WBBSE Solutions For Class 9 Physical Science Chapter 3 Matter Structure And Properties Topic A Pressure Of Liquids And Air Archimedes Principle

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.

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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).

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

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

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.

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

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

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).

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.

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 = \(\frac{\text { mass }}{\text { volume }}\)

= \(\frac{200}{25}=8 \mathrm{~g} / \mathrm{cm}^3\)