyancy > weight of the body
3. Buoyancy < weight of the body
4. Buoyancy changes depending on the depth of the liquid

Answer: 2. Buoyancy > weight of the body

Question 11. A wooden block is floating on water. In this condition

1. Volume of displaced water is equal to the volume of the body
2. Mass of displaced water is equal to the mass of the body
3. Mass of displaced water is equal to the mass of immersed portion of the body
4. Volume of displaced water is equal to the volume of the remaining portion of the body in air

Answer: 2. Mass of displaced water is equal to the mass of the body

Question 12. In the definition of standard atmospheric pressure, the latitude mentioned is

1. 30°
2. 45°
3. 60°
4. 90°

Answer: 2. 45°

Question 13. Value of standard atmospheric pressure is

1. 1.013 bar
2. 1 bar
3. 1.1 bar
4. 1.2 bar

Answer: 1. 1.013 bar

Question 14. Weight of different materials of the same volume are different because different materials have different

1. Elasticity
2. Density
3. Young’s modulus
4. Concentration

Answer: 2. Density

Question 15. The reading of a barometer taken on the surface of the moon is

1. 76 cm
2. 38 cm
3. 0 cm
4. 19 cm

Answer: 3. 0 cm

Question 16. While calculating specific gravity, temperature of water is taken as

1. 0°C
2. l°c
3. 4°C
4. 10°C

Answer: 3. 4°C

Question 17. Three liquids have their densities in the proportion 2:3:4. Density of the first liquid is 1 g/cm³. If the three liquids are mixed together in equal volumes, then the density of the mixture is

1. 1.2 g/cm³
2. 2.5 g/cm³
3. 2 g/cm³
4. 1.5 g/cm³

Answer: 4. 1.5 g/cm³

Question 18. Two circular wheel like structures are floating in a liquid at the same depth. Radii of these two wheels are 2 cm and 3 cm, respectively. If P₁ and P₂ are the pressures on the wheels respectively, then

1. P₁ : P₂ = 2:3
2. P₁ : P₂ = 4:9
3. P₁ : P₂ = 8:27
4. P₁ : P₂ = 1:1

Answer: 4. P₁ : P₂ = 1:1

Question 19. Buoyancy always acts

1. Downward in a perpendicular direction
2. In a parallel direction
3. Upward in a perpendicular direction
4. In upward or downward direction

Answer: 3. Upward in a perpendicular direction

Question 20. Two circular wheel like structures are floating in a liquid at the same depth. Area of these two wheels are 20 cm² and 30 cm², respectively. If thrust of the liquid on them are F₁ and F₂ respectively, then F₁ : F₂ is

1. √20:√30
2. 2:3
3. 4:9
4. 1:1

Answer: 2. 2:3

Question 21. A solid body floats in water keeping 3/5 th portion of its entire body immersed. Its density is

1. 0.4 g/cm³
2. 0.5 g/cm³
3. 0.6 g/cm³
4. 0.8 g/cm³

Answer: 3. 0.6 g/cm³

Question 22. The density of a solid material is d₁. If it floats in a liquid of density d₂ keeping its n portion immersed, then

1. \(n=\frac{d_2}{d_1}\)
2. \(n=\frac{d_1}{d_2}\)
3. \(n=\frac{d_1}{d_2-d_1}\)
4. \(n=\frac{d_2-d_1}{d_1}\)

Answer: 2. \(n=\frac{d_1}{d_2}\)

Question 23. Sudden increase of barometer reading indicates that

1. Amount of water vapour in air is increasing slowly
2. Amount of water vapour in air has reduced considerably
3. Air pressure has come down very fast
4. Amount of water vapour in air is decreasing slowly

Answer: 2. Amount of water vapour in air has reduced considerably

Question 24. Keeping the applied force unchanged, if area is reduced by 10%, then pressure increases by

1. 10%
2. 11.11%
3. 21%
4. 20%

Answer: 2. 11.11%

Question 25. Liquid is kept in a vessel and it is lowered with an acceleration 2. Pressure at a depth h inside the liquid is

1. 1/2 hdg
2. hdg
3. 3/2 hdg
4. 2 hdg

Answer: 1. 1/2 hdg

Question 26. The height of a cone is 10 cm. Area of the base is 10 cm². The cone is filled up with water and kept on the table. Thrust of water on the table is [g = 10 m/s²]

1. 0.5 N
2. 1 N
3. 1.5 N
4. 2 N

Answer: 2. 1 N

Question 27. Densities of two liquids are d₁ and d₂. If the two liquids are mixed together in equal masses, density of the mixture is

1. \(\frac{d_1+d_2}{2}\)
2. \(\frac{2 d_1 d_2}{d_1+d_2}\)
3. \(\frac{d_1+d_2}{2 d_1 d_2}\)
4. \(\sqrt{d_1 d_2}\)

Answer: 2. \(\frac{2 d_1 d_2}{d_1+d_2}\)

Question 28. 1 torr is equal to

1. 0.5 cm Hg
2. 1 mm Hg
3. 2 mm Hg
4. 1cm Hg

Answer: 2. 1 mm Hg

Question 29. When two liquids are mixed in equal volumes, density of the mixture is d_v and when they are mixed in equal masses, density of the mixture becomes d_m. So

1. d_v = d_m
2. d_v > d_m
3. d_v < d_m
4. Cannot be determined

Answer: 2. d_v > d_m

Question 30. A piece of wood is floating in water in a closed container. If some amount of air is pumped out from the container, then

1. The piece of wood sinks a bit more
2. The piece of wood floats up a bit further
3. Volume of the immersed portion of the piece of wood remains unchanged
4. It may float a bit further up at first but later gets completely submerged

Answer: 1. The piece of wood sinks a bit more

Question 31. A piece of wood is floating in water in a – closed container. If some amount of air is pumped into the container, then

1. The piece of wood sinks a bit more
2. The piece of wood floats up a bit further
3. Volume of the immersed portion of the piece of wood remains unchanged
4. It may sink a bit more at first but floats up later

Answer: 2. The piece of wood floats up a bit further

Question 32. 1 N/m = x dyn/cm, value of x is

1. 1
2. 10
3. 100
4. 1000

Answer: 4. 1000

Question 33. Spheres of Iron and Zinc having same mass are completely immersed in water. Density of Zinc is more than that of iron. Apparent loss of weight is W₁ for Iron and W₂ for Zinc spheres. Then \(\frac{w_1}{w_2}\) is

1. 1
2. 0
3. <1
4. >1

Answer: 4. >1

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle Answer In Brief

Question 1. What is the relationship between N/m² and Pa?

Answer:  1 N/m² = 1Pa

Question 2. Keeping the force unchanged, if area of the base is reduced, then does the pressure increase or decrease?

Answer: Keeping the force unchanged, if area of the base is reduced, then pressure also increases.

Question 3. Among solid, liquid and gaseous materials, which are fluids?

Answer: Liquid and gaseous materials are fluids.

Question 4. What is the relationship between pressure and thrust?

Answer: Pressure = \(\frac{thrust}{area}\)

Question 5. What is the reading of a barometer if it is taken on the surface of the moon?

Answer: As there is no atmosphere on the moon, pressure of air is zero and hence, reading of a barometer becomes zero if it is taken on the surface of the moon.

Question 6. Does a siphon work in an air-free place?

Answer: Siphon does not work in an air-free place because in the siphon process, liquid goes up in the pipe due to the atmospheric pressure. According to modern explanation cohesive force of liquid molecules can causes siphon to work; Therefore presence of atmospheric pressure is not essential.

Question 7. If the density of iron in CGS unit is 7.8g/cm³, what is its value in SI?

Answer: Density of iron in SI = 7.8 X 1000 =7800 kg/m³

Question 8. If the density of a liquid is 1.2 g/cm3, what is the specific gravity of that liquid?

Answer: The numerical value of density of any material in CGS system is its specific gravity and hence, in this case, specific gravity and hence of the liquid is 1.2.

Question 9. What is the dimension of specific gravity?

Answer: Specific gravity is a dimensionless physical quantity.

Question 10. What is the dimensional formula of buoyant force?

Answer: Dimensional formula of buoyant force is MLT^-2.

Question 11. Specific gravity of a liquid in CGS system is 1.5. What is its specific gravity in SI?

Answer: Value of specific gravity is the same in any type of measurement. So, specific gravity of the liquid in SI is also 1.5.

Question 12. What is the apparent weight of a floating body?

Answer: The apparent weight of a floating body is zero.

Question 13. What is the condition of floatation of a floating body?

Answer: The weight of the floating body must be equal to the weight of the liquid displaced by the body.

Question 14. Does Archimedes’ principle apply to a freely falling body?

Answer: No, Archimedes’ principle is not applicable. since weight of a freely falling body is zero.

Question 15. What is the relationship between specific gravity and relative density?

Answer: Relative density is the ratio of the density of a substance to the density of a given substance taken as reference, and when the reference substance is pure water at,4°C, then the ratio is called specific gravity.

Question 16. What is a barometer?

Answer: Barometer is an instrument by which atmospheric pressure is measured.

Question 17. Force applied on a plane in a perpendicular direction is 10 N and its area is 0.1 m². Find the pressure.

Answer: Pressure = \(\frac{10 \mathrm{~N}}{0.1 \mathrm{~m}^2}\) = 100 N/m².

Question 18. A vessel contains a liquid. If the vessel falls freely, what is the pressure at a depth of h inside the liquid?

Answer: If the density of the liquid is d, then pressure at a depth h is p = hd(g -g) = 0.

Question 19. What is the value of standard pressure in bar unit?

Answer: Standard pressure = 1.013 bar.

Question 20. There is a mercury barometer inside a lift. If the lift starts moving up with acceleration a(a < g), does the reading of the barometer increase or decrease?

Answer: If the lift goes up with acceleration a(a < g), reading of the barometer decreases.

Question 21. There is a mercury barometer inside a lift. If the lift starts coming down with acceleration a(a < g), does the reading of the barometer increase or decrease?

Answer: If the lift comes down with acceleration a(a < g), reading of the barometer increases.

Question 22. What happens if there is a hole in the siphon?

Answer: If there is a hole in any place of a siphon tube, atmospheric pressure acts on the liquid at that place and as a result, activity of the siphon stops.

Question 23. Give an example where density increases with increase of temperature.

Answer: Density of water increases if temperature increases in the range 0°C to 4°C.

Question 24. 1 torr = how many Pa?

Answer: 1 torr = 133.28 Pa.

Question 25. 1 bar = how many dyn/cm²?

Answer: 1 bar = 106 dyn/cm².

Question 26. Where is centre of buoyancy located?

Answer: Centre of buoyancy of a floating body is located at the centre of gravity of the volume of the displaced liquid.

Question 27. When does a body apply buoyant force on another body?

Answer: When a body is partly or fully immersed in a liquid or gaseous material, then that liquid or gaseous material applies buoyant force on the body.

Question 28. Which body applies reaction of buoyancy on another body?

Answer: When a body is partly or fully immersed in a liquid or gaseous material, then that body applies reaction of buoyancy opposite to the buoyancy on the liquid or gaseous material.

Question 29. How much portion of a solid body of density 0.8 g/cm3 remains immersed, when it floats in a liquid of density 1.2 g/cm³?

Answer: The body keeps \(\frac{0.8}{1.2}\) = \(\frac{2}{3}\)rd of its volume immersed in the liquid while floating.

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle Fill In the Blanks

Question 1. Pressure of a liquid is constant everywhere in any ______ plane inside a stationary liquid.

Answer: Horizontal

Question 2. Specific gravity is the ratio of the mass of a body and the mass of an equal volume of water at ________

Answer: 4◦C

Question 3. In SI, density = _________ x specific gravity.

Answer: 1000

Question 4. If the depth of a body inside a liquid of uniform density increases, value of the ________ of liquid on the body also increases.

Answer: Pressure

Question 5. A solid material A is fully immersed in a liquid material B. This means density of B is ________ than the density of A.

Answer: Less

Question 6. The weight of a body is ________ to the value of buoyant force when it floats in a partly immersed state in a liquid.

Answer: Equal

Question 7. In vacuum, 1 kg of iron is _______ than 1 kg of cotton.

Answer: Lighter

Question 8. The density of material B compared to the density of material A is the ________ of B.

Answer: Relative density

Question 9. Due to the upthrust of a liquid, a body immersed in the liquid suffers an apparent _________ in weight.

Answer: Loss

Question 10. The density of silver is 10500 kg. m^-3 means it relative density is ___________

Answer: 10.5

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle State Whether True Or False

Question 1. Density is defined as the pressure per unit volume.

Answer: False

Question 2. Barometer is the instrument used to measure atmospheric pressure.

Answer: True

Question 3. The force applied by the liquid tangentially on the surface adjacent to the liquid is known as the thrust of the liquid.

Answer: False

Question 4. Archimedes’ principle is also applicable for a freely falling body or for an artificial satellite.

Answer: False

Question 5. The ratio of mass of a material to the mass of an equal volume of water at 4°C the specific gravity of that material.

Answer: True

Question 6. During construction of a dam, the base of the dam wall is usually made wider.

Answer: True

Question 7. Siphon is used in intermittent flush system.

Answer: True

Question 8. In any unit system the value of relative density have the same value.

Answer: True

Question 9. Pressure of 1 mm of mercury column is 1 torr.

Answer: True

Chapter 3 Topic A Pressure Of Liquids And Air Archimedes Principle Numerical Examples

Useful formula

1. If F force is applied on a surface of area A normally, then the pressure at any point on the surface, P = \(\frac{\text { force }}{\text { area }}\) or, P = \(\frac{F}{A}\)
2. Pressure of a liquid of density d at any point at a depth h below the surface of the liquid is P = hdg where g is acceleration due to gravity.
3. Upthrust or buoyant force = volume of the displaced liquid x density of the liquid x acceleration due to gravity = Vdg
4. Specific gravity or relative density = \(\frac{\text { density of a substance }}{\text { density of water at } 4^{\circ} \mathrm{C}}\)
5. Standard atmospheric pressure = 1.013 x 10⁵ Pa

Question 1. Water pressure at the ground floor of a multi storey building is 5 x 10⁵ Pa. What is the water pressure in the second floor at a height of 8 m? (g = 10/s²)

Answer:

Density of water, d = 1000 kg/m³

Height of seconcf floor, h = 8 m

Water pressure in the second floor,

P = 5 x 10⁵ – hdg

= 5 x 10⁵ – 8 x 1000 x 10 = 5 x 10⁵ – 0.8 x 105 = 4.2 x 10⁵ Pa

Question 2. At what depth inside a lake, total .pressure is double of the atmospheric pressure?

Answer:

1 atmospheric pressure P₀

= 76 cm pressure of mercury column

= 76 x 13.6 x 980 dyn/cm²

Suppose, at a depth h inside the lake, pressure (P) is double of the atmospheric pressure, i.e., P = 2P₀.

If density of water, d = 1g/cm³, then total pressure, P = P₀ + hdg.

So, 2P₀ = P₀ + hdg

or, hdg = P₀

or, h x 1 x 980 = 76 x 13.6 x 980

or, h = 76 x 13.6 = 1033.6

Hence, at a depth of 1033.6 cm or 10.336 m inside the lake, pressure is double of the atmospheric pressure.

Question 3. what is the pressure at the bottom surface of a clean lake of depth 10 m? Atmospheric pressure is equivalent to the pressure 76 cm of a mercury column and density of mercury is 13.6 g/cm³.

Answer:

Atmospheric Pressure,

P_a = pressure of 76 cm of mercury column

= 76 x 13.6 x 980

= 1.013 X 106 dyn/cm²

Depth of lake, h = 10 m = 10 x 100 cm

Density of water, d = 1 g/cm³

So, pressure at the bottom surface of the lake,

P = hdg + P_a

= 10 x 100 x 1 x 980 + 1.013 x 10⁶

= 0.98 x 10⁶ + 1.013 x 10⁶ = 1.993 x 10⁶ dyn/cm²

Question 4. The height of a cone is 50 cm and the area of its base is 20 cm². The cone is filled with water and kept on the table. What is the thrust of this cone on the table?

Answer:

Height of the cone, h = 50 cm

Density of water, d = 1 g/cm³

Acceleration due to gravity g = 980 cm/s²

∴ Water pressure at the base of cone,

P = hdg

Area of the base of the cone, A = 20 cm²

So, thrust of water on the table,

F = PA = hdgA = 50 x 1 x 980 x 20 = 9.8 x 10⁵ dyn = 9.8 N

Question 5. A man of moss 50 kg is standing with support on his left ankle. If the area of his ankle is 4 cm², what amount of pressure is he exerting on the ground?

Answer:

Mass of the man, m = 50 kg

∴ Weight of the man,

W = mg = 50 x 9.8 N

Area of the ankle of the man,

A = 4 cm² = 4 x 10^-4 m²

So the pressure exerted by the man on the ground,

P = \(\frac{W}{A}=\frac{50 \times 9.8}{4 \times 10^{-4}}\) = 1.225 X 10⁶ N/m²

Question 6. Area of the base of a right circular cylinder is 50 cm². Water is filled up to a height of 30 cm inside the cylinder. Calculate pressure and thrust of water at the base of the cylinder.

Answer:

Height of water in cylinder, h = 30 cm

Density of water, d = 1 g/cm³

∴ Pressure at the base of the cylinder due to water, P = hdg = 30 x 1 x 980

= 2.94 x 10⁴ dyn/cm²

If the area of the base of the cylinder, A = 50 cm², then thrust of water at the base of the cylinder is given by

F = P x A = 2.94 x 10⁴ x 50

= 14.7 x 105 dyn = 14.7 N

Question 7. Equal weights of mercury and water are taken in a beaker. Total height of these two liquids is 43.8 cm. What is the total pressure at the base of the beaker? (Density of mercury = 13.6 g/cm³)

Answer:

Suppose, depth of mercury in the beaker and depth of water is equal to h cm and (43.8-h) cm, respectively.

Density of mercury = 13.6 g/cm³

Density of water = 1 g/cm³

Weight of mercury and water are same.

Now if the area of a cross section of the beaker is A, then Ah x 13.6 x g = A (43.8 – h) x 1 x g [g = acceleration due to gravity]

or, 13.6 h = 43.8 – h

or, 14.6 h = 43.8

∴ h = \(\frac{43.8}{14.6}\) = 3 cm

So the depth of mercury in the beaker = 3 cm and depth of water = (43.8 – 3) cm = 40.8 cm.

Hence, total pressure at the base of the beaker,

P = pressure of a mercury column of 3 cm + pressure of a water column 40.8 cm = (3 X 13.6 X 980 + 40.8 X 1 x 980)

= 79968 dyn/cm²

Question 8. Three liquids have densities in the proportion of 1:2:3. Density of the first liquid is 1 g/cm³, If these three liquids are mixed in the same volume, what is the density of the mixture?

Answer:

If the density of first liquid, d₁ = d = 1 g/cm³, then the density of the second liquid, d₂ = 2d= 2 g/cm^3, and the density of the third liquid, d₃ = 3d = 3 g/cm³.

Suppose, a mixture is prepared by taking V cm³ of each liquid.

So, density of the mixture is given by \(D=\frac{V d_1+V d_2+V d_3}{V+V+V}\)

= \(\frac{V+2 V+3 V}{3 V}=2 \mathrm{~g} / \mathrm{cm}^3\)

Question 9. The mass of a metallic alloy made up of iron and aluminium is 588 g and its volume is 100 cm³. Specific gravities of iron and aluminum are 8 and 2.7, respectively. Calculate the ratio of

1. The volume and
2. The mass of the components present in the metallic alloy.

Answer:

Suppose, the volume of iron in the metallic alloy, V₁ = Vcm^3, and volume of aluminium,

V₂ = (100 – V) cm³.

Specific gravity of iron = 8.

So the density of iron, d1 = 8 g/cm³.

Now, specific gravity of aluminium = 2.7.

So, the density of aluminium, d₂ = 2.7 g/cm³.

1. According to the question, mass of the metallic alloy = 588 g

So, V₁d₁ + V₂d₂ = 588

or, V X 8 + (100 – V)x 2.7 = 588

or, 8V + 270 – 2.7V = 588

or, 5.3V = 318

∴ V = \(\frac{318}{5.3}\) = 60

Now volume of iron, V₁ = 60 cm³ and volume of aluminium, V₂ = 100 – 60 = 40 cm³.

So, \(\frac{V_1}{V_2}=\frac{60 \mathrm{~cm}^3}{40 \mathrm{~cm}^3}=\frac{3}{2}\)

Hence, ratio of volumes of iron and aluminium in this metallic alloy is 3: 2.

2. In the metallic alloy, mass of iron, m₁ = 60 x 8 g = 480 g, and mass of aluminium, m₂ = (588 – 480)g = 108 g.

So, \(\frac{m_1}{m_2}=\frac{480 \mathrm{~g}}{108 \mathrm{~g}}=\frac{40}{9}\)

Hence, the ratio of masses of iron and aluminium in the metallic alloy is 40: 9.

Question 10. If two materials are mixed in equal volumes, specific gravity is 4.5. But if they are mixed in equal masses, the specific gravity is 4. Calculate the specific gravity of these two materials.

Answer:

Suppose, specific gravities of these two materials are s₁ and s₂.

∴ Densities of these two materials are s₁ g/cm³ and s₂ g/cm³.

If 1 cm³ volume of each material is taken to produce a mixture, specific gravity of the mixture is 4.5. Therefore, density is 4.5 g/cm³.

∴ \(\frac{V s_1+V s_2}{V+V}=4.5 \quad \text { or, } \frac{s_1+s_2}{2}=4.5\)

Again, if m g of mass of each material is taken to produce a mixture, specific gravity of the mixture is 4, i.e., density is 4 g/cm³.

∴ \(\frac{m+m}{\frac{m}{s_1}+\frac{m}{s_2}}=4 \quad \text { or, } \frac{2}{\frac{1}{s_1}+\frac{1}{s_2}}=4\)

or, \(\frac{1}{\frac{s_1+s_2}{s_1 s_2}}=2 \text { or, } \frac{s_1 s_2}{s_1+s_2}=2\)

or, \(\frac{s_1\left(9-s_1\right)}{9}=2\)

or, \(s_1^2-9 s_1+18=0\)

or, \(\left(s_1-3\right)\left(s_1-6\right)=0\)

If (s₁ -3) is zero, then s₁ = 3

∴ s₂ = 9-3 = 6

Again if (s₁ – 6) = 0, then s₁ = 6

∴ s₂ = 9 – 6 = 3

∴ Specific gravities of the materials are 3 and 6.

Question 11. If the densities of iron, mercury, and pure water are 7870 kg/m³, 13546 kg/m³, and 1000 kg/m³ (at 4°C temperature) respectively, then

1. What are the values of specific gravity of iron in CGS system and SI?
2. What is the relative density of iron in comparison with mercury?
3. By discussing the value of relative density, how can it be said that a nail of iron sinks in water but floats in mercury?

Answer:

1. The specific gravity of a material is the ratio of two quantities, similar in nature. So the value of specific gravity is the same in CGS system or SI.

Specific gravity of iron = \(\frac{7870}{1000}\) = 7.87.

2. Relative density of iron in comparison with = \(\frac{7870}{1000}\) = 0.58 (approx)

3. In comparison to water, relative density of iron = specific gravity of iron = 7.87, which is greater than 1.

But relative density of iron in comparison to mercury = 0.58, which is less than 1.

Therefore, an iron nail sinks in water but floats in mercury.

Question 12. 108 g of sulphuric acid with specific gravity 1.8 is taken in a vessel. If 100 g of water is mixed with it, specific gravity of the mixture becomes 1.5. Calculate the contraction of volume of the mixture.

Answer:

Specific gravity of sulphuric acid = 1.8

Density of sulphuric acid = 1.8 g/cm³

So the volume of 108 g of sulphuric acid

= \(\frac{108}{1.8 \mathrm{~g} / \mathrm{cm}^3}\) = 60 cm³

And Volume of 100 g of water

= \(\frac{100 \mathrm{~g}}{1 \mathrm{~g} / \mathrm{cm}^3}\) = 100 cm³

So the total volume of sulphuric acid and water before mixing,

V₁ = 60 + 100 = 160 cm³

Mass of the mixture = 108 g + 100 g = 208 g

Now, specific gravity of mixture = 1.5.

So, density of mixture = 1.5 g/cm³.

∴ Volume of mixture,

\(V_2=\frac{208 \mathrm{~g}}{1.5 \mathrm{~g} / \mathrm{cm}^3}=138.67 \mathrm{~cm}^3\)

Hence, contraction of volume of mixture ΔV = V₁ – V₂ = (160 – 138.67) cm³ = 21.33 cm³.

Question 13. Densities of two liquids are 1.1 g/cm³ and 1.3 g/cm³, respectively. What is the density and specific gravity, if equal volumes of the two liquids are mixed?

Answer:

Suppose, a mixture is prepared by taking V cm³ of each liquid.

Mass of first liquid = 1.1 V g and mass of second liquid = 1.3 V g.

So, mass of the mixture,

m = (1.1 V+ 1.3 V) g = 2.4 Vg

And volume of the mixture, V₁ = (V + V) cm³ = 2 V cm³

∴ Density of the mixture = \(\frac{m}{V_1}=\frac{2.4 \mathrm{~V}}{2 \mathrm{~V}}\) = 12 g/cm³

In CGS system, numerical value of density is its specific gravity. Hence, specific gravity of the mixture = 1.2.

Question 14. On a particular day, the reading of a mercury barometer at a place is 76 cm. What is the reading of a water barometer in that place at that time?

Answer:

According to the question, we know that reading of a mercury barometer is 76 cm. The atmospheric pressure on that day and at that place (P) is equal to the product of height of mercury, density of mercury and acceleration due to gravity.

So, P =76 x 13.6 x g dyn/cm²

Suppose, reading of a water barometer at that place on that day = h cm.

∴ Atmospheric pressure at that place on that day, (P’) = height of water x density of water x acceleration due to gravity

= h x 1 x g = hg dyn/cm²

Now according to t a condition,

P’ = P

or, hg = 76 x 13.6 x g

∴ h = 76 x 13.6 = 1033.6 cm = 10.3 m

Question 15. A piece of metal weighs 1000 gf in air and weighs 900 gf when fully immersed in water. What is the density of the material?

Answer:

Weight of the piece of metal in air = 1000 gf

And when fully immersed in water, its weight = 900 gf

∴ Weight of displaced water by the piece of metal = (1000-900) = 100 gf

Suppose, volume of the piece of metal be V cm³.

Now, in the CGS system, density of water, d = 1 g/cm³.

So, Vdg = 100 g or, V x 1 = 100

∴ V = 100 cm³

Now, mass of the piece of metal, m = 1000 g

Hence, density of metal = \(\frac{m}{V}=\frac{1000 \mathrm{~g}}{100 \mathrm{~cm}^3}\) = 10 g/cm³

Question 16. A solid body floats with 4/5th of its volume immersed in water. What is the density of the body?

Answer:

Suppose the volume of the body = V and density = d.

Density of water, d₁ = 1 g/cm³

So the weight of the body = Vdg, where g is the acceleration due to gravity.

Volume of water displaced by the body = 4/5 V

∴ Weight of water displaced by the body = 4/5 Vd₁g

As per the condition of floatation,

Vdg = 4/5Vd₁g

or, d = 4/5 x 1

∴ d = 0.8 g/cm³

Hence, Density of the body is 0.8 g/cm³.

Question 17. A body floats in water with 40% of its volume outside the water. When the same body floats in an oil-like material, 60% of its volume remains outside the oil. What is the density of the oil-like material?

Answer:

Suppose, volume of the body = V, its density = d, and density of the oil-like material = d₁.

Density of water = 1 g/cm³

The body floats with (100-40)% or 60% of its volume immersed in water.

As per the condition of floatation,

Vdg = 60/100 V x 1 x g

or, d = 0.6 g/cm³

Again, the body floats with (100 – 60)% = 40% of its volume immersed in oil-like material.

Now as per the condition of floatation,

Vdg = \(\frac{40}{100} V \times d_1 \times g\)

or, \(0.6=0.4 d_1\)

∴ \(d_1=\frac{0.6}{0}=1.5 \mathrm{~g} / \mathrm{cm}^3\)

Hence, density of the oil-like material is 1.5 g/cm³.

Question 18. A brass weight of mass 5 kg is hanging from a spring balance. In this condition, aportion of the weight is immersed in water in a beaker and it is seen that a reading of 4 kg is shown in the spring balance. Now, when the weight is immersed completely, a reading of 3 kg is shown.

1. What is the buoyancy acting on the weight in the above two cases? {Take g = 9.8 m/s²)
2. If the weight is taken further inside water, is there any change of buoyancy?

Answer:

1. According to this question, mass of the brass weight = 5 kg.

That means, its weight in air, W = 5gN

Reading of the spring balance with the body in partially submerged condition = 4 kg.

That means, its apparent weight, W₁ = 4g N.

So the buoyancy of water,

B₁ = W-W₂ = 5g – 4g = g = 9.8 N

Again, reading of the spring balance when the weight is immersed completely = 3 kg.

So its apparent weight, W₂ = 3g N.

Hence, buoyancy of water,

B₂ = W- W₂ = 5g -3g = 2g

= 2 X 9.8 N = 19.6 N

2. For a completely immersed body, value of buoyancy does not depend on the depth of the body inside the liquid. As water inside the heater has same density everywhere, so if the weight is taken further inside the water of the beaker, there is not any change in buoyancy, and reading of the spring balance shows only 3 kg.

Question 19. A body, weighed in a spring balance, has a weight of 100 g in air and 70 kg when immersed fully in water. What is the apparent weight of the body, if immersed in a liquid of density 1.2 g/cm³? {Density of water = 1 g/cm³)

Answer:

Mass of water displaced by the body = (100 – 70) g = 30 g

If volume of the body is V,

V x 1 x g = 30 x g or, V = 30

∴ volume of body, V = 30 cm³

Density of liquid, d = 1.2 g/cm³

So, apparent weight of the body when fully immersed in a liquid of density 1.2 g/cm³

= (100 – Vd) = (100 – 30 X 1.2) = 64 g.

Question 20. A glass sphere is coated fully with wax. If this glass sphere floats while fully immersed in water, what is the ratio of the volumes of wax and glass?
(Density of glass = 2.6 g/cm³; Density of wax = 0.8 g/cm³)

Answer:

Suppose, volume of wax = Vr cm³

Volume of glass = V2 cm³

So, total mass of glass and wax,

W = (V₁ x 0.8 + V₂ x 2.6)g

Now mass of water displaced by glass and wax, W₁ = V₁ x 1 + V₂ x 1= (V₁ + V₂)g

Since this glass sphere coated with wax floats while fully immersed in water,

W = W₁ or, 0.8 V₁ + 2.6 V₂ = V₁ + V₂

or, 1.6V₂ = 0.2 V₁

or,\(\frac{V_1}{V_2}=\frac{1.6}{0.2}\)

or, \(\frac{V_1}{V_2}\) = 8

∴ V₁ :V₂ = S:1

Thus, ratio of volume of wax and glass is 8 :1.

Question 21. Some amount mercury is kept in a vessel. Rest of the vessel is filled by water. If a cubical body is released in this vessel, half of the body remains immersed in mercury and half in water. What is the density of the body?

Answer:

Density of mercury, d₁ = 13.6 g/cm³

Density of water, d₂ = 1 g/cm³

Suppose, volume of the body = V and its density = d

According to the condition of floatation,

Weight of the body = weight of mercury displaced by body + weight of water displaced by body.

or, Vdg = \(\frac{V}{2}\) x 13.6 xg + \(\frac{V}{2}\) x 1 x g

∴ d = 6.8 + 0.5 =s 7.3 g/cm³

Question 22. The inner and outer diameter of a hollow sphere are 8 cm and 10 cm respectively. The sphere remains floating completely immersed in a liquid of density 1.5 g/cm³. Find the density of the material of the sphere.

Answer:

The volume of the displace liquid (V) = \(\frac{4}{3} \times \pi \times\left(\frac{10}{2}\right)^3=\frac{4}{5} \times \pi \times 5 \times 5 \times 5 \mathrm{cc}\)

Volume of the material of the sphere (V’)

= \(\frac{4}{3} \times \pi\left[\left(\frac{10}{2}\right)^3-\left(\frac{8}{2}\right)^3\right]\)

= \(\frac{4}{3} \pi(125-64)=\frac{4}{5} \pi 61 \mathrm{~cm}^3\)

Suppose, the density of the material of the sphere is d g/cm³.

For equilibrium, weight of the.sphere = upthrust on the sphere

or, \(\frac{4}{3} \pi 61 \times d \times g=\frac{4}{5} \pi \times 5 \times 5 \times 5 \times 1.5 \times g\)

d = \(\frac{125 \times 1.5}{61}=3.07\)

Question 23. A body of mass 200 g is immersed completely in water as shown in figure. Find the volume and density of the body.

Answer:

The volume of the body = Final reading of the volume – initial reading of the volume

= 75 – 50 = 25 cm³

And Density of the body = \(\frac{\text { mass }}{\text { volume }}\)

= \(\frac{200}{25}=8 \mathrm{~g} / \mathrm{cm}^3\)

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Synopsis:

Newton’s Third Law Of Motion:

To every action, there is an equal and opposite reaction.

Forces can be divided primarily into two types

1. Contact forces and
2. Non-contact force.

A forces that require being in contact with another object are contact forces.

Examples: Frictional force, tension force, etc.

A forces that can be exerted without requiring any contact with any object are non-contact forces.

Examples: Gravitational force, magnetic force, electrostatic force, etc.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

Law Of Conservation Of Linear Momentum:

If no external force is applied on a system of particles, the total linear momentum of the system remains unchanged.

Explanation:

Suppose two bodies of masses m₁ and m₂ moving along a straight line with velocities u₁ and u₂ (u₁ > u₂) respectively collide with each other.

After the collision, the velocities of these two bodies are v₁ and v₂ respectively.

law of conservation of momentum

The total linear momentum of the system before the collision is m₁u₁ + m₂u₂ and total linear momentum of the system after the collision is m₁v₁ + m₂v₂.

According to the law of conservation of linear momentum,

m₁u₁ +m₂u₂ = m₁v₁ +m₂v₂

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Short And Long Answer Type Questions

Question 1. Write down Newton’s third law of motion and explain it.

Answer:

To every action, there is an equal and opposite reaction.

Suppose, block P in a moving state applies a force \(F_{Q P}\) on the block Q.

At the same time, block Q also applies a force \(F_{P Q}\) on block P.

If force \(F_{Q P}\) is action, then \(F_{P Q}\) is reaction.

According to Newton’s third law of motion,

\(F_{Q P}\) = –\(F_{P Q}\)

As long as action exists, reaction also continues to exist.

If there is no action, there is also no reaction. This means that action and reaction take place simultaneously, it never happens that one occurs earlier and the other occurs later.

NEET Biology Class 9 Question And Answers	WBBSE Class 9 History Notes	WBBSE Solutions for Class 9 Life Science and Environment
WBBSE Class 9 Geography And Environment Notes	WBBSE Class 9 History Multiple Choice Questions	WBBSE Class 9 Life Science Long Answer Questions
WBBSE Solutions for Class 9 Geography And Environment	WBBSE Class 9 History Long Answer Questions	WBBSE Class 9 Life Science Multiple Choice Questions
WBBSE Class 9 Geography And Environment Multiple Choice Questions	WBBSE Class 9 History Short Answer Questions	WBBSE Solutions For Class 9 Maths
WBBSE Solutions for Class 9 History	WBBSE Class 9 History Very Short Answer Questions

 

Question 2. With the help of an experiment, show that action and reaction are equal and opposite.

Answer:

Two spring balances are taken. The hook of one balance is attached to the hook of the other. Now the remaining two ends of the balances are pulled by two hands in such a way that there is no displacement of the two spring balances.

It is seen that the reading shown by the left spring balance is equal to the reading shown by the right spring balance.

Now, the right spring balance is fixed with the wall and the end of the left spring balance is pulled. In this case also, the reading of the left spring balance is the same as the reading of the right spring balance.

“derivation of conservation of momentum “

In the first case, if the force of the left hand is designated as action, then the force of the right hand is reaction.

In the second case also, if the force applied on the spring is action, then the force applied by the wall is reaction. Hence, it is found that action and reaction are equal and opposite.

Question 3. Opposite poles of two magnets attract each other—explain this phenomenon with the help of Newton’s third law of motion.

Answer:

If the north pole of a bar magnet is brought near the south pole of another bar magnet, it is seen that the two magnets move towards each other, that is, the two poles attract each other.

If the magnet on the left side attracts the magnet on the right side with force F₁ and the magnet on the right side attracts the magnet on the left side with force F₂, then F₁ = -F₂. Therefore, if force F₁ is called action, then F₂ is the reaction.

Since action and reaction are equal and opposite, hence this type of attraction is called mutual attraction force.

Question 4. When a shell is fired from a cannon, the cannon recoils. Explain this phenomenon with the help of Newton’s third law of motion.

Answer:

When a shell is fired from a cannon, the shell moves forward with high velocity and the cannon recoils immediately. The force applied by the cannon on the shell is taken as action, then the shell moves forward due to this action. The equal and opposite reaction exerted by the shell on the cannon causes the cannon to recoil.

Question 5. Give a scientific explanation of our walking with the help of Newton’s third law of motion.

Answer:

When we walk, we obliquely exert a force (F) on the ground. The ground also exerts an equal reaction (R) on us in the opposite direction. The horizontal component of this reaction force (H) helps us to move forward.

derive the mathematical formula of conservation of momentum

Question 6. Why is a boat pushed backward, if a passenger jumps from it towards the shore?

Answer:

When a passenger jumps from a boat, he applies a force on the boat with his legs. Due to the influence of this force, the boat moves backward. At the same moment, the boat also applies a reaction force of equal magnitude, opposite to the direction of force applied by the passenger.

This force helps the passenger to reach the shore.

Question 7. Though action and reaction forces are equal and work in opposite directions, why cannot they establish equilibrium?

Answer:

Two equal and opposite forces applied on a body with the same line of action, keep the body in equilibrium. But though action and reaction are equal and in opposite directions, they always act on two different bodies.

Hence, action and reaction forces cannot establish equilibrium.

Question 8. What do you mean by tension of a rope?

Answer:

One end of a rope is firmly fixed and the other end is hanged with a wooden block attached to it.

The weight (W) of the block is acting downward in a perpendicular direction, so the block applies a force of equal amount of its weight on the rope which acts downward.

The rope also applies a force on the block equal to the weight of the body, acting in an opposite direction. This force is called tension of a rope.

conservation of momentum equation

Question 9. According to the third law of motion, if a person pushes a book, then the book also pushes the person in the opposite direction. In which direction does the book move?

Answer:

As the person pushes the book, then according to Newton’s third law of motion, the book also pushes the person in the opposite direction with equal magnitude of force.

The force applied by the person is action force whereas the force applied by the book is reaction force. Action and reaction are always of equal magnitude working in opposite directions but they always act on separate bodies.

That is why they cannot establish equilibrium. So the book moves in the direction of the force applied by the person. But the amount of force applied should be greater than the frictional force applied by the book to put it in motion.

Question 10. What do you mean by attraction and repulsion?

Answer:

If the opposite poles of two-rod magnets are brought nearby, it is seen that due to mutual action-reaction, they come close to each other. This type of force is called attraction.

Again, if the like poles of these two-rod magnets are brought nearby, it is seen that due to mutual action-reaction, they recede away. This type of force is called repulsion.

This means if two bodies come nearby due to mutual action-reaction, then that force is called attraction and if two bodies move away from each other, then that force is called repulsion.

Question 11. In the following given cases, what is the acting force?

1. A stone is attached to a string and the system is suspended.
2. A rolling iron ball strikes another ball.
3. A man is standing on the floor.
4. An iron nail kept in front of a magnet moves towards the magnet.
5. When two rod magnets are kept In front of each other, they repulse each other and move away In opposite directions.
6. A book when pushed on a table, traverses some distance and then stops.

Answer:

Incidents and acting forces are mentioned below:

Question 12. Among the forces mentioned below, Which are responsible for acceleration, deceleration, and establishment of equilibrium:

1. Tension,
2. Force due to collision,
3. Normal force,
4. Attraction,
5. Repulsion,
6. Friction.

Answer:

1. Tension: force that establishes equilibrium.
2. Force due to collision: Force that creates acceleration or deceleration.
3. Normal force: Force that establishes equilibrium.
4. Attraction: Force that creates acceleration or deceleration.
5. Repulsion: Force that creates acceleration.
6. Friction: Force that creates deceleration.

Question 13. Give an example to show that attraction force may create acceleration as well as deceleration.

Answer:

If a body is thrown upwards from the earth’s surface, its velocity gradually decreases due to force of gravity acting on it. This means it decelerates. Here, gravity is the attraction force.

Again, when the body moves downwards after reaching the highest point, its velocity gradually increases. This means it accelerates. This acceleration also takes place due to gravity.

Hence, attraction force may create acceleration as well as deceleration.

“law of conservation of momentum class 9 formula ”

Question 14. A sand clock is weighed twice using a sensitive common balance, once when sand particles are slowly falling from upper chamber to lower chamber and the other time when all particles have accumulated in the lower chamber. Are the two weights equal?

Answer:

In the first case, when some sand particles keep falling downward from the upper chamber, they do not apply any force on the common balance.

 

But in the second case, when all the particles have accumulated in the lower chamber, all the sand particles apply force on the common balance. So, reading in the first case is somewhat lower than the reading in the second case.

Question 15. What do you mean by thrust?

Answer:

Suppose, a man is sitting on a floor. He applies a force on the floor acting downward and equal to his own weight and the floor also applies an equal reaction force on the man, but in the opposite direction. This type of force is called thrust.

In other words, if a body is kept on another body, then the mutual action-reaction force applied by them is called thrust.

Question 16. What do you mean by push?

Answer:

A tennis ball is thrown toward the wall. When the ball touches the wall, it applies a force on the wall and the wall also applies an equal and opposite force on the tennis ball as reaction. As a result, the tennis ball and the wall try to move away from each other. This type of action-reaction is called push.

In other words, if two bodies want to move away from each other during the period of their contact due to mutual action-reaction, then this action- reaction is called push.

Question 17. Establish the law of conservation of third law of motion.

Answer:

Suppose two bodies of masses m₁ and m₂ are moving in the same straight line with velocities u₁ and u₂. There is a collision between these two bodies if u₁ > u₂.

After collision, these bodies move with velocities v₁ and v₂ in the same straight line. During collision, force applied by the body with mass m₁ on the body with mass m₂ is F₁ and the force applied by the body with mass m₂ on the body with mass m₁ is F₂.

Clearly, F₁ and F₂ are action and reaction,

∴ according to Newton’s third law of motion,

F₁ = -F₂ ……(1)

Now, force F₁ is applied on the body with mass m₂.

∴ \(F_1=\frac{m_2 v_2-m_2 u_2}{t}\) [t= time of collision]

Again, force F₂ is applied on the body with mass m₁.

“momentum conservation equation “

∴ \(F_2=\frac{m_1 v_1-m_1 u_1}{t}\)

So from equation (1), we get

\(\frac{m_2 v_2-m_2 u_2}{t}=-\frac{m_1 v_1-m_1 u_1}{t}\)

or, \(m_2 v_2-m_2 u_2=-m_1 v_1+m_1 u_1\)

or, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)

∴ total linear momentum remains unchanged after collision of two bodies, unless an external force acts on the system.

Question 18. Establish Newton’s third law of motion with the help of law of conservation of linear momentum.

Answer:

Suppose two bodies of masses m₁ and m₂ are moving in the same straight line with velocities u₁ and u₂ (u₁ > u₂). After collision, these bodies will move with velocities v₁ and v₂ in the same straight line.

From the law of conservation of linear momentum, we get

\(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)

or, \(m_2 u_2-m_2 v_2=m_1 v_1-m_1 u_1\)

or, \(-m_2\left(v_2-u_2\right)=m_1\left(v_1-u_1\right)\)

or, \(\frac{m_2\left(v_2-u_2\right)}{t}=-\frac{m_1\left(v_1-u_1\right)}{t}\) [t= time of collision]

Now, it is known that \(F_1=\frac{m_2\left(v_2-u_2\right)}{t}\) is the force applied on the second body by the first body and \(F_2=\frac{m_1\left(v_1-u_1\right)}{t}\) is force applied on the first body by the second body.

∴ F₁ = -F₂

Hence, if F₁ is action, F₂ is reaction. Further, F₁ and F₂ are equal and act in opposite directions. This is Newton’s third law of motion.

Question 19. If a bullet is fired from a gun, explain this phenomenon with the help of the law of conservation of linear momentum.

Answer:

When a bullet is fired from a gun, the bullet moves forward with high velocity, and the gun also recoils or kicks back immediately. We can explain this phenomenon with the help of the law of conservation of linear momentum.

Suppose, a bullet of mass m is fired with velocity u in the forward direction from the gun of mass M. As a result, the velocity of the gun becomes V.

Just before firing the bullet, total linear momentum of the system = 0, and the moment after the firing the bullet, total linear momentum = MV+ mu.

Since no external force is acting here. So from the law of conservation of linear momentum, we get 0 = MV + mu or,V = –\(\frac{m}{M}\)u…..(1)

“prove conservation of linear momentum using newton’s third law “

Negative sign in equation (1) indicates that directions of velocities of bullet and gun are opposite, i.e., if the bullet moves forward, the gun recoils backward.

Question 20. A body is at rest in a floor. After the explosion, it divided into two parts of equal mass. A parts moves towards north direction with velocity u. In which direction and with what velocity the other part of the body will move?

Answer:

Let, mass of the each parts of the body be m and the second part moves with velocity v.

Now, according to conservation of linear momentum we can write, (m + m) x 0 = mu + m • V m

or, \(v=-\frac{m}{m} \cdot v=-u\)

∴ The second part of the body will move with velocity u and in the direction opposite to the first part i.e., the second part will move towards south direction with the velocity u.

Question 21. Explain the working principle of a rocket with the help of law of conservation of linear momentum.

Answer:

The working principle of a rocket can be explained with the help of law of conservation of linear momentum. Due to the burning of fuel in the combustion chamber of a rocket, gas produced is emitted out through the nozzle at one end with high velocity, producing thrust.

Gas emitted through the nozzle has a high value of linear momentum. As a result, the rocket also attains a linear momentum of equal value in opposite direction and moves forward with high velocity.

Question 22. Five balls of equal masses are kept in a straight line, touching each other. The balls are suspended from separate strings. Now, the ball at the extreme left is displaced a little and then let loose. It is seen that when this ball strikes the next one, only one ball of the other end is budged (displaced). Again, if two balls of the left side strike in the same way, two balls of the right side are displaced. What is the reason behind these phenomena?

Answer:

According to the question, the balls have the ass. When one of them is displaced and released, it strikes the next ball. Now according to the law of conservation of momentum, the total linear momentum of the system should remain conserved.

Consequently, only one ball of the other end is deflected with the same velocity (the velocity with which the left ball struck).

If two balls of the left side strike in the same way, then two balls from the right side are deflected with the same velocity so that the total linear momentum of the system remains conserved.

Question 23. Why is it difficult to hold a hose pipe properly when water emits through the pipe with high velocity?

Answer:

When water emits out with high velocity through a hose pipe, the emitted water has a definite momentum in the forward direction. So, according to the law of conservation of linear momentum, the pipe would also have a momentum of equal value but in the opposite direction.

For this reason, force has to be applied to hold the pipe, making it a bit difficult.

 

Question 24. Why a cricket player lowers his hand while catching a cricket ball? Explain.

Answer:

When a cricketer catches a ball, its momentum reduces to zero in his hands. If initial momentum of the cricket ball be P, t be the time duration in which the ball stops in his hand and F be the average force applied by the cricketer to stops the ball, then,

\(F=\frac{\text { change of momentum }}{\text { time }}=\frac{0-P}{t}=\frac{P}{t}\)

(-ve sign shows that force is applied against the motion of the ball).

“prove conservation of linear momentum using newton’s third law “

By lowering his hands while catching the ball, the player allows a longer time for change in momentum and stops the ball slowly in hands, and protects his hands from getting injury.

Question 25. A wooden block of mass m is kept on a table. What is its normal reaction?

Answer:

A is a wooden block of mass m kept on a table. The earth extras downward force on the block which is its weight, w = mg.

The table also extras an equal and opposite reaction force of magnitude W = mg on the block. This is the normal reaction force (N) on the block that directed vertically upward.

Question 26. A body is placed on an inclined plane such that it is in rest without slipping. In which direction normal reaction force acts? Explain with proper diagram.

Answer:

If a body is kept at rest in an inclined plane then normal reaction force acts on the body in upward direction perpendicular to the inclined plane.

[If mass of the body be m and angle of inclination of the inclined plane be θ, then weight of the body W = mg, component of weight perpendicular to the plane = W cosθ.

Therefore normal reaction force is R = Wcosθ.]

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Multiple Choice Questions

Question 1. The basis of working principle of a jet engine is

1. Principle of conservation of mass
2. Principle of conservation of energy
3. Principle of conservation of linear momentum
4. Principle of conservation of angular momentum

Answer: 3. Principle of conservation of linear momentum

Question 2. The motion of a rocket is established on the principle of conservation of

1. Energy
2. Kinetic energy
3. Linear momentum
4. Mass

Answer: 3. Linear momentum

Question 3. Action and reaction

1. Are applied on the same body
2. Are applied on different bodies
3. Are equal and unidirectional
4. Are in opposite directions but their values are not

Answer: 2. Are applied on different bodies

“prove conservation of linear momentum using newton’s third law “

Question 4. A gun recoils if a bullet is fired from it

1. According to Newton’s second’law of motion
2. According to Newton’s first law of motion
3. Due to reaction force
4. Due to action force

Answer: 3. Due to reaction force

Question 5. The principle of conservation of linear momentum states that the linear momentum of a system

1. Can not be changed
2. Can not remain constant
3. Can be changed only if internal forces act
4. Can be changed only if external forces acts

Answer: 4. Can be changed only if external forces acts

Question 6. A boy of mass 40 kg jumps from a boat of mass 200 kg at a velocity 10 m/s. What is the velocity of the boat?

1. 2 m/s
2. 1 m/s
3. 4 m/s
4. 5 m/s

Answer: 1. 2 m/s

Question 7. A bullet of mass m hits a wooden block of mass M with velocity v and attached with it. Velocity of the system after collision is

1. \(\frac{M}{M+m} v\)
2. \(\frac{M-m}{M} v\)
3. \(\frac{m}{M+m} v\)
4. \(\frac{M+m}{M} v\)

Answer: 3. \(\frac{m}{M+m} v\)

Question 8. A gun fires N bullets per second, each of mass m with velocity v. The force exerted by the bullet on the gun is

1. \(\frac{m v}{N}\)
2. \(\frac{m v^2}{N}\)
3. mNv
4. mNv²

Answer: 3. mNv

Question 9. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio of 3:2. The ratio of their mass will be

1. 3:2
2. 2:3
3. 9:4
4. 4:9

Answer: 2. 2:3

Question 10. The difference between the ,nature of a rocket and a jet is

1. Rockets can only move in air but jets can not
2. Rockets can move out side the atmosphere but jets can not
3. Rockets obtain oxygen from air but jets carries its own oxygen
4. Both of them carries own oxygen

Answer: 2. Rockets can move out side the atmosphere but jets can not

Question 11. If 10 N force acts on a body, then reaction force will be

1. 10 N along the direction applied force
2. 20 N along the direction of applied force
3. 10 N opposite to the direction of the applied force
4. 20 N opposite to the direction of the applied force

Answer: 3. 10 N opposite to the direction of the applied force

Question 12. A and B are two spring balances of negligible small weight. Mass M is hang from lower end of B. Reading of any of the two spring balances is

1. M kgf
2. Mg kgf
3. 2M kgf
4. 2 Mg kgf

Answer: 1. M kgf

Question 13. Conservation of linear momentum is applicable when external force will be

1. Very small in magnitude
2. Zero
3. Very large in magnitude
4. Constant in magnitude

Answer: 2. Zero

Question 14. An iron ball collides with another iron ball at rest. Here the force comes into play is

1. Friction force
2. Normal reaction force
3. Tension force
4. Collision force

Answer: 4. Collision force

Question 15. When a book lying on a table is pushed, it moves some distance and then stops. The external force which affects the motion of the body is

1. Collision force
2. Tension
3. Friction
4. Normal reaction

Answer: 3. Friction

Question 16. In the case of a game of tug-of-war, if both the parties exert a force T from either side, what will be the tension in the rope?

1. T
2. 2T
3. T/2
4. 3

Answer: 1. T

Question 17. According to Newton’s third law of motion, the angle between action force and its reaction force

1. 0°
2. 180°
3. 90°
4. 360°

Answer: 2. 180°

Question 18. Which is not a contact force?

1. Friction
2. Gravitational force
3. Tensile force
4. Collision force

Answer: 2. Gravitational force

Question 19. Recoil velocity of a gun

1. Less than the velocity of the bullet
2. Grater then the velocity of the bullet
3. Equal to the velocity of the bullet
4. A and B Both

Answer: 1. Less than the velocity of the bullet

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Answer In Brief

Question 1. What is the force due to which it is difficult to push a body on the ground?

Answer: It is difficult to push a body on the gro due to the force of friction between ground and the body.

Question 2. “Action-reaction can establish equilibrium” is the statement, true or false?

Answer: The statement is false.

Question 3. The working principle of a rocket depends on which basic principle?

Answer: The working principle of a rocket depends on the principle of conservation of linear momentum.

Question 4. What is repulsion?

Answer: Due to action-reaction, if two bodies move away from each other, then it is called repulsion.

Question 5. Write down the law of conservation of linear momentum.

Answer: The law states that if no external force is applied, the total linear momentum of a body or system of bodies always remains unchanged.

Question 6. What do you mean by friction?

Answer: When a body moves or tends to move on another body or on any surface, then the opposing force that acts against this movement or tendency of movement is called friction.

Question 7. What change in total linear momentum does occur in the case of collision between two bodies?

Answer: If no external force is applied total linear momentum of the system will remain unchanged.

Question 8. Give one example of contact force.

Answer: Force of friction or simply friction is an example of contact force.

Question 9. Why it is difficult to move a body over the earth’s surface?

Answer: Due to the arise of frictional force it is difficult to move a body over the earth’s surface.

Question 10. Give one example of a non contact force.

Answer: Gravitational force of attraction is an example of non contact force.

Question 11. What do you mean by friction?

Answer: When a body moves or tends to move on another body or any surface, then the opposing force that acts against this movement or tendency of movement is called friction.

Question 12. Why do not the forces of action and reaction cancel each other?

Answer: Action and reaction forces act on different bodies which is why they do not cancel each other.

Question 13. Name the principle on which a rocket works.

Answer: Conservation of linear momentum is the principle on which a rocket works.

Question 14. A bucket full of water is suddenly pushed. State whether the water splashes forward or backward?

Answer: Water splashes backward.

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Fill In The Blanks

Question 1. Every action has an _________ and opposite _________

Answer: Equal, reaction

Question 2. Rocket propulsion is possible even in an __________ place.

Answer: Air-free

Question 3. Action-reaction cannot ________ mutually.

Answer: Balance

Question 4. The theory of flight of a jet plane can be explained by Newton’s _______ law of motion.

Answer: Third

Question 5. Action and reaction forces acts on ________ bodies.

Answer: Different

Question 6. If a body A slips over another body B the force comes into play that resist this motion is called ________

Answer: Friction

Question 7. The linear momentum of a system remains constant if no _______ force acts on it.

Answer: External

Question 8. Normal reaction force is a ________ force.

Answer: Contact

Question 9. Magnetic force is an example of _______ force.

Answer: Non-contact

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum State Whether True Or False

Question 1. The total linear momentum of a system of bodies can be changed without the application of an external force.

Answer: False

Question 2. A spring can pull an object as well as push an object.

Answer: True

Question 3. A ball moving on a horizontal surface stops because of the force of friction.

Answer: True

Question 4. Any pair of equal and opposite forces forms an action-reaction pair.

Answer: False

Question 5. Magnetic force is contact force.

Answer: True

Question 6. Birds can fly in air-free space.

Answer: False

Question 7. If the magnitude of action force be 15 N then the magnitude of the reaction force will be 15 N.

Answer: True

Question 8. A boat is floating on still water. A man walks from one end of the boat moves backwards.

Answer: True

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Numerical Examples

Useful relations

Two bodies of masses m₁ and m₂ respectively, moves in a straight line with velocities u₁ and u₂ (u₁ > u₂) respectively, collides with each other.

1. If after collision these two bodies moves along the same straight line with velocity v₁ and u₂ respectively, then, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
2. If after collision they coalesce with each other and moves with velocity V in a straight line then,

\(m_1 u_1+m_2 u_2=\left(m_1+m_2\right) V\)

 

or, \(v=\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\)

If a bullet of mass m is fired with a velocity u from a gun of mass M and the recoil velocity of the gun be V, then,

(M + m) x 0 = mv + M • V

or, V = -(\(\frac{m}{M}\))v

Question 1. A bullet of mass 20 g is fired from a gun of mass 5 kg with a velocity of 400 m/s. What is the recoil velocity of the gun?

Answer:

Mass of the gun, M = 5 kg; mass of the bullet, m = 20 g = 0.02 kg; velocity of the bullet, u = 400 m/s

Let the velocity of the gun be V.

So, from the law of conservation of linear momentum, we get

mu + MV = 0 or, MV = -mu

or, \(V=-\frac{m u}{M}=\frac{0.02 \times 400}{5}=-1.6 \mathrm{~m} / \mathrm{s}\)

∴ recoil velocity of the gun = 1.6 m/s .

Question 2. A bullet of mass 20 g moving with a velocity of 500 m/s strikes a stationary wooden block of mass 3.98 kg. Calculate the velocity of the wooden block.

Answer:

Mass of the bullet, m = 20 g = 0.02 kg; velocity of the bullet, u = 500 m/s; mass of the wooden block, M = 3.98 kg

Let us assume that the bullet is stuck in the wooden block and both, as a single system, move with a velocity V.

So, from the law of conservation of linear momentum, we get mu = (M + m)V

∴ \(V=\frac{m u}{(M+m)}=\frac{0.02 \times 500}{(3.98+0.02)}=2.5 \mathrm{~m} / \mathrm{s}\)

Question 3. A piece of stone of mass m is sliding over a smooth surface of ice with a velocity v. The stone is collected by a boy of mass M who is standing on the surface. What is the gain in velocity of the boy standing on the surface of ice?

Answer:

Since the surface of the ice is smooth, it is frictionless. So, no external resistant force is applied on the boy or the stone.

Then, according to the law of conservation of momentum, the sum of initial velocity of the moving stone and initial momentum of the boy is equal to the final velocity of the boy with stone in his hand.

So, mv + 0 x M = (m + M) x V where V = velocity attained by the boy.

∴ \(V=\frac{m v}{(m+M)}\)

Question 4. A body (A) of mass 100 g and another body (B) of mass 400 g are approaching each other with velocities 100 cm/s and 10 cm/s, respectively. After a head on collision, they coalesce with each other. This coalesced mass will move in which direction after the collision and how much distance will it cover in 10s?

Answer:

As no external force is being applied, so the law of conservation of momentum is applicable here. Suppose, velocity of the coalesced mass after collision = V.

So, sum of individual momentums of A and B before collision is equal to the total momentum after collision.

So, 100 X 100 + 400 X (-10) = (100 + 400) x V

[As velocities of the two masses are in opposite directions, if the velocity of one body is taken as positive, the other must be negative]

So, 10000 – 4000 = 500 V

Positive sign of the velocity of 12 cm/s signifies that after collision, the coalesced mass will move in the direction of the mass of 100 g.

∴ distance traversed by this coalesced mass in 10s, s = vt = 12 x 10 = 120 cm .

Question 5. A rocket consumes fuel at the rate of 100 kg • s^-1. Gas ejects out of it with a velocity of 5 x 10³ m/s. Find the force experienced by the rocket.

Answer:

Here, rate of consumption of fuel is = 100 kg/s.

Velocity of ejected gas = 5 x 10³ m/s.

∴ Upward impulsive force experienced by the rocket is

= rate of consumption of fuel x velocity of ejected has = 100 x 5 x 10³ N = 5 x 10⁵ N

Question 6. A body of mass 5 kg while moving with velocity 15 m/s explodes in two parts of mass 3 kg and 2 kg. if the first part moves with velocity 5 m/s find velocity of the second.

Answer:

Initial mass of the body M = 5 kg and initial velocity V = 15 m/s.

Mass of the first part m₁ = 3 kg and mass of the second part m₂ = 2 kg.

Velocity of the first part v₁ = 5 m/s

Let, velocity of the second part be v² m/s.

∴ From conservation of linear momentum,

M • V = m₁v₁ + m₂v₂

or, 5 x 15 = 3 x 5 + 2 x v₂ or, 2v₂ = 75 – 15

∴ v₂ = 60/2 = 30 m/s

∴ Velocity of the second part is 30 m/s.

Question 7. From a gun of mass M, a bullet of mass m is fired with a velocity v. If recoil of the gun stops after time t second by applying an average force F, show that F = \(\frac{m}{v}{t}\).

Answer:

Here, initial mass of the gun bullet system is = M and initial velocity of the system = 0.

Velocity of the bullet = v and mass of the bullet = m.

Let, recoil velocity of the gun be V

∴ According to conservation of linear momentum,

0 = MV + mv

∴ MV = -mv

∴ Average force applied on the gun to stop its recoil

\(F =\frac{\text { change of momentum of the gun }}{\text { time }}\)

= \(\frac{0-M V}{t}=-\frac{M V}{t}=\frac{m v}{t}\) (proved)

Chapter 2 Topic C Newtons Third Law Of Motion And Law Of Conservation Of Linear Momentum Miscellaneous Type Questions

Match The Columns

Question 1.

Answer: 1. B, 2. A, 3. D, 4. C

Question 2. 

Answer: 1. C, 2. A, 3. D, 4. B

Chapter 2 Topic B Newton’s First And Second Laws Of Motion Synopsis

Newton’s First Law Of Motion:

Everybody continues to be in its state of rest or of uniform motion in a straight line unless an external force is applied to it.

Inertia is the property of a body existing in a state of rest or of uniform motion due to which it continues to remain in that state and opposes any effort or attempt to change that state.

The tendency of a moving body to continue to exist in its state of motion with uniform velocity in a straight line is called inertia of motion.

The tendency of a stationary body to continue its state of rest is called inertia of rest.

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A single force, which represent the result of the summation of a number of forces both in magnitude and in direction, is called the resultant of those forces.

The Law Of Parallelogram Of Forces:

If a particle is acted on by two forces represented in magnitude and direction by the two sides of a parallelogram drawn from a point simultaneously then they are equivalent to a force represented in magnitude and direction by the diagonal of the parallelogram passing through the point.

Explanation: If P and Q be two forces acting simultaneously on point A are represented in magnitude and direction by two adjacent sides AB and AD of a parallelogram ABCD respectively, then their resultant force R is given in magnitude and in direction by the diagonal of the parallelogram.

Resolution of a force refers to the division of a force in two given directions in such a way that the resultant of the two components is equal to the given force.

Resolution Of A Force Into Components At Right Angles To Each Other:

Resolution of a force into components at right angles to each other means that the force is resolved into two components which are at right angles to each other, so that the resultant of the two components is equal to the given force.

Explanation: A force R is resolve into two components P and Q perpendicular to each other.

Here, \(R^2=p^2+Q^2 \text { or, } R=\sqrt{p^2+Q^2}\)

Where 9 is the angle between R and P.

When more than one force acts on a body such that the resultant of the forces is zero, then the forces are called balanced forces. Balanced forces acting on a body do not produce acceleration of the body but the body becomes strained.

When one or more than one force acts on a body, the resultant of the acting forces is non-zero, then these forces are called effective forces or unbalanced forces. Acceleration of a body arises mainly due to the influence of effective force.

During application of force on a body, if the applicator is a part of the whole system, then the applied force cannot change the velocity of the particle but only attempts to change it. This force is called internal force.

During application of force on a body, if the applicator of the force and the body are different systems, then the applied force changes the velocity of the body or attempts to change the velocity. This force is called external force.

The ratio of the applied force on a body and the acceleration of the body is called, its inertial mass.

Linear Momentum is defined as the property of motion of a body that is produced due to the combination of velocity and mass of the moving body. Linear momentum is a vector quantity. Unit of linear momentum in SI is kg • m • s^-1 and its dimensional formula is MLT^-1.

Newton’s Second Law Statement:

The time rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.

Mathematical Expression: Suppose an external constant force F is applied on a body of constant mass m.

If a be the acceleration of the body, then according to Newton’s second law of motion

F = ma, i.e., Force = mass x acceleration

The force applied on a body of unit mass and produces an acceleration of one unit is called unit force.

Newton’s Second Law Of Motion For A Constant Mass:

If the mass unchanged, then the effective force applied on the body is equal to the product of its mass and acceleration. Acceleration takes place in the direction of the applied force.

Variable Mass System is a system whose mass increases or decreases during its state of motion but its mass is never created nor destroyed.

Newton’s Second Law Of Motion For A Variable Mass:

The rate of change of momentum of a body is equal to the external working force applied on the body. Change of momentum takes place in the direction in which the force is applied.

Units Of Force in CGS system and SI are dyne and newton respectively.

The amount of force that acts on a mass of 1 g and produces an acceleration of 1 cm/s² is called 1 dyne.

The amount of force that acts on mass of 1 kilogram and produces an acceleration of 1 m/s² is called 1 newton.

Relation Between Newton And Dyne:

I N = 1 kg x 1 m • s^-2

= 1000 g x 100 cm • s^-2  = 10⁵ g • cm • s^-2 = 10⁵ dyn

Dimensional Formula Of Force:

Dimensional formula of force = Dimensional formula of mass x Dimensional formula of acceleration

= M x LT^-2 = MLT^-2

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Chapter 2 Topic B Newtons First And Second Laws Of Motion Short And Long Answer Type Questions

Question 1. What is resultant force?

Answer:

A resultant force is a single force obtained by combining several active forces. The important describing feature of a resultant force is that it has the same effect on the body as the original system of forces.

Question 2. What are the effects of an applied force on a body.

Answer:

By applying a force,

1. A static body may be set in motion,
2. Magnitude or direction or both of a moving body may be changed,
3. A moving body may be brought to a stationary state or
4. Form or shape of a body may be changed.

Question 3. What is balanced force?

Answer:

When more than one force acts on a body and the resultant of the forces is zero, then the forces are called balanced forces.

Question 4. State Newton first law of motion.

Answer:

Every body continues to be in its state of rest or of uniform motion in a straight line unless the body is compelled to change its state by a net external force i.e. there will be no acceleration of the object.

Question 5. Give an example of balanced force.

Answer:

In a tug-of-war game, four boys on the left side and four boys on the right side pull the rope in such a way that the rope does not move in either direction. Here, a total of eight forces are working on the rope.

But as the forces applied by the boys on the left side are equal to the forces applied by the boys on the right side and are in the opposite directions, resultant force on the rope is zero.

Hence, the rope does not move in any direction. These eight forces working on the rope are called balanced forces.

Key Questions on Newton’s First and Second Laws for Class 9

Question 6. What do you mean by effective force? Which property of the body changes due to its influence?

Answer:

When the resultant of the forces due to the application of one or more than one force on a body is not zero, then those forces are called effective forces. The property of motion (rest or motion) changes due to the influence of effective forces.

Question 7. Two friends are pulling a table without being able to move it. How is it possible?

Answer:

If two friends are Ruling a table from two sides by applying two opposite and equal forces, then it will not move. This type of application of force means that balanced force is being applied on the table.

Due to this reason, the value of the resultant force on the table is zero.

Question 8. Two friends are moving a table by pulling it with equal forces. Application of forte in which direction does make this possible?

Answer:

If the table is to be moved by the two friend by pulling it with equal force, the forces can never be applied opposite to each other.

Applying the forces in any other direction budges the table.

Question 9. What is the influence of effective force on a body?

Answer:

The action of effective force on a body generates acceleration, i.e., change of velocity.

Change of velocity may take place in different ways:

1. A static body may start moving,
2. There may be a change in the velocity of a moving body,
3. There may be a change of direction of the moving body or
4. Both the magnitude and the direction of the moving body may change.

Question 10. The action of balanced force on e body creates strain—explain with example.

Answer: A spring shrinks when it is compressed with equal force by two hands. In this case, equal and opposite forces are applied on the spring by the hands.

This is a balanced force. No acceleration is produced in the spring due to this. But the spring gets compressed. This compression or shrinkage of the spring is known as strain.

Concepts Related to Mass, Acceleration, and Force for Class 9 Solutions

Question 11. What is internal force and external force?

Answer:

Internal Force:

During application of a force on a body, if the applicator is a part of the system applying force, then the applied force cannot change the velocity of the particle but only attempts to change it. This force is called internal force.

External Force:

During application of a force on a body, if the applicator of the force and the body are different systems, then applied force changes the velocity of the body or attempts to change the velocity. This force is called external force.

Question 12. Which force is responsible for change of the state of a body? Answer with the help of Newton’s first law of motion.

Answer: It is obvious from Newton’s first law of motion that it is possible to change the state of rest or state of motion of a body with the help of only external effective force. Internal force or external balanced force cannot do it.

Question 13. Give an idea of force from Newton’s first law of motion.

Answer:

We get a qualitative definition of force from Newton’s first law of motion. The velocity of a body does not change if any external effective force is not applied on the body.

This means that there is no change in velocity of the body if it is zero (body at rest) or of any definite magnitude (with uniform velocity).

Question 14. What are the things known from Newton’s first law of motion?

Answer:

The following things are known from Newton’s first law of motion:

1. Property of inertia of matter.
2. Qualitative definition of force.
3. Change of velocity of a body is possible only by external unbalanced force and not by any internal force.
4. A clear statement that there is some sort of similarity or resemblance between the state of rest and state of motion with uniform velocity.

Question 15. What is inertia? How many types of inertia are there and what are those?

Answer:

The property of a body due to which it exists in a state of rest or of uniform motion and continues to remain in that state and opposes any attempt to change that state, is called inertia.

It is of two types, inertia of rest and inertia of motion.

Question 16. What Is inertia of rest and inertia of motion?

Answer:

Inertia Of Rest:

The tendency of a stationary body to always continue in its state of rest is called inertia of rest.

Inertia Of Motion:

The tendency of a moving body to always continue in its state of motion with uniform velocity in a straight line is called inertia of motion.

Question 17. A stationary bus starts moving suddenly. Why do the passengers standing inside lean in the backward direction?

Answer:

When the bus is at rest, the bodies of the passengers in contact with it also remain at rest.

When the bus starts moving suddenly, the lower parts of the bodies of the passengers in contact with the floor of the bus move forward while the upper parts tend to remain at rest due to inertia and the passengers lean in the backward direction.

Study Guide for Class 9 Force and Motion Questions

Question 18. What happens if a bucket full of water is suddenly pushed?

Answer:

When a force is applied suddenly to a bucket full of water in the forward direction, the bucket moves forward. But water inside the bucket continues to remain at rest due to inertia of rest. So, water splashes in the backward direction.

Question 19. If a running bus stops suddenly, passengers standing inside lean forward. Why? Or, It is not prudent to get down from a running bus, because if one does it carelessly he may fall in the forward direction. Explain.

Answer:

If a running bus with passengers stops suddenly, passengers within the bus stumble or lean in the forward direction. The reason is that when the bus was in motion, the passengers were also moving with the same velocity.

When the bus stops by the application of brakes, lower parts of the bodies of passengers remaining in contact with the floor stop immediately.

But the upper parts move forward due to inertia of motion and as a result, passengers lean forward. This is the reason why one should not get down from a running bus.

Question 20. Why cannot the participant of a sprinting competition stop after reaching the end point?

Answer:

In a sprinting competition, the goal of the competitor is to reach the end point before all.

With this goal, he does not reduce his speed even at the moment before reaching the end. Now, if he stops suddenly after reaching the end point, the upper part of his body continues to move forward due to inertia of motion.

As a result, there are chances of his stumbling in the forward direction. In order to save himself from injury due to the fall, he does not stop suddenly. Rather, he gradually reduces his speed and runs for some distance before stopping.

“numericals of physics class 9 chapter 2 “

Question 21. When a blanket is to be dusted, why is rope and then beaten with a stick?

Answer:

To dust a blanket, it is hung on a support (say, a rope) and then beaten vigorously with a stick. The reason is that when beaten vigorously, the blanket moves in the direction of the applied force.

But the dust particles attached lightly to the blanket prefer to continue in stationary state due to inertia of rest. As a result, dust particles get separated from the blanket, thereby cleaning it.

Question 22. Is it possible for a passenger sitting inside a stationary car to move it by pushing from the inside?

Answer:

No, it is not possible for a passenger sitting inside a stationary car to move it by pushing from the inside. Because a passenger sitting inside may be considered to be a part of th system.

Without application of force from outside, it is not possible to move it by overcoming the frictional force between the car and the road.

Question 23. A passenger sitting inside a bus moving on a horizontal plane with uniform velocity throws a ball upward. After some time, how does the ball come back in the hands of the passenger?

Answer:

The velocities of the passenger and the ball inside the bus moving with uniform velocity are always equal to the velocity of the bus. When the ball is thrown up, an upward motion is also created along with horizontal motion.

Horizontal velocity and perpendicular velocity are in perpendicular direction to each other. So, the velocities do not affect each other. This is the reason why the ball tries to maintain its inertia of motion in the horizontal direction even when it is in air.

As a result, the distance traversed by the passenger in the forward direction is equal to the distance traversed by the ball in the forward direction during the same period.

On the other hand, motion of the ball in the perpendicular direction is accompanied by acceleration due to gravity. So, its velocity gradually decreases to zero and then it falls downwards to reach the hands of the passenger.

Question 24. Write down Newton’s second law of motion without introducing the concept of momentum. Or, Write down Newton’s second law of motion in case of unchanged mass.

Answer:

If the mass of a body remains unchanged, then the magnitude of external applied effective force on the body is the product of its mass and acceleration. Acceleration takes place in the direction of the applied force.

This is Newton’s second law of motion without the concept of momentum or in case of unchanged mass.

Question 25. Explain the equation F = m x o of Newton’s second law of motion with examples.

Answer:

A qualitative definition of force is obtained from Newton’s first law of motion. But a quantitative definition of force is obtained from Newton’s second law of motion. A driver starts his car.

As the engine of the car applies force on the car, it starts moving. The driver increases the velocity with the help of the accelerator to raise it to a certain value. Accelerator is responsible for the acceleration of the car.

Force applied by the engine is responsible for this acceleration. Now, if the applied force by the engine increases, acceleration also increases. So, there is a direct relationship between the applied force and the acceleration produced.

According to the second law of motion, force applied on a body of definite mass and its acceleration are directly proportional and the direction of the acceleration generated is in the direction of applied force, i.e., applied force = mass x acceleration

If applied force = F, mass of the body = m, and acceleration = a, then F = ma.

Question 26. How do you infer that mass is an inherent property of matter? Or, A hanging sand bag and an oscillating pendulum are displaced by the same distance. Which one requires more force to displace and why?

Answer:

A sand bag requires more force for its displacement by the same distance compared to that for an oscillating pendulum. Both of them were initially at rest.

Now, while producing the same amount of acceleration in them by application of force, both the bodies oppose the forces due to their properties of inertia.

This property of inertia of a body is measured by its mass. Since the mass of the sand bag is many times greater than the mass of the oscillating pendulum, hence more force is required in this case.

Question 27. From Newton’s second law of motion, explain mass as an inherent property of a body.

Answer:

According to Newton’s second law of motion, force (F) applied on a body = mass of the body (m) x acceleration (a) due to the application of force.

According to this, we can infer that applying same amount of force on different stationary bodies do not produce the same magnitude of acceleration in them.

“force and law of motion class 9 question answer “

The reason due to which this difference arises is the mass of the body. So, mass is an inherent property of a body. It is not possible to change the mass of a body unless quantity of matter in the body is increased or decreased.

Question 28. According to Newton’s second law of motion, if force is taken as the cause then acceleration may be treated as the effect.Explain the above statement with reasons.

Answer:

From our ordinary experience, we observe that more the amount of force applied on a body, more is the amount of change of its state of motion, that is, the acceleration or deceleration of the body will be more.

For example, a rope is tied to a solid brick kept on the ground and one starts walking by holding the rope in his hand. The brick starts moving with a definite velocity from its state of rest.

This means that the brick has got acceleration. Instead of walking, if one starts running, the brick also starts moving with the same velocity. Acceleration is greater in the second case.

Hence, we may conclude that increase of force on the body increases its acceleration. So, if force is taken as the cause, acceleration may be regarded as the effect.

Question 29. Mass of a body is the measure of its inertia—explain.

Answer:

Suppose, forces F₁ and F₂ are applied on two bodies of masses m₁ and m₂ respectively to produce same acceleration on them.

According to Newton’s second law of motion,

\(F_1=m_1 a \text { and } F_2=m_2 a\)

∴ \(\frac{F_1}{F_2}=\frac{m_1}{m_2}\)

Now as m₁ > m₂, F₁ is greater than F₂. Therefore, to produce same amount of acceleration (that is, to change state of motion by same amount at the same time) in a heavier body, a comparatively greater amount of force has to be applied.

Resistance against the force, i.e., property of inertia will be greater for a heavier body.

Question 30. How is inertial mass measured?

Answer:

Let the acceleration of a body be a when a force of F is applied on it. From Newton’s second law of motion, we may say that mass of the body,

\(m=\frac{F}{a}\)

So, the ratio of the force applied on a body and the acceleration of the body is the inertial mass (m,).

∴ inertial mass, \(m_i=\frac{F}{a}\)

Question 31. Draw applied force-acceleration graph for a body of constant mass.

Answer:

If a body of constant mass m is moving under the action of an external applied force F. Then according to Newton’s second law of motion,

F = ma, where a is the acceleration of the body.

If a is considered along x-axis (horizontal axis) and F is considered along y-axis (vertical axis) then F-a graph is a straight line that passes through the origin and the gradient of the straight line is the mass (m) of the body.

[gradient (m) = tanθ = \(\frac{F}{a}\) = m]

Question 32. Define unit force from Newton’s second law of motion.

Answer:

If a force F is applied on a mass of m to produce acceleration a, then according to Newton’s second law of motion, F = ma

Now, if m = 1 and a = 1, then F = 1.1 = 1.

So, the amount of force applied on a body of unit mass to produce unit acceleration is called a unit force.

Question 33. Write down the definition of units of force in CGS system and SI. Establish a relationship between them.

Answer:

Units of force in CGS system and SI are dyne (dyn) and newton (N), respectively.

The amount of force that is applied on a body of mass of 1 g to produce an acceleration of 1 cm/s² is called 1 dyn.

∴ 1 dyn = 1 g • cm • s^-2

Again, the amount of force that is applied on a body of mass 1 kg to produce an acceleration of 1 m/s^-2 is called 1 N.

∴ 1 N = 1 kg • m • s^-2

Relation between Newton (N) and dyn:

∴ 1 N = 1 kg. m.s^-2

= 1000 g x 100 cm/s²= 10⁵ dyn

Question 34. With your heavy school bag full of books kept in the carrier of the bicycle, you are riding the bicycle to go to the school with uniform velocity. The bag suddenly falls from the carrier.

1. What is the change of motion of the cycle at that moment due to this fall?
2. What is the change of this system-cycle with the bag?
3. Give an idea of momentum through this incident.

Answer:

1. The velocity of the cycle increases at the moment when the bag falls.
2. When the bag falls, the mass of the system decreases and as a result, its velocity increases.
3. It is clear from this incident that the velocity of the body may change simply by a change of mass of the moving body without applying any force from outside. Therefore, a new property of the body emerges by combination of the velocity of the moving body with its mass. This property is called the momentum of the moving body.

Question 35. A cycle and a rickshaw moving with the same uniform velocity are to be stopped within the same interval of time. Greater force has to be applied on which vehicle to achieve the desired result?

Answer: One has to apply a greater force to stop the rickshaw compared to the moving cycle within the same interval of time. This is because the mass of the rickshaw is more than that of the cycle.

Hence, the momentum of the rickshaw is greater even though both of them have the same velocity. To stop within the same interval of time, rate of change of momentum of the rickshaw or force applied on the rickshaw is greater.Sample Solutions from WBBSE Class 9 Physical Science Chapter 2

Question 36. Define linear momentum. How do you measure the momentum of a body? What are the units of linear momentum in CGS system and SI? Write down the dimensional formula of linear momentum.

Answer:

Linear momentum is defined as the property of motion of a body that is manifested due to the combination of velocity and mass of the moving body.

Momentum of a body is measured by the product of its mass and velocity. If v is the velocity of a body of mass m, then its momentum, p = mv.

Units of linear momentum in CGS system and SI are g • cm • s^-1 and kg • m • s^-1, respectively.

“force and laws of motion exercise “

Dimensional formula of linear momentum is MLT^-1.

Question 37. What do you mean by variable mass system? Give examples.

Answer:

Variable mass system is such a system whose mass keeps on increasing or decreasing during its state of motion, but its mass is neither created nor destroyed.

Examples:

1. A rain drop falls towards the earth’s surface when it is attracted by gravity. When it passes through moist air, its mass keeps on increasing due to the blending of water particles with this drop.
2. When a rocket is launched, combustion of rocket’s fuel takes place and the gas produced is emitted with very high velocity. Due to this burning of fuel, the mass of the rocket decreases gradually.

Question 38. Using the concept of linear momentum, write down Newton’s second law of motion.

Answer:

The rate of change of linear momentum of a body is equal to the force applied on it. The direction of change of momentum of a body takes place in the direction of the applied force.

Question 39. Several phenomena are described here. In which, case, the formula \(F=\frac{m_1-m_2}{t} \times u\) applied, and in which case, the formula x if is applied?

1. A body is falling freely from a height.
2. A piece of iron kept in front of a magnet ; is attracted to it.
3. Brakes are applied to maintain the uniform velocity of a cycle when a heavy bag falls from the carrier of a running cycle.
4. A moving wagon is filled up with coal and its uniform velocity maintained with the help of an engine.
5. A bucket full of water, with a hole in its bottom, is tied to a rope and is towered from the second floor with uniform velocity.

Answer:

1. There is no change in the mass of a freely falling body. But due to gravitational attraction, its velocity gradually increases. In this case, the formula F = ma is applicable. Here, F = gravity and a = acceleration due to gravity.

2. A piece of iron moves towards the magnet with uniform acceleration. The formula F = ma is applicable as there is no change of mass.

3. The mass of a cycle is reduced when a heavy bag falls off from the carrier of a moving cycle. Brakes have to be applied to maintain its velocity.

In this case, the formula \(F=\frac{m_1-m_2}{t} \times u\) is applicable.

Here, m₁ = mass of the cycle with the bag,

m₂ = mass of the cycle without the bag,

u = uniform velocity of the cycle,

F = force applied on the wheel of the cycle due to application of brakes,

t = duration of application of brakes at the moment the bag falls off.

4. The mass of a moving wagon gradually increases when it is filled up with coal. As a result, its velocity tends to decrease. Its uniform velocity is maintained by application of force by the engine.

Here, the formula \(F=\frac{m_2-m_1}{t} \times u\) is applicable.

Here, m₂ = mass of the wagon with additional coal, and m₁ = mass of the wagon before coal was dumped in it.

5. Water leaks continuously through the hole at the bottom of the bucket. As a result, mass of the water filled bucket is reduced. To maintain a uniform velocity, magnitude of tension force on the rope in the upward direction is to be increased.

Here, the formula \(F=\frac{m_1-m_2}{t} \times u\) is applicable.

Question 40. Establish the equation F = ma from the concept of linear momentum. Here, m = mass of the body, F = applied force and a = acceleration of the body.

Answer:

Suppose, a body of mass m is moving with a velocity u. Due to the application of a constant force F in the direction of motion of the body for a time t, its velocity becomes v.

Initial linear momentum of the body = mu and after time f, its linear momentum = mv.

∴ Change of linear momentum during time t = mv – mu = m(v – u)

∴  Rate of change of linear momentum

= \(\frac{m(v-u)}{t}=m a\)

Where, a = \(\frac{m(v-u)}{t}\) = acceleration of the body.

According to Newton’s second law of motion, F ∝ ma

or, F = Kma ……(1)

(where K is a constant)

If we assume that unit force produces unit acceleration on a body of unit mass, then m = 1, F= 1, results in F= 1.

From Equation (1), we get

1 = K • 1 • 1 or, K = 1

∴ F = ma

This is Newton’s second law of motion in case of a stationary mass.

Question 41. Explain Newton’s first law of with the help of second law of motion.

Answer:

If an acceleration a is produced by the application of a force F on a body of a mass m, then according to Newton’s second law of motion, F = ma. If no external force is applied on the body, then F = 0 and thus, o = 0 (because m cannot be zero).

This means the body does not have any acceleration. A body is without acceleration only if it is stationary or is moving with uniform velocity.

Hence, if no external force is applied on a body, then a stationary body always remains stationary and a moving body always keeps moving with uniform velocity. This is Newton’s first law of motion.

Question 42. Write down the law of parallelogram of forces and explain it.

Answer:

A particle is acted on by two forces represented in magnitude and direction by the two sides of a parallelogram drawn from a point. The resultant is equivalent to a force represented in magnitude and direction by the diagonal of the parallelogram passing through that point. This is the law of. parallelogram of forces.

Suppose, two forces P and Q are working simultaneously from point A. Forces P and Q have been represented in magnitude and direction by the two adjacent sides AB and AD of the parallelogram ABCD, respectively.

Then, diagonal of the parallelogram through the point A or the line AC will represent the resultant of the two forces.

Question 43. A car and a lorry are moving with the same momentum. Equal and retarding forces are applied to bring them to rest. Determine which one will take less time to stop?

Answer:

Mass of lorry is greater than that of car.

Suppose, mass of car = m₁,

Mass of lorry = m₂,

Initial velocity of car = v₁

And initial velocity of lorry = v₂.

As per the question, both had the same initial momentum,

∴ m₁v₁ = m₂v₂

Suppose, same opposing force F is applied in both the cases to stop them. This means final velocity is zero for both of them.

∴ \(F=m_1 a_1=m_1 \frac{v_1}{t_1}\)

[a₁ = deceleration of car and t₁ = duration of application of force on the car]

and \(F=m_2 a_2=m_2 \frac{v_2}{t_2}\)

[a₂ = deceleration of lorry and t₂ = duration of application of force on the lorry]

∴ \(m_1 \frac{v_1}{t_1}=m_2 \frac{v_2}{t_2}\)

or, \(m_1 v_1 t_2=m_2 v_2 t_1\)

or, \(\frac{t_2}{t_1}=\frac{m_2 v_2}{m_1 v_1}=1\)

or, \(t_2=t_1\)

∴ equal time is required to stop both of them.

Question 44. How does a bird fly in the sky? Can a bird fly In an air-free atmosphere?

Answer:

When a bird intends to fly from ground to sky, it exerts force on air through its wings. As a reaction, air also exerts equal and opposite force on the bird. It flies along the resultant of these two reaction forces.

In an air-free atmosphere, this type of reaction force cannot be produced. So, a bird cannot fly in air-free atmosphere.

Question 45. Explain resolution of forces with an example.

Answer:

As two forces may be added, similarly a single force may be resolved into two forces. But the resolution should be such that these resolved forces when added, would result in the given force.

Suppose, a force of 5 N is to be divided along two angles 30° and 45° on two opposite sides. By taking length 1 cm as 1 N, a straight line OA of length 5 cm is drawn. Lines OD and Of are drawn in two opposite sides of OA with angles 30° and 45°, respectively.

From point A, straight lines parallel to Of and OD are drawn which intersect OD and Of at points B and C, respectively.

“laws of motion class 9th exercise “

As per the diagram, OBAC is a parallelogram. Lengths of OB and OC are measured by a scale.

It is seen that OB = 3.7 cm (approx.) and OC = 2.6 cm (approx).

This means that magnitude of resolved part of force along OD is 3.7 N and magnitude of resolved part of force along Of is 2.6 N.

Understanding Inertia and Force for Solutions

Question 46. Resolve a force R along two opposite angles a and fi.

Answer:

Force R is represented by OC. R is to be resolved along two angles α and β on the opposite sides. Two lines OA and OB are drawn in two opposite sides along angles α and β.

Two straight lines are drawn from point C parallel to OA and OB which intersect OB and OA at points E and D, respectively. ODCE is a parallelogram. If OD = P and OE=Q, then P and Q are two components of R along two opposite angles α and β.

Question 47. What is resolution of a force into components at right angles to each other?

Answer:

The resolution of a force into components at right angles to each other means that the force is resolved into two components in such a way that the resultant of the two components is equal to the given force.

Question 48. Resolve a force R into components Which are at right angles to each other along an angle 0 and its perpendicular.

Answer:

The force R is represented by OC. R is to be divided along an angle θ or along OX and its perpendicular, that is along / into components at right angles to each other.

From point C, two perpendiculars CD and CE are drawn on OX and OY, respectively. So, ODCE is a parallelogram. Hence, OD = P and OE = Q are the two rightangled components of force R.

Question 49. Explain orthogonal resolution Of a force with examples.

Answer:

Suppose a force 4 N has to be resolved into components at an angle 30° and its perpendicular. By taking a length of 1 cm as 1 N, a straight line OA of length 4 cm is drawn. An angle of 30° is drawn at one side of OA along straight line OX and also along OY, its perpendicular direction is drawn.

Two perpendiculars AB and AC are drawn from point A on OX and OY, respectively. OBAC is a rectangle. OB and OC are two right angled components of a force 4 N along OX and OY. Measurement with a scale gives OB = 3.4 cm (approx) and OC = 2 cm

Hence, the orthogonal components of 4 N force along OX and OY are 3.4N (approx.) and 2N, respectively.

Question 50. Resolution of a force Is the opposite process of addition of forces is this statement wholly true?

Answer:

A definite force is obtained by the addition of two forces. Again, resolution of a force means its division in two different directions such that the resultant of these two resolved forces is equal to the given force. So we may get a lot of forces by resolution of a force. Hence, the given statement is not wholly true.

Question 51. A brick attached to a rope is dragged through the ground when one starts walking by holding the other end of the rope. Explain this phenomenon by orthogonal components of a force.

Answer:

The weight of a brick acts downwards in a perpendicular way. If tension force is applied to the rope in a slightly oblique way to the surface of the earth, the vertical component of the force acts in an upward direction.

“laws of motion class 9th exercise “

This vertical component reduces the natural weight of the weight of the brick body to some extent. In this case, the brick is dragged on the ground due to the horizontal component of the tension force.

Question 52. Why is it easier to pull a roller than to push it?

Answer:

Suppose, a roller on the ground is pulled along the line OA with a force F. The force F is resolved into its parallel and perpendicular components. Parallel component H acting along the line OB helps the roller to move forward.

The perpendicular component V along OC that works against the weight (W) of the roller acts in an opposite direction to reduce its weight.

Again, when a roller is pushed with a force F along OA, the force F is resolved into parallel and perpendicular directions at right angle.

The parallel component H working along OB helps the roller to move forward and the perpendicular component V working along OC towards the direction of the weight of the roller increases its effective weight.

For this reason, it is easier to pull a roller than to push it.

Chapter 2 Topic B Newtons First And Second Laws Of Motion Multiple Choice Questions

Question 1. Which law of motion of Newton defines force qualitatively?

1. First law of motion
2. Second law of motion
3. Third law of motion
4. Law of gravitation

Answer: 1. First law of motion

Question 2. Dimensional formula of momentum is

1. MLT^-1
2. MLT^-2
3. ML^-1T^-2
4. ML^-1T²

Answer: 1. MLT^-1

Question 3. Amount of force acting on a body of mass 1 kg moving with an acceleration of lm • s~2 is

1. 1 dyn
2. 1 N
3. 5 dyn
4. 10 N

Answer: 2. 1 N

Question 4. If a force of 50 dyn acts on a body of mass 10 g, its acceleration is

1. 3 cm • s^-2
2. 7cm • s^-2
3. 8 cm • s^-2
4. 5cm • s^-2

Answer: 4. 5cm • s^-2

Question 5. Compared to a light body, inertia of a heavy body is

1. More
2. Less
3. Same
4. Not possible to determine

Answer: 1. More

Question 6. The momentum of a body of mass 100 kg is 8000 kg • m • s^-1. Its velocity is

1. 80 cm • s^-1
2. 8 x 10⁵m • s^-1
3. 8 x 10⁵cm • s^-1
4. 80 cm • s^-1

Answer: 4. 80 cm • s^-1

Question 7. An example of inertia of rest is

1. Dusting of a blanket by a stick
2. Fixing of a nail by a hammer
3. Recoiling of a gun after firing of a bullet
4. Circling of a fan after switching off

Answer: 1. Dusting of a blanket by a stick

Question 8. When a force of 75 N acts on a body to produce an acceleration of 3m• s^-2, mass of the body is

1. 20 kg
2. 30 kg
3. 35 kg
4. 25 kg

Answer: 4. 25 kg

Question 9. An iron balL rolls and strikes another ball. This is an example of a force called

1. Frictional force
2. Colliding force
3. Normal force
4. Tension force

Answer: 2. Colliding force

Question 10. A book which is pushed on the table, goes some distance and stops. The book stops due to a force called

1. Frictional force
2. Colliding force
3. Normal force
4. Tension force

Answer: 1. Frictional force

Question 11. Which of the following is an example of normal force?

1. A stone is attached to one end of a rope and is hung from hand
2. A man is standing on the floor
3. An iron nail is moving towards a magnet
4. A man stumbles forward while getting down from a car

Answer: 2. A man is standing on the floor

Question 12. A body is moving with constant speed. Force is not required in which of the following cases?

1. To reduce speed of the body
2. To increase speed of the body
3. To change the direction of velocity of the body
4. To keep the direction of velocity of the body same

Answer: 4. To keep the direction of velocity of the body same

Question 13. A motor-driven belt is in motion with a uniform velocity of 8 m/s. If sand is poured on the belt at the rate of 2 kg/s, then how much force does the motor apply to keep the same motion?

1. 8N
2. 12N
3. 32N
4. 16N

Answer: 4. 16N

Question 14. The value of a force which has two rectangular components of 15N and 8N is

1. 17N
2. 23N
3. 20N
4. 46N

Answer: 1. 17N

Question 15. A force x has two rectangular components, one of them being y. The other one is

1. x – y
2. x² – y²
3. \(\sqrt{x^2-y^2}\)
4. \(\frac{x}{2}-y\)

Answer: 3. \(\sqrt{x^2-y^2}\)

Question 16. A particle moving with uniform acceleration has an initial velocity of 10 m/s and a final velocity of 20 m/s. Velocity of the particle when it finishes half the total journey time is

1. 15 m/s
2. 18 m/s
3. 12 m/s
4. 16 m/s

Answer: 1. 15 m/s

Question 17. A particle moving with uniform acceleration has an initial velocity of 30 cm/s and a final velocity of 40 cm/s. Velocity of the particle when it crosses half the total distance is

1. 35 m/s
2. 25√2 cm/s
3. 32 cm/s.
4. 28√2cm/s

Answer: 2. 25√2 cm/s

Important Concepts in Newton’s Laws of Motion for Class 9

Question 18. Starting from rest, a particle traversed a certain distance in 100 s with uniform acceleration. How much time does it take to traverse half the distance?

1. 50s
2. 70s
3. 70.7 s
4. 71s

Answer: 3. 70.7 s

Question 19. A bullet loses half of its velocity after entering 3 cm in a wooden block. The distance that the bullet goes further before coming to rest is

1. 3 cm
2. 2 cm
3. 1.5 cm
4. 1cm

Answer: 2. 2 cm

Question 20. A bullet moving with a velocity of 100 m/s can penetrate a 1 cm thick wooden plank. The velocity required by the bullet to penetrate a plank having a thickness of 16 cm is

1. 200 m/s
2. 400 m/s
3. 800 m/s
4. 150 m/s

Answer: 2. 400 m/s

Question 21. Dimensional formula of force is

1. LT^-2
2. MLT^-2
3. MLT^-1
4. ML

Answer: 2. MLT^-2

Question 22. In inertial frame of reference

1. Only Newton’s first law of motion is valid
2. Only Newton’s second law is valid
3. Only Newton’s.third law is valid
4. All the laws of motion are valid

Answer: 4. All the laws of motion are valid

Question 23. SI unit of force is

1. Newton
2. Fermi
3. Pascal
4. Dyne

Answer: 1. Newton

Question 24. Measure of inertia is

1. Mass
2. Velocity
3. Acceleration
4. Displacement

Answer: Mass

Question 25. Momentum of a body of mass 2 kg and velocity 20 cm/s is

1. 0.04 kg • m • s^-1
2. 0.4 kg • m • s^-1
3. 4 kg • m • s^-1
4. 40 kg • m • s^-1

Answer: 2. 0.4 kg • m • s^-1

Question 26. Unit of linear momentum is

1. kg • m • s^-2
2. N/s
3. N • s
4. kg2 • m • s^-1

Answer: N • s

Question 27. From Newton’s first law of motion we get

1. Qualitative definition of force
2. Definition of inertia
3. Measurement of weight
4. Idea about kinetic energy

Answer: 2. Definition of inertia

Question 28. By applying a force 1 N, one can hold a body of mass (g = 10 m • s^-2)

1. 1 kg
2. 0.1 kg
3. 10 kg
4. 0.01 kg

Answer: 2. 0.1 kg

Question 29. Which of the following has the largest inertia?

1. A pencil
2. A needle
3. A brick
4. Your body

Answer: 4. Your body

Question 30. The momentum of a body of given mass is proportional to its

1. Density
2. Velocity
3. Volume
4. Displacement

Answer: 2. Velocity

Question 31. A body of mass 1 kg moves along a straight line with a velocity 1 m • s^-1. Magnitude of the external applied force on the body is

1. 1 N
2. kgf
3. 0
4. 1 dyn

Answer: 3. 0

Question 32. Which of the following relation is not correct?

1. 1 N = 10⁵ dyn
2. 1 N = 1 kg • m/s²
3. 1 N = 981 dyn
4. 1 N = 10⁵ g. cm/s²

Answer: 3. 1 N = 981 dyn

Question 33. Mathematical representation of Newton’s second law is

1. \(F=\frac{m}{a}\)
2. F = ma
3. \(\frac{F}{a}=m^2\)
4. a = Fm

Answer: 2. F = ma

Question 34. Which of the following relation is correct?

1. \(\text { acceleration }=\frac{\text { displacement }}{\text { time }}\)
2. \(\frac{\text { mass } \times \text { displacement }}{\text { time }}=\text { momentum }\)
3. \(\text { velocity }=\frac{\text { distance }}{\text { time }}\)
4. \(\frac{\text { mass } \times \text { displacement }}{\text { time }}=\text { force }\)

Answer: 2. \(\frac{\text { mass } \times \text { displacement }}{\text { time }}=\text { momentum }\)

Question 35. When a bus suddenly starts moving from its rest position, the passengers standing on it lean backwards in the bus. This is because

1. Inertia of rest
2. Inertia of motion
3. Angular momentum
4. Conservation of mass

Answer: 1. Inertia of rest

Question 36. A car is moving with uniform velocity. Resultant force acts on the car F is such that

1. F > 0
2. F ≥ 0
3. F = 0
4. F ≤ 0

Answer: 3. F = 0

Question 37. Velocity of a body of mass 1 kg changes from 20 m/s to 30 m/s in 2 s. Amount of force applied on the body is

1. 25N
2. 4N
3. 2N
4. 5N

Answer: 3. 2N

Question 38. N • kg^-1 is unit of the physical quantity

1. Acceleration
2. Velocity
3. Time rate of change of velocity
4. Speed

Answer: 3. Time rate of change of velocity

Question 39. Resultant force applied on a body under the action of balanced forces is

1. Zero
2. Unit force
3. Undefined
4. None of these

Answer: 1. Zero

Question 40. Under the action of a constant force a body moves with

1. Constant velocity
2. Constant momentum
3. Constant acceleration
4. Constant kinetic energy

Answer: 3. Constant acceleration

Chapter 2 Topic B Newtons First And Second Laws Of Motion Answer In Brief

Question 1. What is the shape of the path of a particle when it is in uniform motion?

Answer: The shape of the path of a particle when it is in uniform motion is straight line.

Question 2. What is the change in total linear momentum of two bodies due to their collision?

Answer: In this case, if there is no applied force from the outside, the total linear momentum of the two bodies does not change.

Question 3. Two friends are displacing a table by application of unequal forces. Application of force in which direction does make this possible?

Answer: When two friends displace the table by application of unequal forces, then application of force in any direction makes this possible.

“laws of motion class 9th exercise “

Question 4. Under the influence of which force on a mass at rest, no acceleration is generated?

Answer: No acceleration is generated on a mass at rest under the influence of balanced force.

Question 5. Under the influence of which force on a mass at rest, acceleration is generated?

Answer: Acceleration is generated on a mass at rest under the influence of effective force.

Question 6. Strain of a body is produced under the influence of which force?

Answer: Strain of a body is produced under the influence of balanced force.

Question 7. Which force can change the state of rest or state of motion of a body?

Answer: External effective force can change the state of rest or the state of motion of a body.

Question 8. Write down Newton’s first law of motion.

Answer: Everybody continues to be in its state of rest or uniform motion in a straight line unless an external force is applied to it.

Question 9. What are the forces active on a body?

Answer: Two types of forces can be active on a body.

These are:

1. Internal forces and
2. External forces.

Question 10. Which law of motion of Newton gives a qualitative definition of force?

Answer: Newton’s first law of motion gives a qualitative definition of force.

Question 11. A bucket full of water is suddenly pushed. State whether the water splashes forward or backward.

Answer: If a bucket full of water is suddenly pushed, water splashes backward.

Question 12. Why do the blades of a rotating electric fan continue to rotate for some time even after switching it off?

Answer: Due to inertia of motion, the blades of a rotating electric fan rotates for a couple of times before it comes to rest.

Question 13. Which law of motion of Newton gives a quantitative definition of force?

Answer: Newton’s second law of motion gives a quantitative definition of force.

Question 14. What is the relationship between Newton and Dyne?

Answer: 1 N = 10⁵ dyn

Question 15. What is the dimensional formula of linear momentum?

Answer: Dimensional formula of linear momentum is MLT^-1.

Question 16. Which one is more convenient—pushing a roller or pulling a roller?

Answer: Pulling a roller is more convenient than pushing it.

Practice Questions for Chapter 2 Newton’s Laws

Question 17. What is attraction?

Answer: If two bodies come nearer to each other due to action-reaction, then that is called attraction.

Question 18. What do you understand by resolution of a force into two components?

Answer: A force is said to be resolved into two components in two given directions such that the resultant of these two components is equal to the applied force.

Question 19. What is the magnitude of a force if its two rectangular components are 3 N and 4 N?

Answer: Magnitude of the force = \(\sqrt{3^2+4^2}\) = 5 N

Question 20. A force of 13 N has two rectangular components 5 N and x. What is the magnitude of x?

Answer: The component,

\(x =\sqrt{13^2-5^2}\)

= \(\sqrt{169-25}=\sqrt{144}\)

= \(12 \mathrm{~N}\)

Question 21. Write the relationship between force applied on a body and change of momentum of the body due to this force.

Answer: The relationship is force = rate of change of momentum i.e. change of momentum

\(\begin{gathered}
\text { Force }=\frac{\text { change of momentum }}{\text { duration of time of application }} \\
\text { of the force }
\end{gathered}\)

Question 22. In which case Newton’s laws of motion are not applicable?

Answer: In non-inertial frame of reference Newton’s law of motion is not applicable.

Question 23. In which type of reference frame Newton’s laws of motion are applicable?

Answer: Newton’s laws of motions are applicable in inertial frame of reference.

Question 24. Find the mass of the body on which force of attraction of the earth is 1 N. [g = 10 m.s-2]

Answer:

Mass of the body = \(\frac{\text { force }}{\text { acceleration }}\)

= \(\frac{1 \mathrm{~N}}{10 \mathrm{~m} \cdot \mathrm{s}^{-2}}=\frac{1}{10} \mathrm{~kg}\)

= \(\frac{1000}{10} \mathrm{~g}=100 \mathrm{~g}\)

Question 25. Is motion possible without force?

Answer: Yes, uniform motion is possible without external force.

Question 26. What will be the direction of motion of a body if more than one forces are applied on it?

Answer: The body will move in the direction of the resultant of those applied forces.

Question 27. Force = mass x y. Write unit of x in CGS system.

Answer: Here y is acceleration. Therefore CGS unit of y is cm/s-2.

Question 28. lgf = ______ N.

Answer: 1 gf = 980 dyn = 980 x 10-5 N = 0.0098 N

Question 29. A particle is moving with uniform velocity and another particle is moving with uniform acceleration. On which particle force is acting?

Answer: Force is acting on that particle which is moving with uniform acceleration.

Question 30. Under the influence of balanced force can a body have acceleration?

Asnwer: No. A body under the influence of balanced force have no acceleration.

Question 31. Velocity of a body of variable mass is fixed. Is there any force acting on the body?

Answer: Here velocity is fixed. But mass of the body is variable. Therefore momentum of the body changes with time. Thus force is acting on the body.

Question 32. Two bodies X and Y have mass 20 kg and 90 kg respectively. Which have greater inertia?

Asnwer: Mass is the measurement of inertia. Here Y is heavier than X and hence it have greater inertia.

Chapter 2 Topic B Newtons First And Second Laws Of Motion Fill In The Blanks

Question 1. In spite of application of a ______ force on a moving body of fixed mass, it may move with uniform velocity.

Answer: Balanced

Question 2. In case of addition of ______ quantities, law of parallelogram is applied.

Answer: Vector

Question 3. The two resolved parts of any force are the two _____ of that force.

Answer: Components

Question 4. Any body at _____ cannot move on its own and any _____ body can not stop on its own.

Answer: Rest, moving

Question 5. Mass is the ________ property of a body.

Answer: Inherent

Question 6. Mass is the measure of __________ of a body.

Answer: Inertia

Question 7. 1 N = ________ dyn.

Answer: 10⁵

Question 8. The total linear momentum of a system of bodies is conserved if the total external force is ________

Answer: Zero

Question 9. Dimensional formula of force is ______

Answer: MLT^-2

Question 10. An electric fan rotates due to _______, even after the current is switched off.

Answer: Inertia of motion

Question 11. More the mass of a body is increased, more its ________ increases.

Answer: Property of inertia

Question 12. Force = mass x ______

Answer: Acceleration

Question 13. We know about the property of inertia from Newton’s ________ law of motion.

Answer: First

Question 14. If the amount of _______ applied to a body increases, its acceleration also increases.

Answer: Force

Question 15. Absolute unit of force in CGS system is _______

Answer: dyn

Question 16. Unit of momentum in SI is ________

Answer: kg.m.s^-1

Question 17. When a bus at rest suddenly starts moving, its passengers fall backward due to ________

Answer: Inertia of rest

Question 18. A body moves with uniform velocity when no ________ act on it.

Answer: Force

Question 19. Direction of momentum is directed along the ________ of the body.

Answer: Velocity

Question 20. Momentum of two bodies of masses m and 4m are equal ratio of their velocities is ________

Answer: 4: 1

Question 21. To accelerate a body an _________ must act on it.

Answer: Unbalanced force

Question 22. Acceleration is not produced under the action of ________ force.

Answer: Balanced

Chapter 2 Topic B Newtons First And Second Laws Of Motion State Whether True Or False

Question 1. If more than one force acts on a body such that the resultant of the forces is non-zero, then the forces are called balanced forces.

Answer: False

Question 2. Variable mass system is a system whose mass remains constant during its state of motion.

Answer: False

Question 3. The ratio of the force applied on a body and the acceleration of the body is called its inertial mass.

Answer: True

Question 4. Dimensional formula of linear momentum is MLT^-2.

Answer: False

Question 5. Newton’s second law of motion states that to every action, there is an equal and opposite reaction.

Answer: False

Question 6. The tendency of a stationary body to continue to exist in its state of rest is called inertia of rest.

Answer: True

Question 7. Amount of force can be calculated from Newton’s second law of motion.

Answer: True

Question 8. Inertia of a body increases with the increase of acceleration of the body.

Answer: False

Question 9. Amount of force required to stop a body depends on the momentum of the body.

Answer: True

Question 10. Force is necessary to maintain the uniform velocity of a moving object having variable mass.

Answer: True

Question 11. CGS unit of linear momentum is g • cm/s.

Answer: True

Question 12. Qualitative definition of force is known from Newtons first law of motion.

Answer: True

Question 13. 1 N = 10⁶ dyn .

Answer: False

Question 14. Two rectangular components of a force are 5 N and 12 N. The magnitude of the force is 13 N.

Answer: True

Question 15. According to Newton’s second law of motion the direction of change of linear momentum is along the direction of applied force.

Answer: True

Question 16. Definition of unit force is known from Newton’s first law.

Answer: False

Chapter 2 Topic B Newtons First And Second Laws Of Motion Numerical Examples

Linear momentum of a body of mass m and velocity v is p = mv.

If initial and final velocity of a body of mass m are u and v respectively and time interval be t, then

Initial momentum = mu, final momentum = mv. Change of linear momentum = mv-mu.

Rate of change of linear momentum \(p=\frac{m(v-u)}{t}=m \cdot a\)

If a be the acceleration of a body of mass m under the influence of force F then,

\(F=m a \quad \text { or, } a=\frac{F}{m}\)

If two rectangular components of a force F be F₁ and F₂ then,

∴ \(F^2=F_1^2+F_2^2 therefore F=\sqrt{F_1^2+F_2^2}\)

Question 1. A motor-driven belt is moving with a uniform velocity of 10 m/s. If sand from above falls on the belt at the rate of 2 kg/s, then what amount of force has to be applied by this motor to maintain the same motion?

Answer:

Here, velocity of the belt remains unchanged but the mass of the belt increases by 2 kg per second due to the sand falling at the rate of 2 kg/s on the belt.

Velocity of the belt,u = 10 m/s

∴ to maintain the same velocity, force required,

\(F=\frac{m_2-m_1}{t} \times u=\frac{2}{1} \times 10=20 \mathrm{~N}\)

Question 2. Starting from rest, a body of mass m attains a velocity V at a distance of x by application of a force F for time t. Show that, \(t=\frac{m v}{F} \text { and } x=\frac{m v^2}{2 F}\)

Asnwer:

Acceleration of the body, a = \(\frac{F}{m}\)

Velocity of the body after time t,

v = at or, v = \(\frac{Ft}{m}\)

∴ t = \(

Again, v² = lax or, v² = [latex]\frac{2F}{m}\)x

∴ \(x=\frac{m v^2}{2 F}\)

Question 3. A car of mass 100 kg is running with a velocity of 2 m/s. What amount of forte is to be applied to stop the car within a distance of 10 m?

Answer:

The car is running with a velocity of u = 2 m/s.

A force is applied on the car to stop it within s = 10 m .

Suppose, deceleration of the car = a.

Final velocity of the car, v = 0

So, the equation v² = u² – 2as gives 0 = u² – 2as or, 2as = u²

∴ \(a=\frac{u^2}{2 s}\)

Now, mass of the car, m = 100 kg

∴ applied force

\(F=m a=\frac{m u^2}{2 s}=\frac{100 \times 2^2}{2 \times 10}=20 \mathrm{~N}\)

Question 4. A body of mass 2 kg is moving with a velocity of 10 m/s. It Is stopped after a time of 10s. What amount of force is applied to the body?

Answer:

Initial velocity of the body, u = 10 m/s When the body is stopped after t = 10 s, its final velocity, v = 0

If deceleration of the body is a, then from the equation v = u – a t, we get

0 = 10 – a x 10 or, l0a = 10

∴ a = \(\frac{10}{10}\) = 1 m/s²

Now, mass of the body, m = 2 kg

∴ Force applied against the motion of the body,

F = ma = 2 kg x 1 m/s² = 2 N

Question 5. A tennis ball of mass 100 g comes running with a velocity of 10 m/s. It is sent back in the opposite’direction by a racket with a velocity of 15 m/s. If the duration of collision is 0.01s, what is the average force applied by the racket?

Answer:

If the direction of return of the ball is considered as positive, initial velocity of the ball, u = -10 m/s and final velocity, v = 15 m/s .

Duration of collision, t = 0.01s and mass of the ball, m = 100 g = 0.1 kg

∴ Average force applied on the tennis ball,

\(F=\frac{m(v-u)}{t}=\frac{0.1\{15-(-10)\}}{0.01}=250 \mathrm{~N}\)

Question 6. A force acts on a body of mass 100 g for 2 s and then stops acting. In the next 2 s, the body moves 100 cm. What amount of force is applied on the body?

Answer: After the force stops acting, the body moves a distance, s = 100 cm in t = 2 s.

If the velocity of the body is v, then from equation s = vt , we get

\(v=\frac{s}{t}=\frac{100 \mathrm{~cm}}{2 \mathrm{~s}}=50 \mathrm{~cm} / \mathrm{s}\)

The force acted on the body for t₁ = 2 s. If the acceleration of the body during that time is a, then

\(v=a t_1\)

∴ \(a=\frac{v}{t_1}=\frac{50 \mathrm{~cm} \cdot \mathrm{s}^{-1}}{2 \mathrm{~s}}=25 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

Now, mass of the body, m = 100 g applied force,

F = ma = 100 g x 25 cm/s² = 2500 dyn

Question 7. A force of 20 N acts on a mass of 10 kg for 4 s. What is the change in velocity of the body?

Answer:

Mass of the body, m = 10 kg; force applied, F = 20 N; force acted for a time t = 4s; initial velocity of the body = u and final velocity = v.

Change in velocity =(v-u) = V (say).

So, \(F=\frac{m(v-u)}{t}=\frac{m V}{t}\)

∴ V = \(\frac{F \cdot t}{m}=\frac{20 \times 4}{10}=8 \mathrm{~m} / \mathrm{s}\)

Question 8. A man of mass 50 kg is standing on a lift. Write down the reaction force on the man by the lift (or apparent weight of the man) in each of the following cases:

1. When the lift is at rest;
2. The lift is moving up with an acceleration of 30 cm/s² [acceleration due to gravity, g = 980 cm/s²]

Answer:

1. The weight of the man when the lift is at rest = 50 kg x g cm/s²

= 50 kg x \(\frac{980}{100}\) m/s2 = 50 x 9.8 N = 490 N

2. The effective acceleration of the man when the lift moves upwards with an acceleration of 30 cm/s2

= (g + 30) cm/s²

= (980 + 30) cm/s² = 1010 cm/s²

= 10.1 m/s²

∴ The apparent weight of the body

= 50 x 10.1 N = 505 N

Question 9. A bullet of mass 50 g moving with a velocity of 400 m/s penetrates 20 cm in a wall and comes to rest. What is the average resistance of the wall?

Answer:

In this case, initial velocity of the bullet, u = 400 m/s; final velocity of the bullet, v = 0;

mass of the bullet, m = 50 g = \(\frac{50}{1000}\)kg

Distance traversed by the bullet inside the wall,

s = 20 cm = \(\frac{20}{100}\) m

Suppose, deceleration of the bullet = a

∴ according to the formula, v² = u² – 2as or, 2as = u² – v²

\(v^2=u^2-2 a s\) \(2 a s=u^2-v^2\)

∴ \(a=\frac{u^2-v^2}{2 s}=\frac{(400)^2-0}{2 \times \frac{20}{100}}\)

= \(\frac{400^2 \times 100}{2 \times 20}=100000 \mathrm{~m} / \mathrm{s}^2\)

So, the average resistance of the wall,

∴ \(F=m a=\frac{50}{1000} \times 100000=5000 \mathrm{~N}\)

Question 10. Two blocks A and B have masses 2 kg and 3 kg, respectively. According to the figure, two blocks are kept on a smooth horizontal table such that they are in contact with each other. A force of 10 N is horizontally applied on block A. How much force will be applied on B by A?

 

Answer:

A force of amount 10 N acts on a combined system of (2 + 3) or 5 kg mass and produces an acceleration of a = \(\frac{F}{m}\) = \(\frac{10}{5}\) =2 m/s².

This means that acceleration of 2 m/s² is produced on each of the two blocks A and B.

Therefore, force applied on block 6 by block A,

F= mass of block B x acceleration of the block = 3 x 2 = 6N

Question 11. Two bodies of masses 3 kg and 5 kg are placed in rest. If same amount of force is applied to them then find the ratio of time required to acquire Same velocity.

Asnwer:

Here, initial velocities of the two bodies are zero.

Let, F be the same force applied on both of the bodies through time t₁ a t₂ respectively to acquire same velocity v.

According to Newton’s second law, for the 1st body,

\(F=\frac{3(v-0)}{t_1}=\frac{3 v}{t_1}\)

And for the second body,

\(F=\frac{5(v-0)}{t_2}=\frac{5 v}{t_2}\)

∴ The required ratio \(t_1: t_2=\frac{3 v}{F}: \frac{5 v}{F}=3: 5\).

Question 12. Two forces of magnitude 10 dyn each acts at an angle 120° with each other. Find resultant of this two forces using the law of parallelogram of vectors.

Answer:

ABCD is a parallelogram, whose AB and AD sides represent 10 dyn force both in direction and magnitude.

∠BAD = 120° and AB = 10 dyn, AD = 10 dyn .

The diagonal of the parallelogram, AC represents the resultant of the two forces.

Here, AB = BC= CD = AD

Thus ABCD is a rhombus and AC is the bisector of the angle ∠BAD.

∴ \(\angle B A C=\frac{120^{\circ}}{2}=60^{\circ}\)

AB = BC

∴ ∠BAC = ∠ACB = 60°

or ∠ABC = 180° – 60° – 60° = 60°

∴ ΔABC is a equilateral triangle.

So, AC – AB = 10 dyn

∴ The magnitude of the resultant of the two forces is 10 dyn and it inclined at an angle of 60° with any of the two forces.

Question 13. A hammer of mass 2 kg falling from a height 5 m, hit a nail partially fixed on a surface, and stops in 1/10 s. Find the average force exerted on the nail, [g = 10 m • s^-2]

Answer:

Mass of the hammer m = 2kg, distance covered by the hammer in 1/10 s is h = 5 m and its velocity,

v = \(\sqrt{2 g h}=\sqrt{2 \times 10 \times 5}=10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Total force exerted on the nail = Impulsive force exerted by the hammer + weight of the hammer

= \(\left(\frac{m v-m u}{t}\right)+m g=\frac{2 \times 10}{\frac{1}{10}}+2 \times 10\)

= 200 + 20 = 220 N

[If a large force acts on a body for a very short interval of time, it is called an impulsive force.

When a nail is hammered the force acts for a very short period of time. Hence it is impulsive force.

\(\text { Impulsive force } \left.=\frac{\text { change of momentum }}{\text { time }}\right]\)

Question 14. A force produces an acceleration of 2 m/s² in a block. Four such block are tied together and the same force is applied on the combination. What will be the acceleration of the combination?

Answer:

Let, mass of a single block = m kg and the force is = F N

∴ Acceleration produced in a single block, a = \(\frac{F}{m}\) = 2 m • s^-2

∴ The acceleration of the combination

= \(\frac{F}{4 m}=\frac{2}{4}\) = 0.5 m • s^-2

Chapter 2 Topic A Rest Motion And Equation Of Motion Synopsis

The state of rest or motion of a body is always observed with respect to some other bodies in the surrounding. This some other bodies provide the frame of reference or reference frame.

There are two types of reference frame:

1. Inertial frame and
2. Non-inertial frame.

If a body does not change its position with respect to a neighboring object with the change of time, then the body is said to be a static body and its state is known as state of rest.

If a body changes its position with respect to a neighbouring object with change of time, then the body is said to be a dynamic body and its state is known as state of motion.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

The motion of a particle is generally of two types:

1. Translational or linear motion and
2. Rotating motion.

Translational motion is the motion of a body along a straight line.

If a body rotates around a fixed axis or a fixed point, its motion is called rotational motion. If a body undergoing periodic motion moves repeatedly along the same straight line at equal intervals of time, then its motion is called rectilinear oscillatory motion.

If a particle rotates around an axis or a point in a circular path, its motion is said to be a circular motion.

Speed of a moving particle is the distance traversed by it in unit time. Speed is a scalar quantity. Unit of speed in SI is m/s. Dimensional formula of speed is LT^-1.

If a moving particle traverses equal distances in equal time intervals, its corresponding speed is said to be uniform speed.

If a moving particle traverses different distances in equal time intervals, its corresponding speed is said to be non-uniform.

For a particle moving with non-uniform speed, its average speed is obtained by dividing the total distance traversed by the total time required to traverse that distance.

If a moving particle traverses s₁ distance in the first t₁ time, s₂ distance in the next t₁ time, and s₃ distance in the last t₃ time, then. average speed,

\(v_a=\frac{s_1+s_2+s_3}{t_1+t_2+t_3}\)

If the magnitude and direction of the velocity of a particle always remain unchanged with time, then the velocity of that particle is called uniform velocity.

If the magnitude and direction or both of a particle change with respect to time, then the velocity of that particle is called non-uniform velocity.

class 9 chapter 2 physics

If a particle moves along a circular path with uniform speed, then the motion of the particle is called uniform circular motion.

It is the motion with uniform speed. In this case, acceleration of the particle (centripetal acceleration) acts towards the centre along the radius of the circular path.

Rate of change of velocity of a particle with respect to time is called its acceleration. Since acceleration has both magnitude and direction, hence it is a vector quantity. Unit of acceleration in SI is m/s² and the dimensional formula is LT^-2.

The initial velocity of a particle moving in a straight line is u. If its velocity is v after time t, then its acceleration, a = \(\frac{v-u}{t}\).

Deceleration is negative acceleration.

The directions of velocity and acceleration are the same if the velocity of a particle moving in a straight line increases but if the velocity of the particle decreases, velocity and acceleration are in the opposite directions.

Equations Of Dynamics: If initial velocity of a particle moving in a straight line = u, uniform acceleration = a, velocity after time t = v, and distance traversed in time t = s, then,

1. \(v=u-a t\)
2. \(s=u t-\frac{1}{2} a t^2\)
3. \(v^2=u^2-2 a s\)

Case 1: If the initial velocity of the particle, u = 0, that is, the particle started its journey from rest,

1. v = at
2. s = 1/2 at²
3. v² = 2as

Case 2: If the particle is moving with uniform deceleration of a, starting with initial velocity of u,

1. v = u – at
2. s = ut- \(/frac{1}{2}\)at²
3. v² = u² – 2as

Case 3: If the particle falls under gravity with an initial velocity of u, it comes down with an uniform acceleration of a = g. Here, if s = h,

1. \(v=u+g t\)
2. \(h=u t+\frac{1}{2} g t^2\)
3. \(v^2=u^2+2 g h\)

Case 4: If the particle falls from rest under gravity,

1. \(v=g t\)
2. \(h=\frac{1}{2} g t^2\)
3. \(v^2=2 g h\)

Case 5: If the particle is thrown upwards in a perpendicular direction with u velocity,

1. \(v=u-g t\)
2. \(h=u t-\frac{1}{2} g t^2\)
3. \(v^2=u^2-2 g h\)

NEET Biology Class 9 Question And Answers	WBBSE Class 9 History Notes	WBBSE Solutions for Class 9 Life Science and Environment
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WBBSE Solutions for Class 9 Geography And Environment	WBBSE Class 9 History Long Answer Questions	WBBSE Class 9 Life Science Multiple Choice Questions
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WBBSE Solutions for Class 9 History	WBBSE Class 9 History Very Short Answer Questions

Chapter 2 Topic A Rest Motion And Equation Of Motion Short And Long Answer Type Questions

Question 1. What do you understand by rest and motion?

Answer:

Rest: If a body does not change its position with change of time and with respect to a object, then that body is said to be a stationary body. In this condition, it is said to be at rest.

Motion: If a body changes its position with change of time and with respect to a neighbouring object, then that body is said to be a moving body. In this condition, it is said to be in motion.

Question 2. All rest and motion are relative- explain.

Answer:

All objects like houses, trees, etc. surrounding us do not change their positions with respect to the earth’s surface and also with change of time. So these are called stationary objects with respect to the surface of the earth.

But the earth itself is moving around the sun. So, houses, trees, etc. which are stationary with respect to the earth are not at absolute rest. In reality, every object of this universe is in motion with respect to every other object.

So, there is no state of absolute rest in nature. Since there is no existence of absolutely stationary bodies, so there is also no absolutely moving body. Thus, all rest and motion are relative.

Key Questions on Rest and Motion for Class 9

Question 3. How many types of motion a body may possess? What are those types?

Answer:

A body may possess two types of motion.

These Two Types Of Motion Are

1. Rotational motion and
2. Translational motion or linear motion.

Question 4. What do you understand by rotational motion? Explain with examples.

Answer:

If a body rotates with respect to one of its axes, then the motion of the body is called rotational motion.

Example: The motion of blades of a running electric fan is rotational motion. Again, the diurnal motion of the earth around its axis throught day and night like a top is also a rotational motion.

“numericals of physics class 9 chapter 2 “

Question 5. What is translational motion? Explain with examples.

Answer:

If a body moves in a straight line, its motion is called translational motion.

Example: If a stone is dropped from a roof top, it falls straight downward. In this case, motion of the stone is a translational motion.

Practice Questions for Chapter 2 Rest and Motion

Question 6. What is compound motion? Give example.

Answer:

If a body is moving in such a way that its motion is neither a pure translational motion nor a rotational motion but a mixture of both, then this motion is called a compound motion.

Example: Let us consider the motion of wheel of a running car. Except the particle at the centre point of the wheel, all other particles are undergoing compound motion.

Question 7. Write down the differences between translational motion and rotational motion.

Answer:

Differences between translational motion and rotational motion:

Question 8. What do you mean by rotational plane and axis of rotation?

Answer:

Rotational Plane: When a particle rotates around a fixed axis or point, then during the state of rotation the particle makes a circle. The horizontal plane on which the circle lies is called rotational plane.

Axis Of Rotation: When a particle rotates in a circular path, the straight line perpendicular to the rotational plane and passing through the centre of the circle is called the axis of rotation.

Question 9. What is the difference between a rotational motion and a circular motion? Explain with examples.

Answer:

A rigid body consists of more than one particle. When such a rigid body rotates around a fixed axis of its own, then this motion is called rotational motion.

In this case, the body does not change its place with respect to its axis but simply changes its orientation. The axis about which the body
rotates is called the rotational axis.

The motion of the blades of a fan is rotational motion. In this case, the blades rotate with respect to the axis situated at the centre of the fan.

In the same way, the motion of a top remaining stationary at one place on the floor, motion of a disc on CD drive, dlurnal motion of the earth (due to which the phenomenon of day and night occurs) etc. are examples of rotational motion.

“force and law of motion class 9 question answer ”

On the other hand, the motion of a particle along a circular path is the circular motion of the particle. Since the particle has the shape of a point, hence it is not possible for the particle to have an axis. So the particle has no rotational motion.

The revolving motion of an electron around the nucleus is circular motion. Again, if the earth is considered to be a point in comparison to the sun, its motion around the sun (due to which seasonal changes occur) is the earth’s circular motion.

Question 10. Diurnal and annual motion of the earth fall under which category?

Answer:

If we assume that the earth rotates in a circular path around the sun (actually it is an elliptical path), annual motion of the earth is a circular motion.

While moving around the sun, the earth rotates with respect to its own axis. This is called diurnal motion which falls under rotational motion.

Question 11. What do you understand by periodic motion? Give examples.

Answer:

If a body traverses the same path repeatedly after equal intervals of time, then the motion of the body is known as periodic motion.

Example: Diurnal motion of the earth around the sun, motion of the hand of a clock, motion of a simple pendulum etc. are examples of periodic motion.

Question 12. What is displacement? What type of quantity is it?

Answer: The change of position of a moving body in a particular direction is the displacement of the body. Displacement is a vector quantity because it has both magnitude and direction.

Question 13. Can the displacement of a particle be greater than the traversed distance?

Answer:

The displacement of a moving particle can never be greater than the traversed distance. If the particle travels from one point to another in a straight line path, the value of displacement becomes equal to that of the traversed distance but if the particle moves in a curved path, value of displacement is less than that by distance traversed.

Question 14. What are the differences between displacement and distance traversed?

Answer:

Differences between displacement and distance traversed:

Question 15. What is speed? What type of quantity is speed? What are the units of speed in CGS system and SI?

Answer:

1. The amount of distance a moving particle traverses in unit time is called its speed.
2. Speed has magnitude but no direction, hence it is a scalar quantity.
3. Units of speed in CGS system and SI are cm/s and m/s respectively.

Question 16. Define uniform speed and non-uniform speed.

Answer:

Uniform Speed: A moving particle has a uniform speed if it travels equal distances in equal intervals of time.

Non-Uniform Speed: A moving particle has a non-uniform speed if it travels unequal distances in equal intervals of time.

Question 17. What do you mean by average speed?

Answer:

Average speed is the ratio of the distance traversed and the time taken to cover that distance by a moving particle traveling with non-uniform speed.

Suppose, while going to your friend’s house, you cover the first s₁ distance by rickshaw in time t₁, the next s₂ distance by auto-rickshaw in time t_2, and the last s₃ distance by walking in time t₃. In this case, you cover the whole journey with non-uniform speed.

Here, the average speed is defined as

\(v_a=\frac{s_1+s_2+s_3}{t_1+t_2+t_3}\)

Question 18. How do you define velocity? What type of quantity is velocity?

Answer:

1. Rate of change of displacement of a body with respect to time in a particular direction is called the velocity of a body.
2. Velocity is a vector quantity because it has both moving in a circular path, the velocity of the magnitude and direction.

Important Concepts in Equations of Motion for Class 9

Question 19. Define uniform velocity and non-uniform velocity.

Answer:

Uniform Velocity: A particle is said to have uniform velocity if it covers equal distance in equal intervals of time in a given direction.

Non-Uniform Velocity: A particle is said to have non-uniform velocity if the magnitude or direction or both of the particle change with respect to time.

“force and laws of motion exercise ”

Question 20. Write down the differences between speed and velocity.

Answer:

Differences between speed and velocity:

Question 21. Explain how a body with uniform speed may not have uniform velocity.

Answer:

A body is said to have uniform speed if it traverses equal distances in equal intervals of time in a curved path. As the body is moving in a curved path, the direction of its velocity changes every moment so it is not moving with uniform velocity.

Hence, a body moving with uniform speed need not have uniform velocity, if it traverses equal distances in equal intervals of time by changing its direction.

Question 22. What do you mean by uniform circular motion?

Answer:

Suppose, a particle is moving in a circular path with uniform speed as shown in the figure. While particle at any moment at a point on the circumference always acts in a tangent to that point.

As a result, the velocity of the particle changes every moment. This type of motion of a particle is called uniform circular motion.

Question 23. What do you mean by acceleration?

Answer:

The rate of change of velocity of a moving particle with respect to time is called its acceleration.

Suppose the initial velocity of a particle moving in a straight line = u and after time t, its velocity becomes v.

∴ according to the definition of acceleration of the particle, \(a=\frac{v-u}{t}\)

Question 24. Determine the dimensional formula of acceleration. What is the dimension of acceleration? What are the units of acceleration in CGS system and SI?

Answer:

Dimensional formula of acceleration

= \(\frac{\text { dimensional formula of velocity }}{\text { dimensional formula of time }}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^{-1}}=\mathrm{LT}^{-2}\)

Dimension of acceleration is 1 in length and -2 in time.

Units of acceleration in CGS system and SI are cm/s² and m/s², respectively.

Question 25. Displacements of a particle in time t, s = 4t + 6t². If displacement is expressed in unit m and time in unit s, what is the acceleration of the particle?

Answer:

If we compare the two equations, s = 4t + 6t² and s = ut + \(\frac{1}{2}\)at²,

we find, \(\frac{a}{2}\) = 6 or, a = 12 m/s²

∴ acceleration of the particle = 12 m/s².

Question 26. What are uniform acceleration and non-uniform acceleration?

Answer:

Uniform Acceleration: A moving particle is said to have uniform acceleration if it travels in a straight line and its velocity increases by equal amount in equal intervals of time.

Non-Uniform Acceleration: A moving particle is said to have non-uniform acceleration if its velocity increases by unequal amounts in equal intervals of time.

Question 27. A moving particle starting from rest traverses a distance x in 1 s with uniform acceleration. How much distance does it cover in 2 s time?

Answer:

Distance traversed by a particle moving with uniform acceleration starting from rest in time t, s ∝ t².

∴ The particle covers a distance of 4x in 2 s time.

Question 28. If the initial velocity of a moving particle with uniform acceleration is zero, what is the nature of its velocity-time curve?

Answer:

The graph of a moving particle with uniform acceleration is always a straight line. Again, since the initial velocity of the particle is zero, hence the graph will always pass through the origin.

Question 29. What is instantaneous acceleration?

Answer: Instantaneous acceleration is the limiting value of the ration of the change in velocity of a moving particle during a very small change of time at any particular instant.

Understanding Types of Motion for Solutions

Question 30. Is it possible for a moving particle to have acceleration but constant magnitude of its velocity?

Answer:

If a particle moves in such a way that its acceleration always acts perpendicularly to the direction of its motion, the magnitude of velocity remains unchanged even though the moving particle has acceleration.

For example, the acceleration of a particle moving in a circle with uniform speed is acting towards the centre along the radius of the circle and at any moment, velo- city is acting at that point along the tangent to the circle.

In this case, angle between acceleration and velocity is always 90°. So for this type of motion, magnitude of velocity always remains unchanged even though acceleration is present.

Question 31. Deceleration is negative acceleration- explain.

Answer:

If the velocity of a moving particle reduces with the change of time, then the rate of change of velocity or acceleration is negative. This negative acceleration is called deceleration or retardation.

In this case, deceleration and velocity act in opposite directions.

Question 32. Why is the phrase ‘per second’ used twice in the unit of acceleration?

Answer:

The word second is used twice in the unit of acceleration because it has two significances. It signifies change of velocity as well as rate of change of velocity.

Question 33. Is it possible for a moving particle to have velocity but no acceleration?

Answer:

It is possible for a particle moving with uniform velocity to have velocity but no acceleration. If a particle remains stationary or moves with uniform velocity, it does not have any acceleration.

“laws of motion class 9th exercise “

Question 34. Is it possible for a body to have acceleration but no velocity?

Answer: It is possible that a particle has acceleration but no velocity. A body is thrown upwards in a perpendicular direction. It moves up and at the highest point, it comes to rest for a moment.

At that time, velocity is zero but acceleration due to gravity still works on the body.

Question 35. Establish the equation s = ut+1/2at² by algebraic method.

Answer: Let the initial velocity of a particle moving in a straight line with uniform acceleration a and its velocity after time t be u and v, respectively.

∴ v = u + at

Since the particle is moving with uniform 0.38 Show that the average velocity of a acceleration, rate of increase of velocity is the same.

∴ Average velocity of the particle at time t,

\(v_a=\frac{u+v}{2}\)

As the particle is moving with uniform acceleration, distance traversed by the particle at time t is equal to the distance traversed in time t with average velocity V_a.

If s is the distance traversed by the particle in time t, then

s = \(v_a \cdot t=\frac{u+v}{2} \cdot t=\frac{u+u+a t}{2} \cdot t\)

= \(\frac{(2 u+a t) t}{2}=\frac{2 u t+a t^2}{2}\)

∴ \(s=u t+\frac{1}{2} a t^2\)

Question 36. Establish the equation v² = u² + 2as by algebraic method.

Answer:

Let the initial velocity of a particle moving in a starigh line with uniform acceleration a and velocity after time t be u and v respectively.

∴ v = u + at ……(1)

If s is the distance traversed by the particle in time t, then

s = ut + \(\frac{1}{2}\)at² …..(2)

Squaring both sides of euqation (1), we get

\(v^2=(u+a t)^2=u^2+2 u a t+a^2 t^2\)

= \(u^2+2 a\left(u t+\frac{1}{2} a t^2\right)\)

Now from equation (2), we get

v² = u² + 2as

Question 37. The equation \(s=u t+\frac{1}{2} a t^2+s_0\) signifies what type of motion?

Answer:

If we put t=0 in the equation \(s=u t+\frac{1}{2} a t^2+s_0\), we get s = S₀

∴ The particle starts its movement from position s = s₀ with initial velocity u and uniform acceleration a.

Question 38. Show that the average velocity of a particle in case of motion with uniform acceleration is half of the sum of initial and final velocities.

Answer: Let the initial velocity of a particle moving with uniform acceleration be u and its velocity after time t be v.

Distance traversed by the particle in time t,

\(s=u t+\frac{1}{2} a t^2\)

∴ Average Velocity,

\(v_a =\frac{s}{t}=\frac{u t+\frac{1}{2} a t^2}{t}=u+\frac{1}{2} a t\)

= \(\frac{2 u+a t}{2}=\frac{u+u+a t}{2}=\frac{u+v}{2}\)

Question 39. The initial velocity of a particle moving with uniform acceleration is u and after travelling a distance s, its velocity becomes v. What is its velocity after it covers half of the total path?

Answer:

If the acceleration of the particle is a, then v² = u² + 2as.

Let the velocity of the particle is v₁ after it traverses half of the total distance or \(\frac{s}{2}\).

∴ \(v_1^2=u^2+2 a \cdot \frac{s}{2}=\frac{2 u^2+2 a s}{2}\)

= \(\frac{u^2+u^2+2 a s}{2}=\frac{u^2+v^2}{2}\) \([because v^2=u^2+2 a s]\)

∴ velocity of the particle after it traverses distance \(\frac{s}{2}\),

\(v_1=\sqrt{\frac{u^2+v^2}{2}}\)

Question 40. Initial velocity of a moving particle with uniform acceleration is u, after time tits velocity is v. What is the velocity of the particle after half the time \(\frac{t}{2}\)?

Answer:

If the velocity of the particle is a, then v = u + at.

Velocity after time \(\frac{t}{2}\),

\(v_1=u+\frac{a t}{2}=\frac{2 u+a t}{2}=\frac{u+u+a t}{2}=\frac{u+v}{2}\) \([because v=u+a t]\)

Sample Solutions from WBBSE Class 9 Physical Science Chapter 2

Question 41. 41 Distances traversed by a particle moving with uniform acceleration and starting from rest in the first, second, and third seconds are s₁, s₂, and s₃, respectively. Determine s₁ : s₂ : s₃.

Answer: 

If the acceleration of the particle is a, then \(s_1=\frac{1}{2} a \times 1^2=\frac{a}{2}\)

If the particle traverses a distance s’₂ in 2s, then \(s_2^{\prime}=\frac{1}{2} a \times 2^2=2 a\)

∴ \(s_2=s_2^{\prime}-s_1=2 a-\frac{a}{2}=\frac{3}{2} a\)

Again, if the particle traverses a distance s’₃ in 3s, then

\(s_3^{\prime}=\frac{1}{2} \times a \times 3^2=\frac{9}{2} a\)

∴ \(s_3=s_3^{\prime}-s_2^{\prime}=\frac{9}{2} a-2 a=\frac{5}{2} a\)

So, \(s_1: s_2: s_3=\frac{a}{2}: \frac{3}{2} a: \frac{5}{2} a=1: 3: 5\)

Question 42. Establish the equation v=u+at with the help of a velocity-time graph.

Answer:

Let the initial velocity of a particle moving with uniform acceleration along a straight line be u and after time t, let its velocity be v.

OX and OY are two mutually perpendicular axes.

A velocity-time line graph of the particle is drawn by taking time (t) along x-axis and velocity (v) along y-axis. Straight line AB represents the velocity-time line graph of the particle.

Here, initial velocity, OA = u and at any definite time, OC=t, velocity of the particle, BC = v. A straight line AD is drawn parallel to OC.

From the graph, we may write BC = BD + DC = BD + OA [because OA=DC]

∴ v = BD + u or, BD = v – u …(1)

As per the graph, acceleration of the particle,

\(a=\frac{\text { change of velocity }}{\text { time }}=\frac{B C-A C}{O C}\)

= \(\frac{B C-D C}{O C}=\frac{B D}{O C}=\frac{B D}{t}\) [because OC = t]

∴ BD = at…(2)

By comparing equations (1) and (2), we get at=v-u or, v=u+at

Question 43. Establish the equation s = ut + \(\frac{1}{2}\) at² with the help of a velocity-time graph.

Answer:

Let a particle with initial velocity u is moving along a straight line with uniform acceleration a. Now suppose v is the velocity of the particle after time t and distance traversed by the particle during that time is s.

OX and OY are two mutually perpendicular axes. If time is expressed along x- axis and velocity is expressed along y-axis, straight line AE expresses velocity-time graph of the particle.

Here, OA = u is the initial velocity of the particle. Straight line AF parallel to the time axis and passing through the point A expresses velocity-time graph of the particle in a state without acceleration.

At any specific time, OB = t, velocity of the particle, BC = v = u + at.

Distance traversed by the particle in time t,

s = area of trapezium OACB

= \(\frac{1}{2}\)(OA + BC) X OB

∴ s = \(\frac{1}{2}\)(u+u+at) xt

= \(\frac{1}{2}\)(2ut + at²)

= \(\frac{1}{2}\)ut + at²

Question 44. Establish the relationship s=vt by using velocity-time graph.

Answer: Suppose, a particle is moving with uniform velocity v.

OX and OY are two mutually perpendicular axes. If time is represented by x- axis and velocity by y-axis, then length OA represents uniform velocity v of the particle.

As the particle is moving with uniform velocity, there would not be any change in its velocity with change of time.

So, velocity-time graph is a Istraight line AB parallel to the time axis. Suppose, length OC represents a definite time t. A perpendicular CD is drawn from C to AB.

Distance traversed by the particle in time t, s = vt = OA x OC = area of the rectangle OADC.

Hence, the area formed by the velocity-time graph with the time axis in a definite time gives the distance traversed by the particle.

Question 45. Establish the equation s = \(\frac{1}{2}\) at² with the help of a velocity-time graph.

Answer:

Let a particle starting from rest, travels along a straight line with uniform acceleration and after time t, its velocity is v.

OX and OY are two mutually perpendicular axes. If time is expressed along x-axis and velocity is expressed along y-axis, straight line OA which passes through the point of origin will express velocity- time graph of the particle.

According to the line graph, velocity of the particle, BC = v at any definite time OB = t.

Distance traversed by the particle at time t,

s = area of ΔOBC

= \(\frac{1}{2}\) x OB x BC = \(\frac{1}{2}\) x t x v = \(\frac{1}{2}\) at²

Question 46. Establish the equation v²=u²+2as with the help of a velocity-time graph.

Answer:

Let a particle with initial velocity u is moving along a straight line with uniform acceleration a.

Now suppose v is the velocity of the particle after time t and distance traversed by the particle during that time is s.

OX and OY are two mutually perpendicular axes. If time is expressed along x-axis and velocity is exposed along y-axis, straight line AE expresses velocity-time graph of the particle.

Here, OA = u is the initial velocity of the particle. Straight line AF parallel to the time axis and passing through the point A expresses velocity- time graph of the particle in a state without acceleration.

At any specific time, OB =t, velocity of the particle, BC = v = u + at.

Distance traversed by the particle in time t,

s = area of trapezium OACB

= \(\frac{1}{2}\)(OA + BC) X OB

= \(\frac{1}{2}\)(OA + BC) X OB X \(\frac{CD}{CD}\)

= \(\frac{1}{2}\)(OA + BC) x OB x \(\frac{BC-BD}{CD}\)

= \(\frac{1}{2}\)(OA + BC) x (BC -OA) x \(\frac{OB}{CD}\)

= \(\frac{1}{2}\)(u + v) x (v – u) x \(\frac{t}{at}\)

[because CD = BC = v – u = at]

or, \(s=\frac{v^2-u^2}{2 a}\)

or, \(v^2-u^2=2 a s\)

∴ \(v^2=u^2+2 a s\)

Question 47. Draw the displacement-time graph of a particle moving with uniform velocity. What is the nature of this graph?

Answer:

Suppose, a particle is moving with uniform velocity v.

∴ Distance traversed by the particle in time t, s = vt

If time (t) is expressed along x-axis and displacement is expressed along y-axis in the line graph, s-t graph (OA) of the particle is a straight line passing through the origin.

Question 48. Draw the displacement-time graph of a particle moving with uniform acceleration. What is the nature of this line graph?

Answer:

Suppose, a particle is moving with uniform acceleration a. If the initial velocity of the particle is u, distance traversed in time t,

s = ut + \(\frac{1}{2}\) at²……(1)

In the line graph, if time (t) is expressed along x-axis and displacement (s) is expressed along y-axis, s-t graph (OA) of the particle will be a parabola passing through the origin.

Question 49. Velocity-time graphs of four different particles are shown in the given figure. Out of these, no. 1 and no. 2 curves are parallel to each other. Answer the following questions according to the image:

1. Which particles have zero initial velocity?
2. Which particle is moving with uniform velocity?
3. Which particles have same acceleration?
4. Among the particles moving with uniform acceleration, which particle or particles have lowest acceleration?

Answer:

1. Curves no. 2 and no. 3 are passing through the origin. So the initial velocity of both the particles is zero.
2. Curve no. 4 curve is parallel to the time axis. So the corresponding particle is moving with uniform velocity.
3. Curves no. 1 and no. 2 are parallel to each other. Hence the corresponding particles have the same acceleration.
4. Out of the curves numbered 1, 2, 3, and 4, the curve no. 3 is inclined to the x-axis with lowest angle. Hence, the acceleration of the particle corresponds to curve no. 3 is the lowest.

Question 50. Velocity-time graphs of two bikes in a bike competition are shown in the image. At the end of this competition of two hours duration:

 

1. Which bike continues to have greater velocity?
2. Which bike traverses greater distance?
3. Which bike has the greater acceleration?

Answer:

It is seen that after two hours, both the graphs meet at the same point. That is, the velocities of both particles are the same and that is 80 km/h.

According to the graphs, the curve corresponding to the first bike is line OAB and for the second bike, it is line OB.
In this case, area of quadrilateral OABC > area of triangle OBC.

In the velocity-time graph, the area contained within the curve and the time axis at a definite time indicates the distance traversed by the concerned particle. Hence, the first bike covers a greater distance.

The first bike traveled the first one hour with uniform acceleration and the second bike was in motion with uniform acceleration throughout the journey. Since the curve OA is inclined to the x-axis with a greater angle than the curve OB, the acceleration of the first car is greater.

Chapter 2 Topic A Rest Motion And Equation Of Motion Multiple Choice Questions

Question 1. A ball is thrown. What type of motion will the ball follow if rotation of the ball along its own axis is ignored?

1. Compound motion
2. Circular motion
3. Linear motion
4. Oscillatory motion in a straight line

Answer: 1. Compound motion

Question 2. A particle traverses a semi-circular path of 1m radius in 1s time. What is the average velocity of the particle?

1. 3.14 m/s
2. 2 m/s
3. 1 m/s
4. Zero

Answer: 2. 2 m/s

Question 3. Starting from your house, you walk 8 km in 2 hours and come back to the house. What will be your displacement?

1. 4 km
2. 2 km
3. 8 km
4. Zero

Answer: 4. Zero

Question 4. n number of bullets are pumped out per second from a machine gun. If mass of each bullet is m kg and its velocity is v m/s, what is the applied force (in N unit) on this machine gun?

1. mnv
2. \(\frac{mn}{v}\)
3. mn
4. \(\frac{mv}{n}\)

Answer: 1. mnv

Question 5. An example of compound motion is

1. A ball rolling on the ground
2. Rotation of a ball at one place
3. Slippery motion of a ball
4. Perpendicular fall of a ball from a height

Answer: 1. A ball rolling on the ground

Question 6. A plane flies 6000 km eastward and then 8000 km northward. Then the plane comes back to its initial position by the shortest route. If the speed of this plane is 200 km/h, what was its average velocity in the total journey path?

1. 0
2. 120 km/h
3. 200 km/h
4. 220 km/h

Answer: 1. 0

Question 7. Which of the following quantity remains unchanged in rotational motion?

1. Velocity
2. Axis
3. Linear momentum
4. None of these

Answer: 2. Axis

Question 8. The motion of a wheel of a running car is

1. Translational motion
2. Rotational motion
3. Compound motion
4. None of the above

Answer: 3. Compound motion

Question 9. Ratio of traversed distance and displacement of a moving body is

1. <1
2. ≤1
3. ≥1
4. = 1

Answer: 3. ≥1

Question 10. A man travels along the circumference of a semi-circular field of radius 14 m and goes to the other side. The magnitude of displacement of the man is

44 m
28 m
88 m
14 m

Answer: 2. 28 m

Question 11. A particle moves from point A along the circumference of a circle of radius 5√2 cm to point B, so that an angle of 60° is formed at the centre. The displacement of the particle is

1. 5√2 cm
2. 5 cm
3. 10√2 cm
4. 10 cm

Answer: 1. 5√2 cm

Question 12. Initial velocity and deceleration of a particle are 20 m/s and 2.5 m/s² respectively. Time taken by the particle to come to rest is

1. 4s
2. 6s
3. 8s
4. 10s

Answer: 3. 8s

Question 13. For uniformly circular motion

1. Velocity of a particle always remains unchanged
2. Speed and acceleration of a particle remain unchanged
3. Speed changes
4. Magnitudes of speed and acceleration remain unchanged

Answer: 4. Magnitudes of speed and acceleration remain unchanged

Question 14. Velocity-time (v-t) graph of a particle. Average velocity of the particle is

1. 4 m/s
2. 5 m/s
3. 6 m/s
4. 7.5 m/s

Answer: 2. 5 m/s

Question 15. Velocity-time (v-t) graph of a particle is shown in the figure. Acceleration of the particle is

1. 0.2 m/s²
2. 0.3 m/s²
3. 0.4 m/s²
4. 0.5 m/s²

Answer: 3. 0.4 m/s²

Question 16. Dimensional formula of acceleration is

1. LT^-2
2. LT^-2
3. LT^-3
4. L^-1T

Answer: 2. LT^-2

Question 17. A particle moving with uniform velocity has an initial velocity u and a final velocity v. The average velocity of the particle is

1. \(\frac{v-u}{2}\)
2. \(\frac{u+v}{2}\)
3. \(\sqrt{\frac{u^2+v^2}{2}}\)
4. \(\sqrt{\frac{v^2-u^2}{2}}\)

Answer: 2. \(\frac{u+v}{2}\)

Question 18. Velocity-time (v-t) graph of a particle is shown in the figure. Ratio of distances traversed by the particle with uniform acceleration and with uniform velocity is

1. 1:2
2. 2:1
3. 2:3
4. 3:5

Answer: 1. 1:2

Question 19. Which of the following is at absolute rest?

1. Earth
2. Sun
3. Moon
4. None of these

Answer: 4. None of these

Question 20. Motion of all moving bodies is

1. Absolute motion
2. Relative motion
3. Translational motion
4. Rotational motion

Answer: 4. Rotational motion

Question 21. Example of translational motion is

1. Rolling motion of a ball on ground
2. Vertical fall of a ball from a height
3. Rotation of a ball at one place
4. Slippery motion of a ball

Answer: 2. Vertical fall of a ball from a height

Question 22. Direction of motion of particles inside a body which moves with translational motion in a straight line

1. Always remains unchanged
2. Always changes
3. May remain unchanged ormay change
4. Changes at first and then remains unchanged

Answer: 1. Always remains unchanged

Question 23. Direction of motion of particles inside a body which moves with oscillatory motion in a straight line

1. Always remains unchanged
2. Changes at an interval of time
3. Always changes
4. None of the above

Answer: 2. Changes at an interval of time

Question 24. Rest and motion of a body is measured in comparison with another body which is called the

1. Reference frame
2. Relative frame
3. Body at rest
4. Reference body

Answer: 4. Reference body

Question 25. Unit of acceleration in CGS system is

1. m.s^-1
2. cm.s^-2
3. m.s^-2
4. cm.s^-1

Answer: 2. cm.s^-2

Question 26. To express speed

1. Only magnitude is required
2. Only direction is required
3. Both magnitude and direction are required
4. Either magnitude or direction is required

Answer: 1. Only magnitude is required

Question 27. Velocity and acceleration of a body moving in a straight line

1. Always act in the same direction
2. Always act in opposite directions
3. Act in opposite directions in some cases
4. Never act in opposite directions

Answer: 1. Always act in the same direction

Question 28. Which of the following is a constant quantity in case of a freely falling body?

1. Displacement
2. Acceleration
3. Velocity
4. Speed

Answer: 2. Acceleration

Question 29. A particle moves half the distance of a semi- circular path with radius r. Displacement of this particle is

1. r
2. 2r
3. r/2
4. 3r

Answer: 2. 2r

Question 30. The initial velocity and acceleration of a particle are u and a, respectively. After time t, its velocity is v. The relationship between u, v, a, and t is given by

1. u = v + at
2. u + v = at
3. v – u = at
4. v = u + at

Answer: 4. v = u + at

Question 31. A boy walks 3 km eastwards and then 4 km northwards. Displacement of the boy is

1. 10 km
2. 5 km
3. 8 km
4. 14 km

Answer: 2. 5 km

Question 32. Dimensional formula of velocity is

1. LT^-1
2. LT^-2
3. LT
4. MLT^-1

Answer: 1. LT^-1

Question 33. A body at rest starts moving with uniform acceleration of 8 cm s^-2 when a force is applied on it. Its final velocity after 12s is

1. 50 cm.s^-1
2. 75 cm-s^-1
3. 80 cm. s^-1
4. 96 cm.s^-1

Answer: 4. 96 cm.s^-1

Question 34. Area between the curve and time axis in a velocity-time graph indicates

1. Displacement of the body
2. Acceleration of the body
3. Change of velocity of the body
4. None of the above

Answer: 1. Displacement of the body

Question 35. A ball is thrown by a man upward in a perpendicular direction with a velocity of 10m s^-1. After sometime, the ball comes back to the man again. Average velocity of the ball is

1. 15 m.s^-1
2. 10 m.s^-1
3. 20 m.s^-1
4. Zero

Answer: 4. Zero

Question 36. Which one of the four graphs shown below represents a body moving with uniform acceleration?

Answer: 2.

Question 37. The velocity-time graph of a particle moving with uniform velocity is

1. A straight line parallel to the time axis
2. A straight line parallel to the velocity axis
3. A straight line passing through the origin
4. A curved line passing through the origin

Answer: 1. A straight line parallel to the time axis

Question 38. A boy is tossing up a coin perpendicularly inside a running compartment of a train with his face towards the engine. If the coin falls behind the boy, then the train

1. Is moving forward with uniform velocity
2. Is moving backward with uniform velocity
3. Is moving forward with deceleration
4. Is moving forward with acceleration

Answer: 4. Is moving forward with acceleration

Question 39. In a rotational motion

1. Velocity of any particle is constant
2. Velocity of a particle further from the axis of rotation is greater
3. Velocity of a particle further from the axis of rotation is lesser
4. Angular velocity and rectilinear velocity are constant

Answer: 2. Velocity of a particle further from the axis of rotation is greater

Question 40. Which of the following is a simple oscillatory motion?

1. Motion of the hand of a clock
2. Motion of the earth around the sun
3. Motion of a simple pendulum
4. Uniform circular motion

Answer: 3. Motion of a simple pendulum

Question 41. A man goes 4 m eastwards, then 4 m northwards and at the end, 3√2 m south- westwards. Displacement of the man is

1. 2√2m
2. √2m
3. 4m
4. 3√2m

Answer: 2. √2m

Question 42. During its journey, a car covers half of the total distance with a speed of 40 km/h and the rest half distance with a speed by 60 km/h. Average speed of the car is

1. 50 km/h
2. 46 km/h
3. 48 km/h
4. 52 km/h

Answer: 3. 48 km/h

Question 43. A ball is thrown vertically upward. If the rotation of the ball along its own axis is ignored then What type of motion will the ball follows?

1. Compound motion
2. Circular motion
3. Linear motion
4. Simple harmonic oscillation

Answer: 3. Linear motion

Question 44. A particle is revolving in a circular path of radius R. The displacement of the particle, when it completes one complete rotation, is

1. πR
2. 2πR
3. 0
4. 2R

Answer: 3. 0

Question 45. Distance travelled by a free falling body under the action of gravity in first 3s of motion is

1. g
2. \(\frac{3g}{2}\)
3. \(\frac{9g}{2}\)
4. \(\frac{9g}{4}\)

Answer: 3. \(\frac{9g}{2}\)

Question 46. Which of the following displacement-time dimensional graph represents one displacement of a particle?

Answer: 4.

Question 47. The velocity-time graph of a particle moving with uniform velocity is

1. A straight line parallel to the time axis
2. A straight line parallel to the velocity axis
3. A straight line passing through the origin
4. A curve line passing through the origin

Answer: 1. A straight line parallel to the time axis

Question 48. Velocity-time graph of a particle projected vertically upward is

Answer: 2.

Question 49. Initial and final velocity of a particle moving with constant acceleration are 30 cm/s and 40 cm/s respectively. Velocity of the particle at the midpoint of its way is

1. 35 m/s
2. 25√2 cm/s
3. 32 cm/s
4. 28√2 cm

Answer: 2. 25.2 cm/s

Question 50. A particle starts moving from its rest with constant acceleration and travel a certain distance in 100 s. Time required to travel first half of the distance is

1. 50s
2. 70s
3. 70.7s
4. 71s

Answer: 3. 70.7s

Question 51. A particle is projected vertically upward such that it attains maximum height ‘h’ and returns to ground. Here total distance traversed by the particle and total displacement of the particle are

1. h, 0
2. 0, 2h
3. 2h,0
4. 0, h

Answer: 3. 2h,0

Question 52. A body travels along a circular path of radius r and complete one rotation in time t: Here

1. average velocity = \(\frac{2 \pi r}{t}\)
2. total displacement = 2πr
3. total distance travel = 0
4. average speed = \(\frac{2 \pi r}{t}\)

Answer: 4. average speed = \(\frac{2 \pi r}{t}\)

Question 53. A piece of stone is dropped from the peak of a minar of height 20 m. What will be the speed of the stone when it hits the ground? (g = 10 m.s^-2)

1. 10 m.s^-1
2. 40 m.s^-1
3. 20 m.s^-1
4. 5 m.s^-1

Answer: 2. 40 m.s^-1

Question 54. In the velocity-time graph of a car is shown. Here the car moves with

1. Non-uniform velocity
2. Uniform acceleration
3. Uniform deceleration
4. Non-uniform deceleration

Answer: 3. Uniform deceleration

Question 55. Gradient of velocity-time graph of a particle is

1. Velocity
2. Acceleration
3. Distance traveled
4. Displacement

Answer: 2. Acceleration

Question 56. A boat first moves 12 m towards east and then 5 m towards west In a river. What is the total displacement of the boat?

1. 13m
2. 17m
3. 7m
4. 8m

Answer: 3. 7m

Question 57. SI unit of retardation is

1. -m.s^-2
2. cm.s^-2
3. m.s^-2
4. -cm.s^-2

Answer: 3. m.s^-2

Question 58. Rotation of the earth about it’s own axis is

1. Circular motion
2. Rotational motion
3. Translational motion
4. Compound motion

Answer: 3. Translational motion

Question 59. Motion of electron round a nucleus is

1. Circular motion
2. Rotational motion
3. Translation motion
4. Simple harmonic motion

Answer: 1. Circular motion

Question 60. Velocity of a particle moving in a straight line changes from 10 m.s^-1 to 20 m.s^-1 in time t s. If the particle travel 130 m distance in this time interval then value of t is 12

1. 1.8 s
2. 10 s
3. 12 s
4. 8.67 s

Answer: 4. 8.67 s

Question 61. Which of the following velocity-time graph represents the motion of a particle with non zero initial velocity and non uniform acceleration?

Answer: 1.

Question 62. Angle between displacement-time graph and time axis of two particles are 45° and 60° respectively. The Ratio of the velocity of two particles is

1. √3:1
2. 1:√3
3. 3:4
4. 1:1

Answer: 2. 1:3

Question 63. Average velocity of a particle is equal to its instantaneous velocity if the particle moves with

1. Constant acceleration
2. Constant speed
3. Gradually increasing speed
4. Constant velocity

Answer: 4. Constant velocity

Chapter 2 Topic A Rest Motion And Equation Of Motion Answer In Brief

Question 1. Imagining the earth as a point with respect to sun, what type of motion does the earth undergo around the sun?

Answer: In this case, rotation of the earth is a circular motion.

Question 2. What type of motion does a rotating body undergo around its own axis?

Answer: A rotating body undergoes a rotating motion around its own axis.

Question 3. Two bodies of masses 5 kg and 6 kg are falling from rest without any resistance. Which one has more acceleration?

Answer: Both the bodies fall with the same acceleration (acceleration due to gravity).

Question 4. Velocity-time (v-t) curve of a moving body is parallel to the time axis. What is the acceleration of the body?

Answer: Acceleration of the body is zero.

Question 5. Velocity-time (v-t) curves of two moving bodies form inclined angles of 30° and 60°. Which body has more acceleration?

Answer: The curve in respect of the body with inclined angle of 60° has comparatively more acceleration.

Question 6. Is it possible for the velocity-time (v-t) curve of a body to be perpendicular to the time-axis?

Answer: No, because in that case, the curve denotes different velocities of the body at a particular time which is impossible.

Question 7. What is the distance that a moving particle with initial velocity u and uniform deceleration a covers in time t?

Answer: Required distance, s = ut – \(\frac{1}{2}\) at².

Question 8. What is the velocity of a particle moving with initial velocity u and uniform acceleration a after traversing a distance s?

Answer: If v is the required velocity, then

v² = u² + 2at²

or, v = √u²+2as

Question 9. What is the change in acceleration of a body falling freely from a height without any resistance?

Answer: No, there is no change in acceleration of the body as this is acceleration due to gravity.

Question 10. What type of motion is the motion of wheel of a moving bicycle?

Answer: Motion of wheel of a moving bicycle is compound motion.

Question 11. What type of motion is the motion of an arm of a clock?

Answer: Motion of an arm of a clock is rotational motion or uniform circular motion.

Question 12. What type of motion is the motion of a freely falling body?

Answer: Motion of a freely falling body is translational motion.

Question 13. When do the passengers of two moving trains find each other to be stationary?

Answer: When two trains move in parallel with the same velocity, passengers find them to be stationary.

Question 14. What type of motion is the motion of a merry-go-round?

Answer: The motion of a merry-go-round is rotational motion.

Question 15. What is circular motion?

Answer: If a particle rotates around a fixed axis or a point in a circular path, then the motion of the particle is called circular motion.

Question 16. What is rectilinear oscillatory motion?

Answer: If a body moves repeatedly along the same straight line at equal intervals of time, then its motion is called rectilinear oscillatory motion.

Question 17. Can the distance traversed by a moving particle be zero?

Answer: No, the distance traversed by a moving particle cannot be zero.

Question 18. Can the displacement of a moving particle be zero?

Answer: If a particle starting from a particular point comes back to the same point, then the displacement of that particle is said to be zero.

Question 19. When displacement of a moving particle is zero?

Answer: Displacement of a moving is zero when its initial position and final position are the same.

Question 20. Can the ratio of distance traversed and displacement of a moving particle be less than 1?

Answer: No, the ratio of the distance travelled and the displacement of a moving particle is always 1 or more than 1.

Question 21. When is the ratio of the distance traversed and the displacement of a moving particle equal to 1?

Answer: The ratio of the distance traversed and the displacement of a moving particle is 1 when it moves in a straight line.

Question 22. When is the ratio of the distance traversed and the displacement of a moving particle more than 1?

Answer: The ratio of the distance traversed and the displacement of a moving particle is more than 1 when it moves in a curved path.

Question 23. What is the dimensional formula of speed?

Answer: Dimensional formula of speed is LT^-1.

Question 24. What type of physical quantity is speed?

Answer: Speed is a scalar quantity because it has magnitude, but no direction.

Question 25. A particle traverses the circumference of a circle of radius r and comes back to the position from where it started. What is its displacement?

Answer: Displacement of the particle is zero.

Question 26. What are the units of displacement in CGS system and SI?

Answer: Units of displacement in CGS system- centimetre (cm) and in SI-metre (m).

Question 27. A particle traverses the circumference of a circle of radius r and comes back to the position from where it started. What is the distance covered?

Answer: Distance covered = 2πr.

Question 28. Can the average speed of a moving particle be zero?

Answer: No, the average speed of a moving particle cannot be zero.

Question 29. Can the average velocity of a moving 38 What remains unchanged during a particle be zero?

Answer: Yes, the average velocity of a moving particle can be zero.

Question 30. What is negative acceleration called?

Answer: Negative acceleration is called deceleration or retardation.

Question 31. Displacement s of a particle in time t,s = 4t + 5t². If displacement of the particle is expressed in unit m and time in units, what is the initial velocity?

Answer: If we compare the equation s = 4t+5t² with the equation s = ut + 1/2 at², initial velocity is u = 4 m/s.

Question 32. What does the area made by a velocity-time graph with the time axis indicate?

Answer: It indicates the distance traversed.

Question 33. What is the nature of a displacement-time graph of a particle moving with uniform velocity?

Answer: The displacement-time graph of a particle moving with uniform velocity is a straight line passing through the origin.

Question 34. How do you define a reference body?

Answer: A reference body is a body with reference to which states of rest and motion of another body are measured.

Question 35. What is the direction of a particle moving in a circular path at a point?

Answer: The direction of a particle moving in a circular path at a point is along the tangent to the circle at that point.

Question 36. What does the odometer of a car indicate?

Answer: The odometer of a car indicates the instantaneous speed of the car.

Question 37. Can the average and the instantaneous speed of a moving particle be equal at any moment?

Answer: The average speed and the instantaneous speed of a particle moving with uniform speed are equal.

Question 38. What remains unchanged during a rotational motion?

Answer: The axis of rotation remains unchanged during rotational motion.

Question 39. 1 km/h = how many m/s?

Answer: 1 km/h = \(\frac{1000}{60 \times 60}\)m/s = \(\frac{5}{8}\) m/s

Question 40. How many times the words ‘per second’ come in the unit of acceleration?

Answer: The words ‘per second’ come twice in the unit of acceleration.

Question 41. What does the inclined angle of a displacement-time graph indicate?

Answer: Inclined angle of a displacement-time graph indicates the instantaneous velocity.

Question 42. What does the inclined angle of a velocity- time graph indicate?

Answer: Inclined angle of velocity-time graph indicates the instantaneous acceleration.

Question 43. What is the nature of the displacement-time graph of a particle moving with uniform acceleration?

Answer: The nature of the displacement-time graph of a particle moving with uniform acceleration is parabolic.

Question 44. What type of motion does a rotating body undergo around its own axis?

Answer: A rotating body under goes a rotational motion around its own axis.

Question 45. What is the direction of displacement?

Answer: Direction of displacement is considered from the initial position towards the final position of a body.

Question 46. s = 4t + 6t² is the displacement of a particle in time t. Considering displacement in m and time is s find acceleration of the particle.

Asnwer: Comparing s = ut + \(\frac{1}{2}\) at² and s = 4t + 6t² equations,

We get, = \(\frac{a}{2}\) = 6 or, a = 12

Therefore acceleration of the particle is 12 m.s^-2.

Question 47. What is the nature of velocity-time graph of a particle moving with constant acceleration and zero initial velocity?

Answer: Velocity-time graph of the particle moving with uniform acceleration and zero initial velocity is a straight line passing through the origin and inclined by an angle with the time axis.

Question 48. Initial velocity and uniform acceleration of a moving particle are A and B respectively. Find its velocity after time C.

Answer: Velocity of the particle after time C is = A + B X C. (Using v = u + at equation)

Question 49. What is the direction of acceleration of a particle moving in a circular path with uniform speed?

Answer: The direction of acceleration of a particle moving in a circular path with uniform speed is towards the ncentre of the circle (centripetal acceleration).

Question 50. Give an example where velocity and acceleration of a particle are directed opposite to each other.

Answer: Velocity and acceleration are directed opposite to each other for vertical motion of a body. Here gravitational acceleration directed opposite to the velocity of the particle.

Question 51. What are the units of acceleration in SI and CGS system?

Answer: Units of acceleration in CGS and SI system are cm-s-2 and m.s^-2 respectively.

Question 52. What are the units of velocity in SI and CGS system?

Answer: Units of velocity in CGS and SI system are cm.s^-2 and m.s^-2 respectively.

Question 53. Initial and final velocity of a particle moving with uniform acceleration are u and v respectively. what is its average velocity?

Answer: Average velocity of the particle is = \(\frac{u+v}{2}\)

Question 54. Draw velocity-time graph of a particle moving with uniform velocity.

Answer: Velocity-time graph of a particle moving with uniform speed is drawn. Here the graph is parallel to time axis.

Question 55. A train travels 10 km in 10 minute. What is the speed of the train in km/h unit?

Answer: 10 min = \(\frac{10}{60}h=\frac{1}{6}\)h

∴ speed of the train \(\frac{10}{\frac{1}{6}}\) km/h = 60 km/h

Chapter 2 Topic A Rest Motion And Equation Of Motion Fill In The Blanks

Question 1. Average _______ of a body may be zero but its average speed need not be zero.

Answer: Velocity

Question 2. Velocity-time (v-t) curve of a moving body with uniform acceleration starting from rest passes through the ______ of the graph.

Answer: Origin

Question 3. Starting from a particular place for a journey, if one comes back to the same place, total displacement is _______

Answer: Zero

Question 4. Speed is a ______ quantity.

Answer: Scalar

Question 5. If both the magnitude and the direction of the velocity of a particle remain unchanged with time, then velocity of that particle is called ________ velocity.

Answer: Uniform

Question 6. For a particle falling freely under gravity, constant quantity is ________

Answer: Acceleration

Question 7. Rest and motion for a particle are ______

Answer: Relative

Question 8. From a moving train, distant trees appear to be ________

Answer: Moving

Question 9. For pure rotation, the axis of rotation always remains _______

Answer: At rest

Question 10. Diurnal motion of the earth is an example of _______ motion.

Answer: Rotational

Question 11. Unit of ________ is used twice in the unit of acceleration.

Answer: Time

Question 12. The ______ of a body may not be zero even though its is zero.

Answer: Acceleration velocity

Question 13. Increase of velocity of a particle with time is called ________

Answer: Acceleration

Question 14. Velocity-time graph of a body moving from its rest with uniform acceleration is a straight line passing through the _______

Answer: Origin

Question 15. Mass of a body is its ______ property.

Answer: Intrinsic

Question 16. Retardation is also called _______ acceleration.

Answer: Negative

Question 17. The value of average velocity and average speed are same in _______ motion.

Answer: Translation

Question 18. Motion of simple pendulum is _______ in nature.

Answer: Periodic

Question 19. Area bounded by velocity-time graph and time axis divided by total time taken gives the physical quantity _______

Answer: Average velocity

Chapter 2 Topic A Rest Motion And Equation Of Motion State Whether True Or False

Question 1. If a body rotates around a fixed axis or a fixed point, its motion is called rotational motion.

Answer: True

Question 2. Rate of change of velocity of a particle with respect to time is called its acceleration.

Answer: True

Question 3. Average speed can be obtained by dividing the total displacement by the total time required to transverse the total distance.

Answer: False

Question 4. Dimensional formula of retardation is LT^-3

Answer: True

Question 5. Passengers of two trains running side by side with the same speed in the same direction, find them mutually at rest.

Answer: True

Question 6. Average velocity of a body will be zero when average speed of the body is zero.

Answer: False

Question 7. If speed is zero, velocity may not be zero.

Answer: False

Question 8. Speed may not be negative.

Answer: True

Question 9. v-t graph of a particle moving with uniform velocity is parallel to time axis.

Answer: True

Question 10. Free falling of a body is translational motion with constant acceleration.

Answer: True

Question 11. The annual motion of the centre of mass of the earth around the sun is rotational motion.

Answer: False

Question 12. Even if the instantaneous velocity of an object is zero, it may have acceleration.

Answer: True

Question 13. Particle having uniform speed must not have non-uniform velocity.

Answer: False

Question 14. Particle moving in a circular path may have uniform speed.

Answer: True

Question 15. Velocity and acceleration of an object must not have opposite directions.

Answer: False

Question 16. Gradient at any point in a v-t graph represents acceleration at that instant.

Answer: True

Chapter 2 Topic A Rest Motion And Equation Of Motion Numerical Examples

Useful Relations

If d be the distance covered by a particle in time t then speed, u = \(\frac{d}{t}\)

If distance covered by a particle in time \(t_1, t_2, t_3, \cdots, t_n \quad \text { are } \quad d_1, d_1, d_3, \cdots, d_n\) respectively, then

average speed, \(v_{a v g}=\frac{d_1+d_2+d_3+\cdots+d_n}{t_1+t_2+t_3+\cdots+t_n}\)

If s be the displacement of a moving particle in time t then velocity, v = \(\frac{s}{t}\)

If the initial velocity of a particle is u and the final velocity of the particle after time t will be v, then acceleration, a = \(\frac{v-u}{t}\)

If initial velocity of a particle = u, final velocity of the particle = v, acceleration = a, time taken = t, displacement = s then,

1. \(v=u+a t\)
2. \(s=u t+\frac{1}{2} a t^2\)
3. \(v^2=u^2+2 a s\)

If the initial velocity, u = 0,

1. v = at
2. s = \(\frac{1}{2}\)at²
3. v = √2as

If a be the retardation,

1. v = u – at
2. s = ut – \(\frac{1}{2}\)at²
3. v² = u²-2as

Question 1. Two individuals start their journey from C and reach B. One goes through the path CDAB and the other through the path CDB. ABCD is a square whose length of each side is 2 m. What are the displace ments of C and B? How much distance does each one cover?

Answer:

As both of them start from initial position C and reach final position B, both have the displacement, CB = 2 m.

Distance covered by the first individual

= CD + DA + AB = 2 + 2 + 2 = 6m

Distance covered by the second individual

= CD + DB = 2 + 2√2 = 2(1+ √2) m

Concepts Related to Speed, Velocity, and Acceleration for Class 9 Solutions

Question 2. A moving particle goes from one end to the other end of a semicircular path of radius 7 cm. Calculate the total distance cov- ered and the magnitude of displacement.

Answer:

Radius, r = 7 cm

∴ Distance Covered

= Circumference of the semicircle

= πr = \(\frac{22}{7}\) x 7 = 22 cm

∴ displacement 2r = 2 x 7 = 14 cm

Question 3. A particle moves from point A to point B along the circumference of a circle of radius 10√2 cm such that an angle of 90° is formed at the centre. What displacement of the particle?

Answer:

Suppose, centre of circle is at point O.

Radius of the circle = 10√2 cm

According to AOB is a right-angled triangle.

∴ \(A B^2=O A^2+O B^2=r^2+r^2=2 r^2\)

or, AB = \(\sqrt{2} r=\sqrt{2} \times 10 \sqrt{2}=20 \mathrm{~cm}\)

∴ Displacement of the particle is 20 cm.

Question 4. A wheel makes half a rotation on a plane surface. If radius of this wheel is 21 cm, what is the displacement of the point touching the ground?

Answer:

Initial and final positions of the wheel are shown respectively.

Point A touching the ground has moved to the position A₁ after making half a rotation.

At the final position, B₁ is the point touching the ground.

Radius of the wheel, r = 21 cm

∴ \(A B_1=\pi r=\frac{22}{7} \times 21=66 \mathrm{~cm}\)

and \(A_1 B_1=2 r=2 \times 21=42 \mathrm{~cm}\)

So, \(A A_1=\sqrt{\left(A B_1\right)^2+\left(A_1 B_1\right)^2}\)

= \(\sqrt{(66)^2+(42)^2}=\sqrt{6120}=78.23 \mathrm{~cm}\)

∴ Displacement of the point touching the ground is 78.23 cm.

Question 5. During the first half of its journey, a car moves with a speed u and it makes the remaining half of the journey with a speed v. What is the average speed of this car?

Answer:

Suppose, distance to be covered = 2s.

Time taken to cover the first half of the journey, \(t_1=\frac{s}{u}\)

Time taken to cover the second half of the journey, \(t_2=\frac{s}{v}\)

∴ Average Speed,

\(v_a=\frac{2 s}{t_1+t_2}=\frac{2 s}{\frac{s}{u}+\frac{s}{v}}=\frac{2}{\frac{1}{u}+\frac{1}{v}}=\frac{2}{\frac{u+v}{u v}}=\frac{2 u v}{u+v}\)

Question 6. A particle moves along the circumference of a circle of radius 21 cm from point A to point B with uniform speed and forms an angle of 60° at the centre. What is the speed of the particle?

Answer:

Circumference of a circle of radius 21 cm

= \(2 \pi r=2 \times \frac{22}{7} \times 21=132 \mathrm{~cm}\)

After completion of one round along the circumference, the particle forms an angle of 360° at the centre with respect to the line joining the initial position of the particle (A) and the centre of circle (O).

But as the particle forms an angle of 60° at the centre, distance traversed by it is given by

s = \(2 \pi r \times \frac{60^{\circ}}{360^{\circ}}=132 \times \frac{1}{6}=22 \mathrm{~cm}\)

Now, Time Spent, t = 2 s

∴ Speed of the particle, v = \(\frac{s}{t}=\frac{22}{2}\) = 11 cm/s

Question 7. A train travels the first 30 km of its total journey of 100 km at the uniform speed of 30 km/h. What is its speed for the remaining 70 km so that the average speed of the total journey is 40 km/h?

Answer:

Train travels first s₁ = 30 km of the journey at the uniform speed of v₁ = 30 km/h.

Let the train travel remaining s₂ = 70 km distance at uniform speed v₂ so that its average speed for the total journey, v_a becomes 40 km/h.

If time t₁ and t₂ are taken to cover the first 30 km and the last 70 km respectively, then

\(v_a=\frac{s_1+s_2}{t_1+t_2} \quad \text { or, } v_a=\frac{s_1+s_2}{\frac{s_1}{v_1}+\frac{s_2}{v_2}}\)

 

or, \(40=\frac{30+70}{\frac{30}{30}+\frac{70}{v_2}} \quad \text { or, } 1+\frac{70}{v_2}=\frac{100}{40}\)

or, \(\frac{70}{v_2}=2.5-1\)

∴ \(v_2=\frac{70}{1.5}=46.67 \mathrm{~km} / \mathrm{h}\)

Question 8. A train goes from Dum Dum to Naihati with uniform speed u and comes back from Naihati to Dum Dum with uniform speed v. What is the average speed of this train?

Answer:

Let the distance between Dum Dum and Naihati is s.

∴ Time taken to go from Dum Dum to Naihati with uniform speed u, \(t_1=\frac{s}{u}\)

During the return journey, time taken to cover this same path with uniform speed v, \(t_2=\frac{s}{v}\)

∴ Average speed of the train

= \(\frac{\text { total distance }}{\text { total time }}=\frac{s+s}{t_1+t_2}=\frac{2 s}{\frac{s}{u}+\frac{s}{v}}\)

= \(2 s \times \frac{u v}{s u+s v}=\frac{2 s \times u v}{s(v+u)}=\frac{2 u v}{u+v}\)

Question 9. A car picks up a velocity of 45 km/h after 10s of starting. Calculate the acceleration of the car.

Answer:

Velocity of the car after t = 10 s of starting,

v = \(45 \mathrm{~km} / \mathrm{h}=\frac{45 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=12.5 \mathrm{~m} / \mathrm{s}\)

Suppose, acceleration of the car = a

∴ \(v=a t \text { or, } a=\frac{v}{t} \text { or, } a=\frac{12.5}{10}=1.25 \mathrm{~m} / \mathrm{s}^2\)

Example 10. A train stops after 10s of application of brakes. If deceleration of the train is 3 m/s², what is its velocity at the time of application of the brakes?

Answer:

Suppose, velocity of the train at the time of application of brakes is u.

Deceleration of the train, a = 3 m/s²

The train stops after time t = 10 s

From the equation v=u-at, we get

0 = u – at [v=0]

or, u = at = 3 x 10 = 30 m/s

Question 11. A train running with a velocity of 10 m/s attains an acceleration of 2 m/s². What is the distance covered by it in 10s?

Answer:

Initial velocity (u) = 10 m/s and acceleration (a) = 2 m/s².

A train traverses a distnce s in t = 10 s.

∴ \(s=u t+\frac{1}{2} a t^2=10 \times 10+\frac{1}{2} \times 2 \times 10^2\)

= 100 + 100 = 200 m

Question 12. A car travels y km in the first x min and x km in the next y min. What is the average velocity of this car?

Answer:

Total distance traversed, s = (y + x) km

Total time required, t = (x + y) min

= \(\frac{x+y}{60} \mathrm{~h}\)

∴ Average velocity of this car,

\(v_a=\frac{s}{t}=\frac{x+y}{\frac{x+y}{60}}=60 \mathrm{~km} / \mathrm{h}\)

Question 13. A train was running with a velocity of 72 km/h. It took a time of 40 s to stop after the brakes were applied. Calculate the distance traversed by the train and its deceleration after the application of brakes.

Answer:

Initial velocity of the train,

\(u=72 \mathrm{~km} / \mathrm{h}=\frac{72 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}\)

Velocity of train, v = 0 after t = 40 s

Let deceleration of the train = a.

From the equation v = u – at, we get

0 = u – at or, at = u

or, \(a=\frac{u}{t}=\frac{20}{40}=0.5 \mathrm{~m} / \mathrm{s}^2\)

If the train traverses a distance s before it stops, v² = u²-2as gives

\(0=u^2-2 a s \quad \text { or, } 2 a s=u^2\)

∴ \(s=\frac{u^2}{2 a}=\frac{20^2}{2 \times 0.5}=400 \mathrm{~m}\)

Question 14. A particle starts from rest and covers a distance of 200 m with uniform accelera- tion. What percentage of total time is required to cover half the distance?

Answer:

Suppose, acceleration of the particle = a and t time is taken to cover a distance of s = 200 m.

∴ \(s=\frac{1}{2} a t^2\)

If half the disnce \(s_1=\frac{s}{2}=100 \mathrm{~m}\) is traversed in time, t₁ then

∴ \(s_1=\frac{1}{2} a t_1^2 \quad \text { or, } \frac{s}{2}=\frac{1}{2} a t_1^2\)

or, \(\frac{1}{2} \times \frac{1}{2} a t^2=\frac{1}{2} a t_1^2\) [from equation (1)]

or, \(t_1^2=\frac{t_2}{2}\)

or, \(t_1=\frac{t}{\sqrt{2}}=\frac{\sqrt{2}}{2} t=\frac{1.414}{2} t=0.707 t\)

∴ \(t_1=70.7 \% of t\)

Question 15. A bullet loses half of its velocity after entering 6 cm in a wooden block. How much more distance does it penetrate to come to a halt?

Answer:

Suppose the deceleration of the bullet in the wooden block a and its initial velocity = u.

If the velocity of the bullet is \(v_1=\frac{u}{2}\) after it traverses a distance s₁ = 6 cm.

∴ \(v_1^2=u^2-2 a \cdot s_1 \quad \text { or, } \frac{u^2}{4}=u^2-2 \cdot a \cdot 6\)

or, \(12 a=u^2-\frac{u^2}{4}=\frac{3 u^2}{4} \quad \text { or, } a=\frac{u^2}{16}\)

Its final velocity is V = 0 as it comes to rest after traversing a further distance.

So, \(v^2=v_1^2-2 a s\)

or, \(0=\frac{u^2}{4}-2 \cdot a s \quad or, 2 a s=\frac{u^2}{4}\)

or, \(2 \times \frac{u^2}{16} \times s=\frac{u^2}{4} \quad or, \frac{s}{8}=\frac{1}{4}\)

∴ \(s=\frac{8}{4}=2 \mathrm{~cm}\)

Question 16. A bullet running with a velocity enters a wooden block. After moving x distance in it, its velocity becomes v and after moving a further distance y, it comes to rest. Show that \(\frac{y}{y}=\sqrt{\frac{y}{x+y}}\).

Answer:

Suppose the deceleration of the bullet in the wooden block = a and it stops after traversing a total distance of (x + y).

∴ 0 = u² – 2a(x + y)

or, u2 = 2a(x+y) ……(1)

Again, when its velocity is v, it stops after traversing an additional distance y.

∴ 0 = v² – 2ay or, v² = 2ay…..(2)

Dividing equation (1) by (2), we get

\(\frac{v^2}{u^2}=\frac{y}{x+y}\)

∴ \(\frac{v}{u}=\sqrt{\frac{y}{x+y}}\)

Study Guide for Class 9 Force and Motion Questions

Question 17. Velocity-time (v-t) graph of a particle is shown in the following diagram.

1. What is the highest velocity of the particle?
2. What is the acceleration of the particle?
3. How much distance does it cover with uniform acceleration?
4. How much distance does it cover with uniform velocity?
5. What is the deceleration of the particle?
6. How much distance does it cover with uniform deceleration?
7. How much total distance does the particle cover?

Answer:

It is seen from the graph that the particle moves with uniform acceleration for the first 2 s to attain a velocity of 10 m/s. Then it travels with a uniform velocity of 10 m/s for 4s (from 2s to 6s).

It travels with uniform deceleration to come to rest in the last 2s(6 s to 8 s).

1. Highest velocity of the particle = AD = 10 m/s

2. Acceleration during the first 2 s

= \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time }}\)

= \(\frac{(10-0) \mathrm{m} / \mathrm{s}}{2 \mathrm{~s}}=5 \mathrm{~m} / \mathrm{s}^2\)

3. Distance traversed by the particle with uniform acceleration,

s₁ = area of AOAD = \(\frac{1}{2}\) × OD X AD

= \(\frac{1}{2}\) x 2s x 10 m/s = 10 m

4. Distance traversed by the particle with uniform velocity,

s₂ = area of rectangle ABED

=AD X DE = 10 m/s x (6-2)s

= 40m

5. Deceleration of the particle in the last 2s

= \(\frac{\text { initial velocity }- \text { final velocity }}{\text { time }}\)

= \(\frac{(10-0) \mathrm{m} / \mathrm{s}}{2 \mathrm{~s}}=5 \mathrm{~m} / \mathrm{s}^2\)

6. Distance traversed by the particle with uniform deceleration,

\(s_3=\text { area of } \triangle B E C=\frac{1}{2} \times E C \times B E\)

= \(\frac{1}{2} \times(8-6) \times 10=\frac{1}{2} \times 2 \times 10=10 \mathrm{~m}\)

7. Distance traversed by the particle in total 8 seconds,

s = s₁ + s₂ + s₃ = 10 + 40 + 10 = 60 m

Question 18. Velocity-time (v-t) graph of a particle is shown in the following diagram. Determine the average velocity of the particle.

Answer:

Distance traversed by the particle,

s = \(\text { area of } \triangle O A B=\frac{1}{2} \times O B \times A C\)

= \(\frac{1}{2} \times 10 \times 10=50 \mathrm{~m}\)

∴ Average velocity of the particle,

\(v_a=\frac{\text { total displacement }}{\text { total time }}=\frac{50}{10}\)

= 5 m/s

Question 19. A particle moves in a straight line with uniform acceleration. Velocity of the particle at different times is given In the following chart:

Draw The Velocity-Time Graph Of The Particle.

Determine From The Graph:

1. Velocity after 3 s,
2. Acceleration of the particle and
3. Distance traversed during the last 6 s.

Answer:

Velocity-time (v-t) graph of the particle is

1. From the graph, velocity of the particle after 3s, v₁ = BC = 6 m/s

2. Acceleration of the particle, \(a=\frac{B C}{O B}=\frac{6}{3}=2 \mathrm{~m} / \mathrm{s}^2\)

3. Distance traversed during the last 6 s time,

s = area of trapezium DEFG

= \(\frac{1}{2}\)(DE + FG) X DG

= \(\frac{1}{2}\)(8 + 20) X (10 – 4)

= \(\frac{1}{2}\) X 28 X 6 = 84 m

Question 20. A particle starts from rest and attains a velocity of 40 cm/s after traversing for 10s with uniform acceleration. Next, the particle moves for 10s with uniform velocity. Finally, the particle comes to rest while moving for 10s with uniform deceleration. Draw the velocity-time graph for the partide. Calculate the total distance traversed from the graph.

Answer:

Velocity-time (v-t) graph of the particle is shown where OA is the graphical representation of time 0-10 s during uniform acceleration state.

Further, AB and BC are graphical representations of the uniform velocity period during 10s-20s and uniform deceleration period during 20s – 30s, respectively.

∴ Total distance traversed by the particle,

s = area of trapezium OABC

= \(\frac{1}{2}\)(OC + AB) X AE = \(\frac{1}{2}\)(30 + 10) X 40

= \(\frac{1}{2}\) x 40 x 40 = 800 cm = 8 m

Question 21. A ball is thrown vertically upward with a velocity 50 m/s. Calculate the time taken by the ball to return at the point of ejection. [g = 10m-s-2].

Answer:

Initial velocity of the ball is u = 50 m • s^-1.

Let, t be the required time.

∴ Total height covered by the ball in time t is,

h = 0

From equation h = ut-1/2 gt² we get,

0 =(50.t) – (1/2 .10.t¹)

or, t(50 – 5t) = 0

∴ t = 0 or, 50 – 5t = 0 or, t = 50/5 = 10

(t = 0 means initial position of the ball)

Therefore, the required time is 10 s.

Question 22. A car starts moving with an initial velocity and uniform acceleration. It traversed 81 m distance in 5 s. After that it
travels 72 m with uniform velocity in next 4s. Determine the initial velocity and acceleration of the car.

Answer:

Let, the initial velocity and the acceleration of the car be u m/s and a m • s^-2.

Final velocity of the car after 5 s is v m/s .

The car travels 72 m distance in 4 s with uniform velocity.

∴ v = 72/4  = 18 m/s

Now from equation v = u + at we get,

18 = u + a x 5

∴ u = 18 – 5a

Here, t = 5 s, u = 18 m/s

Again from equation v² = u² + 2as we get,

18² = (18- 5a)² + 2 x a • 81 [s = 81 m]

or, 18² = 18² – 2 • 18 • 5a + 25a²+ 162a

or, 25a² -(180-162)a = 0

or, a[25a- 18] = 0

∴ a = 0

or, a = 18/25 = 0.72 [Here, car moves with uniform acceleration.  ∴ a ≠ 0]

∴ Acceleration of the car is 0.72 m • s^-2 and initial velocity,

u = 18 – 5a = 18 – 5 x 0.72

= 18- 3.6 = 14.4 m • s^-2

Chapter 1 Measurement Synopsis

A Physical Quantity is one which is related to a material body or an event that can be measured directly or indirectly.

Example: mass, length, time, velocity, force etc. are examples of some physical quantities.

Physical Quantities Are Of Two Types-

1. scalar quantity and
2. vector quantity.”

1. The physical quantities which have only magnitude but no direction are called scalar quantities.

Example: length, mass, time, temperature, density, volume, work, energy etc.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

2. The physical quantities which have both magnitude and direction are called vector quantities.

Examples: displacement, velocity acceleration, linear momentum, force, electric field etc.

Electric current intensity have both magnitude and direction but it does not obey ‘vector rules’ and hence it is not a vector quantity.

The Fundamental Units are the units of the physical quantities, which are not dependent on any other units, and all other units are derived from them.

In the international system of units (SI system) the Seven Fundamental Units Are-

Units which are derived from one or more, than one fundamental unit are called derived units.

For Example area, volume, speed, velocity, acceleration, force, momentum, etc. are derived units.

Unitless physical quantities are defined as the ratio of the same units of two different physical quantities.

Examples: atomic weight, molecular weight, relative density, refractive index, relative humidity etc.

A dimensionless physical quantity may have a unit but a unitless physical quantity is always dimensionless. (example plane angle = \(\frac{length of arc}{radius}\)  i.e, dimensionless but have a unit (radian in SI)

SI Unit of length is metre-The metre is the distance travelled by light in second in \(\frac{1}{299792458}\) vacuum.

Unit Of Time is second in SI-the second is the duration of 9192631770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground states of the cesium-133 atom.

The amount of space occupied by a body is its volume.

WBBSE Class 9 Measurement Solutions

Unit Of Volume In CGS System: cm³ and in Sl: m³, 1m³ 10⁶ cm³ another unit: litre. The volume of 1 kilogram of pure water at 4°C is called 1 litre.
1L = 1000 cm³ and 1m³ 1000 L

The density of a substance is its mass per unit volume.

M The formula for density is d = \(\frac{m}{v}\) where d is V’ density, M is mass and V is the volume of a substance.

Units Of Density: In CGS system: g/cm³ and in Sl: kg/m³

Units For Measurement Of Very Small Distances Are:

1. Micron (u): 1 micron = 10^-6 metres. The size of microscopic objects is expressed in micron units.
2. Angstrom (Å): 1 angstrom 10^-10 metre. The wavelength of light and distance between atoms in a crystal are expressed in angstrom unit.
3. X-unit: 1 X-unit 10^-13 metre. The diameter of an atom is expressed in X-unit.
4. Fermi: 1 Fermi 10^-15 metre. The diameter of the nucleus of an atom is expressed in fermi.

Units For Measurement Of Very Large Distances Are:

1. Astronomical Unit (AU): The average distance between the sun and the earth is called 1 AU. 1AU 1.496 x 10¹¹ metre.

2. Light Year: The distance that is traversed by light in a vacuum in one year is called one light year. One light year = 9.46 x 10¹²km. The unit light year is used to express the distance between stars, the size of a galaxy etc.

3. Parsec: One parsec is the distance of an astronomical object from the sun that has a parallax angle of one arc second (1 degree = 60 minutes, 1 minute = 60 seconds).

1 parsec 3.26 light year = 3.08 x 10¹³ km. Parsec is the largest unit of length.

Some Very Small And Very Large Units For The Measurement Of Masses Are

1. Atomic mass unit (amu or u): 1 amu or u = 1 Da = 1.66054 x 10^-27kg.

The atomic mass unit is used for the measurement of masses of molecules or atoms.

2. Carat: 1 carat = 200 mg = 0.2 g. Carat is used as the unit of measurement for masses of gold, diamond etc.

3. Quintal: 1 quintal (q)= 100 kilogram.

4. Metric Ton: metric ton = 1000 kilogram.

5. Chandrasekhar Limit (CSL)-1 Chandrasekhar limit = 1.39 x mass of the sun = 2.765 x 10³⁰kg.

Measurement Of Small Area:

Barn (b) is used to measure a very small area. It is approximately equal to the area of the cross-section of the nucleus.

1 b = 10^-28m²

Chapter 1 Topic A Measurement And Units Short And Long Answer Type Questions

Question 1. What is a physical quantity? Give some examples.
Answer: Any natural event or phenomenon that can be measured directly or indirectly is called a physical quantity.

Example: Length, mass, temperature, work, acceleration, velocity, force, displacement etc. are physical quantities.

Question 2. Can all natural phenomena be termed as physical quantities? Explain with examples.
Answer: We observe different natural phenomena in our day-to-day life and feel them. But all of them cannot be quantified by measurement.

Examples: Anger, affection, charity etc.

“measurement questions “

Since all these natural phenomena cannot be measured, they are not called physical quantities.

Question 3. What is a scalar quantity? Give some examples of scalar quantities.
Answer: Those physical quantities which have only magnitude but no direction are called scalar quantities.

Example: Length, mass, work, temperature etc. are scalar quantities.

Question 4. What is a vector quantity? Give some examples of vector quantities.
Answer: Those physical quantities which have both magnitude and direction and whose addition follows the rules of vector addition are called vector quantities.

Example: Displacement, velocity, acceleration, force etc. are vector quantities.

Key Questions on Measurement and Units for Class 9

Question 5. Write the differences between scalar and vector quantities in a tabular form.
Answer: The differences between scalar and vector quantities:

Question 6. If any physical quantity has both magnitude and direction, can it be called a vector quantity? Or, Electric current has both magnitude and direction. Then why is it called a scalar quantity?
Answer: Any physical quantity having magnitude and direction cannot necessarily be called a vector quantity. For example, electric current.

Electric current has both magnitude and direction, but the addition of electric current does not follow the rules of vector addition. Hence, electric current is not a vector quantity but a scalar quantity.

NEET Biology Class 9 Question And Answers	WBBSE Class 9 History Notes	WBBSE Solutions for Class 9 Life Science and Environment
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WBBSE Class 9 Geography And Environment Multiple Choice Questions	WBBSE Class 9 History Short Answer Questions	WBBSE Solutions For Class 9 Maths
WBBSE Solutions for Class 9 History	WBBSE Class 9 History Very Short Answer Questions

 

Question 7. Mention which one is scalar and which one is vector among the following quantities: length, time, mass, weight, force, speed, velocity, acceleration, momentum, work, power, pressure, displacement, frequency, and density.
Answer:

Scalar Quantity:

Length, time, mass, speed, work, power, pressure, frequency, density.

Vector Quantity:

Weight, velocity, acceleration, force, momentum, displacement.

Question 8. What is unit?
Answer: While measuring any physical quantity, some convenient and definite quantity of it is taken as a standard and then the whole physical quantity is measured in terms of the standard. This standard is called one unit.

Question 9. What is the necessity of a unit?
Answer: In a scientific experiment, it is necessary to mention measurements accurately. So, a unit is essential while measuring any physical quantity. Any physical quantity is expressed in terms of a numerical number and its unit.

Measurement is not possible without a unit. Unit is also necessary for establishing relationships among different physical quantities and verifying the correctness of the equations involving physical quantities.

Question 10. What is the primary or fundamental or base unit? What are the primary units in SI?
Answer: Fundamental or primary unit is a set of units used in the measurement of physical quantities from which other units can be derived. Primary units or base units are independent of each other.

In SI, length, mass, time, temperature, electric current, luminous intensity and amount of substance are represented by metre, kilogram, second, kelvin, ampere, candela and mole, respectively.

Practice Questions for Chapter 1 Measurement

Question 11. Why are the units of length, mass and time called fundamental units?
Answer: Units of length, mass and time are called fundamental units because

1. Units of length, mass and time are independent of each other;
2. One cannot reduce them further into more simple units;
3. With the help of these three units, one can fundamental units. form units of other physical quantities.

Question 12 What is a derived unit? Explain with examples.
Answer: A unit of measurement that is formed by combining one or more fundamental units is called a derived unit.

Example: Units of velocity, acceleration, momentum, force, work etc. are derived units. displacement time

 

 

 

 

Therefore, a unit of velocity is made up of a unit of length and a unit of time. So, it is a derived unit.

Question 13. Distinguish the units of the following quantities into fundamental and derived units: area, volume, displacement, velocity, acceleration, force, work, energy, power, momentum, mass, weight, height, density, wavelength, and time period.
Answer:

1. Quantities With The Fundamental Unit: Displacement, mass, height, wavelength, and time period.

2. Quantities With The Derived Unit: Area, volume, velocity, acceleration, force, work, energy, power, momentum, weight, and density.

Question 14. Give an example of a derived unit formed by two fundamental units.
Answer: A unit of speed is a derived unit formed by two fundamental units.

Unit of speed =

 

 

∴ unit of speed is a derived unit formed by the two fundamental units of length and time.

Question 15. Give an example of a derived unit which is formed by three fundamental units.
Answer: A unit of force is a derived unit formed by three fundamental units

 

 

 

 

 

unit of measurement practice problems

∴ unit of force is a derived unit formed by three fundamental units of length, mass and time.

Important Concepts in Measurement for Class 9

Question 16. What is a system of units? What are the advantages of a system of units?
Answer: A system of units is a set of all the fundamental and derived units that constitute the units of all physical quantities.

The introduction of a system of units has not only helped the scientific community but also removed the difficulties of maintenance of accounts in our daily lives.

Saying that a table has a length approximately equal to four cubits does not signify its correct length. But if it is said that its length is 1.5 m, we get a correct idea about its length.

Question 17. Why CGS system or SI is called a metric system?
Answer: CGS system or SI is called a metric system because to convert any physical quantity from any definite unit to another smaller or larger unit in these systems, one has to only shift the decimal point.

Example: 500 cm = 5m = 5000 mm = 0.005 km. No multiplication or division is necessary for this conversion.

Question 18. What are the advantages of the metric system? 
Answer:

The Advantages Of The Metric System Are:

1. The metric system is based on powers of 10. So, it is easier to convert units of any physical quantity simply by moving the decimal points.
2. Once the meaning of the prefixes is remembered, one can easily convert mass, distance, and volume measurements. No further conversion factors are to be memorised except the power of 10.
3. There is a convenient relationship between mass and volume in this system. For example, the mass of 1 cm3 of water is 1 g or the mass of 1 L of water is 1 kg (at 4°C).

Question 19. What is a unitless quantity? Give example.
Answer: If a physical quantity is the ratio of two physical quantities with the same unit, then that physical quantity has no unit. This type of physical quantity is called unitless quantity.

Example: Atomic weight, specific gravity, molecular weight, strain etc. are unitless quantities.

 

 

 

 

 

Hence, atomic weight and specific gravity are both the ratios of two quantities of the same nature (here, mass). Hence, these physical quantities are unitless.

Question 20. Define the unit of length in Sl.
Answer; The Unit of length in SI is metre (m). One metre (m) is the distance which is traversed by light in \(\frac{1}{299792458}\) vacuum in a second.

Question 21. Define the unit of time in SI.
Answer: The unit of time in SI is second (s). One second is the duration of 9192631770 periods of radiation corresponding to the transition between the two hyperfine levels of the ground state of the ¹³³Cs atom.

Question 22. Names of some ‘lengths’ and some ‘units’ are given below. You have to write which unit is used generally for the measurement of a particular length. Lengths: Wavelength of light, the diameter of an atomic nucleus, size of a galaxy, the diameter of an atom, the average distance of the earth from the sun, size of a microscopic object. Units: Astronomical unit (AU), angstrom (A), light year, fermi, X-unit, micron.
Answer: The given lengths and their units are written in the following tabular form.

Question 23. What do you mean by 1 solar mass?
Answer: Solar mass is a large unit of mass used in astronomy. Masses of galaxies, nebulae and stars are expressed with the help of this unit.

 

 

Question 24. What is the difficulty of expressing the mass of an elephant (6000 kg approximately) in terms of atomic mass units?
Answer: Atomic mass unit is generally used to measure the masses of very small particles like molecules and atoms because 1 atomic mass unit (amu or u) = 1.66054 x 10^-27kg.

In this case, the mass of the elephant is 6000 kg. If this is expressed in u, its value becomes

 

 

 

=  6/166054 x 10 ²⁵
 = 0.000036133 x 10²⁵
= 3.5 x 10²⁰ u (approx)

As this figure is a very large amount, it is impractical to work with it. For this reason, the mass of an elephant is not expressed in terms of u.

Question 25. What do you mean by the volume of a body? Is there any difference between this volume and the volume of the material of the body?
Answer: The volume of a body is the amount of space occupied by it.

The volume of the material of the body

If a solid body is made up of a single material, then its volume is the same as the volume of its material. But the volume of a hollow body made up of only one material is greater than the volume of its material.

Question 26. What are the units of volume in the CGS system and SI? Write the relationship between these two units. 
Answer: The unit of volume in CGS system and Sl are cm³ and m³, respectively.

The relationship between them is 1m³ (100 cm)³ = 106 cm³.

Understanding SI Units for Solutions

Question 27. Define litre. 1 litre = how many cm3 ?
Answer: One litre (1 L) is defined as the volume of 1 kilogram of pure water at 4°C or 277 K.

1L = 1 dm³ = (10 cm)³ = 1000 cm³.

Question 28. Why is 4°C or 277K mentioned in the definition of a litre? Establish a mathematical relationship between the density of a body and its mass and volume.
Answer: The temperature of 4°C or 277 K is mentioned in the definition of litre because in general, the volume of a liquid increases and its density decreases with the rise of temperature. But in case of water, there is an exception to this rule for 4°C.

If the temperature between 0°C and the temperature of water is increased from 0°C, its volume decreases and density increases. This process goes on up to a temperature of 4°C. At 4°C, the density of water is 1g/cm³ and is maximum.

After 4°C, if the temperature is increased, the volume of water increases and its density decreases. So at 4°C, volume of 1000 g or 1 kg of water is 1000 cm³ or 1 L.

Let the mass and volume of any body be m and V, respectively.

∴  mass of unit volume = \(\frac{m}{v}\)

∴ density of the body, d = \(\frac{m}{v}\)

This is the required relationship.

Question 29. What is density? What are the units of density in CGS system and SI? Establish a relationship between these two units.
Answer:

The density of a material is defined as its mass per unit volume.

Units of density in CGS system and Si are g/cm³ and kg/m³, respectively.

The relationship between them is

 

 

or, 1 g/cm³ = 1000 kg/m³

Question 30. How does the density of a material change with the increase of temperature?
Answer: The density of a material decreases with an increase of temperature. However, the density of water increases when the temperature is increased from 0°C to 4°C.

Question 31. What is the mean solar day? Write its value in seconds.
Answer: The average of solar days for one year is a mean solar day.

Mean Solar Day

 

 

Value of 1 mean solar day = 86400 seconds.

Chapter 1 Topic A Measurement And Units Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following is the largest unit for measurement of length?

1. Light Year
2. Kilometre
3. Parsec
4. Metre

Answer: 3. Parsec

Question 2. Which of the following is not a physical quantity?

1. Velocity
2. Mass
3. Displacement
4. Anger

Answer: 4. Anger

Question 3. Which of the following is a physical quantity?

1. Anger
2. Affection
3. Speed
4. Frustration

Answer: 3. Speed

Question 4. Unit of which of the following quantities is different from the others?

1. Pressure
2. Stress
3. Coefficient Of Elasticity
4. Force

Answer: 4. Force

Question 5. Which of the following is a scalar quantity?

1. Momentum
2. Work
3. Weight
4. Force

Answer: 2. Work

Question 6. Unit of which of the following quantities is made up of three fundamental units?

1. Velocity
2. Acceleration
3. Speed
4. Momentum

Answer: 2. Acceleration

Question 7. Unit of which of the following quantities is made up of four fundamental units?

1. Force
2. Specific Heat
3. Power
4. Heat Capacity

Answer: 4. Heat Capacity

Question 8. The wavelength of a light ray is 6000Å. What is the value of this wavelength in ‘m’ unit?

1. 6 x 10^-7
2. 6 x 10^-6
3. 6 x 10^-8
4. 6 x 10^-10

Answer: 1. 6 x 10^-7

Question 9. The distance from a point at which a length of 1 AU forms an angle of one second at that point is

1. 4.26 light year
2. 3.26 light year
3. 2.26 light year
4. 5.26 light year

Answer: 2. 3.26 light year

Question 10. The length, breadth and height of a water tank are 3m, 2m and 1 m, respectively and the tank is half-filled with water. What is the volume of water contained in the tank?

1. 6000 L
2. 3000 L
3. 30000 L
4. 60000 L

Answer: 3. 30000 L

Question 11. What is the temperature at which the density of water is maximum?

1. 0°C
2. 4°C
3. 8°C
4. 10°C

Answer: 2. 4°C

Question 12. The density of iron is 7.8 g/cm3. What is the mass of 100 cm3 of iron?

1. 0.78 kg
2. 7.8 kg
3. 0.078 kg
4. 78 kg

Answer: 1. 0.78 kg

Question 13. Select one scalar quantity and one vector quantity from the following whose units are derived from three fundamental units

1. Speed, velocity
2. Work, acceleration
3. Work, force
4. Power, acceleration

Answer: 3. Work, force

Question 14. Steradian is the unit of

1. Angle
2. Solid angle
3. Arc of a circle
4. Circumference

Answer: 2. Solid angle

Question 15. Which of the following quantities is not a physical quantity?

1. Length
2. Water
3. Force
4. Time

Answer: 2. Water

Question 16. Which of the following units is a derived unit?

1. Mole
2. Cubic metre
3. Ampere
4. Light year

Answer: 2. Cubic metre

Question 17. Unit of volume is

1. Mole
2. Kilogram
3. Litre
4. Square Metre

Answer: 3. Litre

Question 18. The temperature at which the volume of 1 kg of water is considered to be 1L is

1. 0°C
2. 10°C
3. 4°C
4. 100°C

Answer: 3. 4°C

Concepts Related to Measuring Instruments for Class 9 Solutions

Question 19. The unit which is convenient for the measurement of the diameter of molecules and atoms is

1. Micron
2. Metre
3. Parsec
4. Light Year

Answer: 1. Micron

Question 20. One nanosecond is equal to how many seconds?

1. 10^-3
2. 10^-5
3. 10^-9
4. 10^-12

Answer: 3. 10-9

Question 21. One part of how many parts of an average solar day is considered to be one solar second?

1. 84600
2. 86400
3. 8640
4. 365

Answer: 2. 86400

Question 22. Which of the following is a larger unit used for the measurement of mass?

1. Kilogram
2. Quintal
3. Carat
4. Gram

Answer: 2. Quintal

Question 23. 1 nm = how many metres?

1. 100
2. 1000
3. 10
4. 10-9

Answer: 4. 10-9

Question 24. Unit of temperature in SI

1. Degree Celsius
2. Candela
3. Kelvin
4. Mol

Answer: 3. Kelvin

Question 25. Mass of 5 cm3 of water at 4°C is

1. 5g
2. 5 kg
3. 0.5 g
4. 50 kg

Answer: 1. 5g

Question 26. The unit which is used to measure the distance between stars is

1. Angstrom
2. AU
3. X-unit
4. Fermi

Answer: 2. AU

Question 27. What is the relationship between the mass m, the density d and the volume V of an object?

1. mdV = 1
2. md = V
3. \(\frac{m}{v}\) = d
4. \(\frac{v}{m}\) = d

Answer: 3. \(\frac{m}{v}\) = d

Question 28. The cross-sectional area of the nucleus of an atom is x barn. Its value in m2 units is

1. X x 10^-24
2. X x 10^-28
3. X x 10^-26
4. X x 10^-30

Answer: 2. X x 10^-28

Question 29. One ring is made up of 10-carat gold. What is the mass of gold in g unit?

1. 4
2. 2
3. 20
4. 0.2

Answer: 2. 2

Question 30. The length of a bacteria is 3 microns. The length of the bacteria in m unit is

1. 3 x 10^-13
2. 3 x 10^-6
3. 3 x 10^-5
4. 3 x 10^-7

Answer: 2. 3 x 10^-6

Question 31. The wavelength of the yellow light of Na has a wavelength of 592 nm. In Å unit the wavelength is

1. 59.2 Å
2. 5920 Å
3. 59200 Å
4. 5.92 Å

Answer: 2. 5920 Å

Chapter 1 Topic A Measurement And Units Answer In Brief

Question 1. What are the types of physical quantities?
Answer:

Physical Quantities Are Of Two Types:

1. scalar and
2. vector.

Question 2. State whether the barn is a fundamental unit or a derived unit.
Answer: Barn is a derived unit.

Question 3. What is a barn?
Answer: 1 barn is equal to 10^-28 m² and this unit is used to measure nuclear cross-section.

Question 4. What type of physical quantity is electric current?
Answer: Electric current is a scalar quantity.

Question 5. What type of physical quantity is area?
Answer: Area is a vector quantity.

Question 6. Give an example of the unit of a scalar quantity which is formed by two different fundamental units.
Answer: A unit of speed is a scalar quantity which is formed by two different fundamental units.

Question 7. Give an example of the unit of a vector quantity which is formed by two different fundamental units.
Answer: A unit of velocity is a vector quantity which is formed by two different fundamental units.

Question 8. What are the two present systems of the unit?
Answer: Two present systems of the unit are:

1. CGS system and
2.  SI.

Question 9. Give an example of one scalar quantity and one vector quantity with the same unit.
Answer: Speed and velocity are the two physical quantities with the same unit, speed being a scalar quantity and velocity being a vector quantity.

Question 10. What are the fundamental units in CGS system?
Answer: In CGS system, unit of length is a centimetre (cm), unit of mass is a gram (g) and a unit of time is a second (s).

Question 11. In SI, how many basic units are there?
Answer: There are seven basic units in Sl.

Question 12. How much distance is meant by 1 AU?
Answer: 1 AU (astronomical unit) is the average distance between the sun and the Earth.

Question 13. Is a light year a fundamental or a derived unit?
Answer: Light Year is a fundamental unit.

Question 14. Which unit is used for measuring the mass of molecules and atoms?
Answer: The unit u (unified atomic mass unit) is used for measuring the mass of molecules and atoms.

Question 15. Which unit is used for measuring the mass of different precious stones?
Answer: Carat is used for measuring the mass of different precious stones. 1 carat = 0.200 g.

Question 16. What is the relationship between the Chandrasekhar limit (CSL) and the mass of the sun?
Answer: 1 Chandrasekhar limit = 1.39 x mass of the sun.

Question 17. Which temperature is mentioned in the definition of a litre?
Answer: 4°C or 277 K is mentioned in the definition of a litre.

Question 18. What is the temperature at which the density of water is maximum?
Answer: At 4°C or 277 K, the density of water is maximum.

Question 19. What is the density of water at 4°C?
Answer: The density of water at 4°C is 1 g/cm³.

Question 20. Which element is mentioned in the definition of second?
Answer: ¹³³Cs is mentioned in the definition of second.

Question 21. Which unit is used to measure interstellar distance?
Answer: AU (astronomical unit) is used to measure interstellar distance.

Question 22. Is it possible to measure 0.4 mm by an ordinary scale?
Answer: With the help of an ordinary scale, a measurement of 0.4 mm is not possible since the smallest constant in an ordinary scale is 1mm.

Question 23. 1 gallon how many litres?
Answer: 1 gallon 4.536 litres.

Question 24. What is the largest unit for measurement of length?
Answer: Parsec is the largest unit for measurement of length.

Chapter 1 Topic A Measurement And Units Fill In The Blanks

Question 1. Specific gravity is the ratio between two quantities having the same unit. So it has ______ unit.
Answer: No

Question 2. At 4°C temperature, a volume of 1 ______of pure water is called a litre.
Answer: Kilogram, 1

Question 3. The unit of the area is a ______ unit.
Answer: Derived

Question 4. The mass of unit ______ of any material is known as the ______ of that material.
Answer: Volume, density

Question 5. The number of fundamental units in SI is ______.
Answer: 7

Question 6. The number of fundamental units in CGS system is ______.
Answer: 3

Question 7. ______ quantities have only magnitude but no direction.
Answer: Scalar

Question 8. ______ quantities have both magnitude and direction.
Answer: Vector

Question 9. ______ K temperature is mentioned in the definition of litre.
Answer: 277

Question 10. ______ of water is minimum at 277 K.
Answer: Volume

Question 11. ______ is the unit of density in CGS system.
Answer: g/cm³

Question 12. Parsec is a unit for measurement of ______.
Answer: Length

Question 13. Light Year is a ______ unit.
Answer: Fundamental

Question 14. ______ is the unit used for the measurement of diameters of atoms and molecules.
Answer: x-unit

Question 15. Micron is used as a unit to express dimension of ______ objects.
Answer: Microscopic

Question 16. It is convenient to express the mass of a star in terms of ______ which is a large unit of mass.
Answer: Solar mass

Question 17. 1 AU is the average distance between the ______ and the Earth.
Answer: Sun

Question 18. 1 light year is equal to ______ kilometre.
Answer: 9.46 X 10¹²

Question 19. ¹³³Cs is mentioned in the definition of ______.
Answer: Second

Chapter 1 Topic A Measurement And Units State Whether True Or False

Question 1. Physical quantities are those natural events or phenomena which can be measured directly.
Answer: False

Question 2. Atomic mass is a unitless physical quantity.
Answer: True

Question 3. Acceleration is a vector quantity.
Answer: True

Question 4. According to CGPM, one metre is that distance which is traversed by light in air in \(\frac{1}{299702458}\)s.
Answer: False

Question 5. The unit of force is a fundamental unit.
Answer: False

Question 6. The unit of momentum is a derived unit.
Answer: True

Question 7. The density of water at 4°C is minimum.
Answer: False

Question 8. Pressure is a scalar quantity.
Answer: True

Question 9. The thickness of a plastic carry bag is measured in micron units.
Answer: True

Question 10. The fundamental units depend on each other.
Answer: False

Chapter 1 Topic A Measurement And Units Numerical Examples

Useful Formula

If V be the volume of a body of mass m, the density of the body will be

 

 

If V be the volume of a sphere of radius r, then V = 4/3πr³

If two liquids of density D₁ and D₂ are mixed in the same proportion of mass, then the density of the mixture,

 

 

 

 

Question 1. if the density of mercury is 13.6g/cm3 In CGS, what is the density of mercury in SI?

Answer:

The density of mercury in CGS, d= 13.6g/cm³

Again, 1g/cm³ 100kg/m³

∴ The density of mercury in SI

= 13.6 x 100kg/m³ = 13600 kg/m³

unit of measurement practice problems

Question 2. If the density of iron in St is 7800 kg/m3, what is its density in CGS?

Answer:

The density of iron in Sl, d= 7800 kg/m³

Again, 1 kg/m³ = \(\frac{1}{1000}\) g/cm³

∴ The density of iron in CGS

= 7800 ×\(\frac{1}{1000}\) g/cm³ = 7.8 g/cm³ 1000

Question 3. The length, breadth and height of a water tank are 1.2 m, 1 m and 0.8 m, respectively. What is the volume of water contained in the tank?

Answer:

Volume of water in the tank

= 1.2 m x 1m x 0.8 m

= 0.96 m³ = 0.96 x 1000 L = 960 L

Question 4. If the density of a liquid is 0.8 g/cm3, what is the mass of 200 cm3 of the liquid?

Answer:

Density of liquid, d=0.8 g/cm³

= Volume, V = 200 cm³

∴ If m is the mass of the liquid,  d=m/v

or, m = d x V = 0.8 g/cm³ x 200 cm³ = 160 g

Study Guide for Class 9 Measurement Questions

Question 5. The diameter of a solid sphere of iron is 21 cm. If the density of iron is 7.8 g/cm3, what is its mass?

Answer:

According to the question, the diameter of the sphere = 21 cm

∴ The radius of the sphere, r = 21/2 cm

Density of iron, d = 7.8 g/cm³

∴ The volume of the sphere,

 

 

∴ Mass of the iron sphere,

m = dxV = 7.8 g/cm³ x 4851 cm³

= 37837.8 g = 37.84 kg (approx.)

Question 6. The densities of two liquids are 1.2 g/cm3 and 1.5 g/cm3, respectively. If the liquids are mixed in the same proportion of volume, what is the density of this mixture?

Answer:

Let the mixture be prepared by taking V cm3 of each liquid.

Mass Of The First Liquid

= Density of first liquid x volume of the liquid

= 1.2 V g

Mass Of The Second Liquid

= Density of second liquid x volume of the liquid

= 1.5Vg

∴ Mass of the mixture (m) = (1.2 V+1.5 V)g = 2.7 Vg;

Volume of the mixture (V1) =(V+V)cm³ = 2V cm³

unit of measurement practice problems

∴ Density of the mixture,

 

 

Question 7. Densities of two liquids are 1.2 g/cm3 and 0.8 g/cm3, respectively. If the liquids are mixed in the same proportion of mass, what is the density of this mixture?

Answer:

Suppose the mixture is prepared by taking mg of both the liquids.

∴ Volume of the first liquid = m/1.2 cm³

Volume of the second liquid= m/0.8 cm³

 

 

 

 

 

 

 

Question 8. it is known that light travels in a vacuum at a fixed velocity of 9.46 x 1012km/year (approx.). The distance of a star alpha cen- tauri from the Earth is nearly 4.24 light years. When did the light emanate from the star which we see today? What is the distance of the earth from that star in a kilometre?

Answer:

As per the definition of a light year, light emanated 4.24 years back from alpha centauri.

Distance of the Earth from this star

= 4.24 light year

= 4.24 years x velocity of light in vacuum

= 4.24 x 9.46 x 1012 km

= 40.1104 x 1012 km (approx.)

Sample Solutions from WBBSE Class 9 Physical Science Chapter 1

Question 9. The largest diameter of the Milky Way galaxy is nearly 9.5 x 1015 m. The average diameter of the earth is 12800 km. The average minimum diameter of all the viruses that have been discovered so far is 20 nanometres. Then, which one is smaller- the Milky Way galaxy as compared to the Earth. or the earth as compared to a virus?

Answer:

Comparison Of The Diameters Of The Milky Way Galaxy And The Earth:

 

 

 

 

 

Comparison Of The Diameters Of The Earth And A Virus:

 

 

 

 

unit of measurement practice problems

Clearly, a virus is much smaller as compared to the earth than the Earth as compared to the Milky Way galaxy.

Question 10. Light takes 8 min 20 s to come from the sun to the earth. Find the distance of the earth from the sun in light year unit.

Answer:

1 light year 9.46 x 1012 km

speed of light in vacuum

(c) = 3 x 10⁸ m/s = 3 x 10⁵ km/s

time (t) = 8 min 20 s = (8 x 60+20) s = 500 s .

The distance of the Earth from the sun is

d = cxt=3x 10³ x 500 km

 

 

=1.586 x 10^-5 lightyear