NEET Foundation

Physics Chapter 5 Sound

Sound is something we hear all the time around us, be it honking of cars on a busy road or chirping of birds in a serene landscape. It plays a major role in our lives where we receive and transmit sound for various purposes in our day-to-day life. Sound is a form of energy which propagates from one place to another through a medium. It is produced by bodies which vibrate. Consider a tuning fork ‘F’ which is excited by hitting on a rubber hammer.

When such a tuning fork is kept near our ears, we hear the sound but are unable to detect the vibrations of the tuning fork. When we speak, sound is produced by the vibration of vocal chords present in a cavity called larynx, in our throat. Sound is transmitted in the form of mechanical waves.

Thus, sound needs a medium to travel, since mechanical waves can propagate only through material medium.

In this chapter, we will learn more about sound and its propagation, its uses and its various forms in detail.

Chapter 5 Sound

Sound is a form of energy like heat energy, light energy, potential energy and kinetic energy. It causes a sensation of hearing in our ears. Sound cannot be created nor destroyed but can be changed from one form to another. For example, when we clap, a sound is produced.

Here, muscular energy is converted into sound energy. This is in accordance with the law of conservation of energy.

Read and Learn More: NEET Foundation Notes

Sound of various varieties is heard around us. We cannot see the sound but it travels in the form of waves and reaches our ears. These waves are formed due to vibrations of particles of the medium. Waves carry energy with them in the form of mechanical energy and produce a sensation of hearing in our ear.

Similarly, in an electric bell, when connected to electricity starts producing sound. Here, again electrical energy is converted into sound energy. Examples of sound discussed below.

1. Horn of a car
2. Alarm clock
3. Music
4. Human voice
5. Barking of dogs
6. Hand clapping

Sound can be described in two different ways.

1. Subjective or psychological sound, which refers to hearing sensation which ceases when the sound sensing organ is withdrawn from the scene.
2. Objective or physical meaning refers to the energy reaching the ear from outside. The energy continues to propagate even if no ear is present to detect it.

Chapter 5 Sound Production of Sound

Sound is produced when an object vibrates. Now, what is vibration? Vibration is the rapid to and fro motion of an object. The motion of materials or objects causes vibration. To understand it better let us illustrate some examples.

Example 1: Take a guitar and pluck its string at the centre . Now pluck it at the centre. It will start vibrating and you can hear a sound. After few minutes, string will stop vibrating and sound will also stop as well.

Example 2: Take a U shaped fork. Hang a ball with a thread. Now strike the ball and bring the vibrating tuning fork near the ball. You will hear the sound of the vibrating tuning fork. As soon as the arm of the fork stops vibrating, the ball will also stop oscillating and there will be no sound.

These examples show that sound is produced when an object vibrates and it stops when the object stops vibrating.

As we said abovem sound is a form of energy. Let us elaborate it further. Mechanical energy is required to start vibrations in an object producing sound. The vibrations of object are transmitted in a medium in waveforms from one point to the next and so on.

These waves on reaching our ears produced vibrations in the eardrum which are perceived as sound by us. Thus, sound is a form of energy.

Chapter 5 Sound Master Your Test Question And Answers

Question 1. Define sound.
Answer:

Sound:

Sound is a form of energy which propagates from one place to another through a medium.

Question 2. How is sound produced?
Answer:

Sound is produced when an object vibrates. Vibration is the rapid to and fro motion of an object.

Question 3. Sound is a form of energy. Explain.
Answer:

Sound is a form of energy:

Mechanical energy is required to start vibrations in an object producing sound. The vibrations of object are transmitted in a medium in waveforms from one point to the next and so on. These waves on reaching our ears produce vibrations in the eardrum which are perceived as sound by us. Thus, sound is a form of energy.

Chapter 5 Sound Track Your Learning Question And Answers

Question 1. _______________ is required to start vibrations in an object producing sound.
Answer. Mechanical energy

Question 2. Sound is a form of _________________.
Answer. Energy

Question 3. The vibrating part of a guitar is

1. Hand
2. Fulcrum
3. String
4. Chords

Answer. 3. String

Question 4. Sound cannot be produced in a body if it’s not vibrating. (True/False)
Answer. True

Chapter 5 Sound Propagation of Sound

The vibrating object produces sound thattravels through a medium to reach the ­listener. The medium through which sound travels can either be solid, liquid or a gas. When an object vibrates, it creates a periodic disturbance in the nearby medium.

The particles around the medium also start vibrating. As a result of it, the particles in contact with the vibrating object is first displaced from its equilibrium position. It then exerts a force on the adjacent particle and the adjacent particle is displaced from its position of rest.

After displacing the adjacent particle the first particle comes back to its original ­position. This process gets repeated in the medium till the sound reaches the listener.

Therefore, the disturbance moves forward in the form of compression. The particles of the medium do not move with the compression.

Sound Needs Medium to Travel

Amedium is necessary forthe propagation of sound from one place to another. The matter or substance through which sound is transmitted is called a medium. The medium can be solid, liquid, or gas. Sound cannot travel in vaccum. A true vaccum refers to the complete absence of matter.

Sound waves can travel only through matter. So, sound needs a physical medium in order to travel anywhere. A vibrating object travels from one place to another through the mechanical vibrations of medium particles in the form of waves.

Activity to Show that Sound Needs a Material Medium for its Propagation

Take an electric bell and suspend it inside an air tight glass bell jar. Connect the bell jar to a vacuum pump. When the key is pressed, the circuit of electric belt is complete. The hammer of the electric bell begins to strike the gong repeatedly due to which sound is heard. Now keeping the key pressed, air is gradually with drawn from the jar by starting the vacuum pump.

It is noticed that the loudness of sound goes on decreasing as the air is removed from the bell jar and finally no sound is heard when all the air from the jar has been drawn out. The hammer of the electric bell is still seen striking the gong repeatedly which means that sound is still produced, but it is not heard.

Illustration

When hammer of the bell hits the gong, sound is produced due to vibrations of gong which travels through air to the wall of jar. This causes the wall of jar to vibrate and the air outside the jar is also set in vibration. Therefore, we can hear the sound.

But when hair has been removed from the jar, sound produced due to vibration of gong could not travel to the wall of jar, so wall could not vibrate, hence no sound is heard. This clearly demonstrates that sound requires a material medium for transmission and it cannot travel through vacuum.

Requisites of the Medium

The medium required for propagation of sound must have the following properties:

1. The medium must be elastic so that its particles may come back to their initial positions after displacement on either side.
2. The medium must have inertia so that its particles may store mechanical energy.
3. The medium should be frictionless so that there is no loss of energy in propagation of sound through it.

We know sound can propagate in all states, i.e., solid, liquid and gas. Some materials like water, air, etc., can easily transmit sound from one place to another while on the other hand materials like blanket, thick curtains, etc., transmit only a small fraction of it because it absorbs most of the sound incident on them.

Sound Propagates as Waves

Sound propagates from one place to another in the form of waves, i.e. because of the disturbances of particle of the medium. Wave is a phenomenon or disturbance in which energy is transferred from one point to another without any direct contact between the points. So, sound is considered as a wave. Particle of medium only vibrate. They do not move from one point to another.

On the basis of direction of propagation, waves can be divided into two types – (a) longitudinal waves and (b) transverse waves

Longitudinal Waves

A wave in which particles of the medium vibrate about their mean positions, in the direction of propagation of the wave is called longitudinal wave (Fig. 5.5). Sound travels in air in the form of longitudinal waves. Longitudinal waves can be produced in all forms, i.e., solid, liquid and gas. At compressions, the density and pressure of the medium are maximum, while at rarefaction the density and pressure of the medium are minimum.

Compression and Rarefaction

Air is the most common medium through which sound travels. When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called compression. It is represented by the symbol C.

When the vibrating object moves backwards, it creates a region of low pressure called rarefaction. It is represented by the symbol R.

In the we see that waves are formed when a stone is dropped on the still water surface. Along with waves we also hear sound of stone as the stone strikes the water surface. This is due to the disturbance produced in water. The disturbance spreads in all directions rapidly outwards in the form of circular waves on the surface of water.

Now keep a small paper boat on the water surface at some distance away from the point where the stone strikes, we notice that the ball doesn’t move ahead, instead it will vibrate in the up and down direction as the wave moves ahead. The reason behind this is that the particles of water start vibrating up and down at the point where the stone strikes.

These particles then transfer their energy to other neighbouring particles and they themselves come back to their mean positions. This process continues and thus the disturbance moves ahead on the water surface in the form of waves. The waves die out as soon as the energy imparted by the stone gets dissipated.

Transverse Wave

A wave in which the particles of the medium vibrate about their mean positions, in a direction perpendicular to the direction of propagation of the wave is called a transverse wave.

Transverse wave is composed of:

1. Crest: The position of maximum upward displacement is called crest.
2. Trough: The position of maximum downward displacement is called trough.

Comparison between transverse and longitudinal waves

Progressive and Stationary Waves

There are some waves that start at the point of origin of the waves and progress endlessly into other parts of the medium. Such waves are known as progressive waves.

Consider a progressive transverse water wave moving from left (point P) to right and striking a hard surface at ‘Q’and is called an ‘incident wave’. It then gets reflected at ‘Q’, and travels towards ‘P’. Thus the two waves, one going from ‘P’ to ‘Q’ and the other going from ‘Q’ to ‘P’ called the ‘reflected wave’ overlap resulting in the formation of ‘nodes’ and ‘antinodes’.

Points, where the displacement of a vibrating particle of the medium is zero or minimum are called ‘nodes’ and points, where the displacement of the vibrating particles is maximum are called ‘antinodes’. The closed ­figures so formed are called ‘Loops’.

These loops are On the whole, the wave appears to be stationary or standing, contained between two positions ‘P’ and ‘Q’ and so called as ‘stationary waves’ or ‘standing’. Thus a ‘progressive wave is a wave which is generated at a point in a medium and travels to all parts of the medium infinitely carrying the energy’ and a ‘stationary wave is a wave which is formed by a superposition of two identical progressive waves traveling in opposite directions’.

Comparative study of progressive and stationary waves

Characteristics of Wave Motion

1. A wave is produced by the periodic disturbance at a point in the medium.
2. Due to propagation of a wave in a medium, the particles of the medium vibrate about their mean positions and energy is transferred with a constant speed from one place of medium to the other.

Characteristics of Sound Wave

Sound can be described in terms of the following physical quantities:

1. Amplitude: The maximum displacement of the particle of a medium on either side of its mean position is called amplitude of the wave. It is denoted by the symbol ‘A’. S.I. Its unit is metre (m).

2. Time period: The time taken by a particle of the medium to complete its one ­vibration is called time period of the wave. It is denoted by the letter T. Its S.I. unit is second (s).

3. Frequency: The number of vibrations made by a particle of the medium in one second is called frequency of wave.

Another definition is number of waves passing through a point in one second.

It is denoted by the symbol n, f, neu (υ)

Its S.I. unit is hertz (Hz) or s^-1.

Relation:

υ = 1/T

where υ = frequency, T = time period.

The frequency of a wave is equal to the frequency of vibrations of its source. It is the characteristic of its source which produces the disturbance. It does not depend on the amplitude of vibration or on the nature of medium in which the wave propagates.

4. Wavelength: The distance travelled by the wave in the time period of vibration of particles in the medium is called its wavelength. It is denoted by the letter λ (lambda). Its S.I. unit is metre (m). It depends on the medium in which the wave travels.

In a longitudinal wave, the distance between two consecutive compressions or two consecutive rarefactions is equal to one wavelength.

In a transverse wave, the distance between two consecutive crests or between two consecutive troughs is equal to one wavelength.

5. Wave velocity: The distance travelled by a wave in one second is called its wave velocity. It is the velocity with which energy is transferred from one place to another by wave motion. It is not the velocity of an individual particle vibrating about its mean position.

Wave velocity is constant for a given medium. It depends on the elasticity and the density of the medium. It changes when same wave passes from one medium to the other medium. It is denoted by the letter v. Its S.I. unit is m/s.

Displacement–Time Graph

The above graph shows the variation of displacement of a certain particle of the medium with time, when a wave propagates through it. It is called displacement–time graph.

Displacement–Distance Graph

The above figure shows the displacement–distance graph of a transverse wave at an instant. It is a snapshot of a wave. Distance AE or CG gives us the wavelength (λ) of the wave.

Relationship between Wavelength, Wave Velocity and Frequency

Let velocity of wave = v

Time period = T

Frequency = υ

Wavelength = λ

As per the definition of wavelength,

λ = Distance travelled by wave in one time period

= Wave velocity × Time period

= v × T

vT = λ

T = 1/υ

v × (1/υ) = λ

Therefore,

v = λυ

Hence,

Wave Velocity = Frequency × Wavelength.

Speed of Sound in Different Media

Sound travels in a medium with a finite speed and it takes some time to reach a destination from the source.

The speed of sound in a medium depends on the following factors:

1. Elasticity of (E) of the medium
2. Density (r) of the medium

The speed of sound in a medium is given by the formula

v = \(\sqrt{\frac{E}{\rho}}\)

where E is the modules of elasticity and ρ is the density of medium.

Sir Isaac Newton assumed, when sound travels in a gas, temperature of the gas does not change. In other words, the propagation of sound is an isothermal change. For isothermal change, modulus of elasticity is equal to the pressure of the gas, i.e. E = P.

Therefore, velocity of sound in gas was given by

v = \(\sqrt{\frac{P}{\rho}}\)

The expression given by Newton for the velocity of sound in a gas medium, later was modified by Laplace, where he introduced ‘γ ’, as a constant for a given gas. It is defined as the ratio of the specific heat capacity of the gas at constant pressure to its specific heat capacity at constant volume.

Therefore, v = \(\sqrt{\left(\frac{\gamma P}{d}\right)}\), where, ‘γ ’(gamma) is a constant.

The speed of sound is different in different media. It is more in solids, less in liquid and least in gas. This is because solids are much more elastic than the liquids and gases. The speed of sound is nearly 5100 m/s in steel, 1450 m/s in water and 330 m/s in air at 0 °C. Table 5.3 below gives the speed of sound in different media at 0 °C.

Speed of sound in different media

Example: If sound is produced at one end of a very long steel bar, two sounds are heard at the other end. One which reaches first is propagated through steel and the other which is heard later is through air.

Factors Affecting the Velocity of Sound in a Air

The speed of sound in a gas is affected by

1. Effect of density: As we know v = \(\sqrt{\frac{\gamma P}{\rho}}\), it shows that \(v \propto \frac{1}{\sqrt{\rho}}\)

The speed of sound is inversely proportional to the square root of density of the gas.

The density of oxygen is 16 times the density of hydrogen; therefore the speed of sound in hydrogen is four times the speed of sound in oxygen other factor being same.

2. Effect of temperature: The speed of sound in a gas increases with the increase in the temperature of the gas. This is because with the increase of temperature, density of gas decreases and consequently the speed of sound increases.

Speed of sound is directly proportional to the square root of temperature of medium.

\(V \propto \sqrt{T}\) , where T is the temperature of the gas on the Kelvin scale.

The speed of sound in air increases by about 0.61 m/s for each degree celsius rise in its temperature.

V_t = V₀ + 0.61 t

where ‘t’ is temperature of air in °C.

3. Effect of humidity: The speed of sound in air increases with the increase in humidity in air.

The density of water vapour is about 5/8th times the density of dry air at ordinary temperatures, therefore the increase of moisture in air tends to decrease the density of air. This is the reason why the speed of sound in the humid air is greater than the speed of sound in the dry air.

4. Effect of direction of wind: The speed of sound increases or decreases according to the direction of wind. If wind is blowing in the direction of propagation of sound, the speed of sound increases but if it is flowing in the direction opposite to that of sound, then speed of sound will decrease.

If v is the speed of sound in still air and W is the speed of wind, the speed of sound becomes v + W when wind blows in the direction in which sound travels and the speed of sound becomes v – W when wind blows in direction opposite to the direction in which sound travels.

Factors not Affecting the Velocity of Sound in a Air

The velocity of sound in air is not affected by the change in

1. Wavelength (λ): The velocity of sound in air or any other medium does not depend on its wavelength (λ).
2. Frequency (n): The velocity of sound in air or any other medium does not depend on its frequency.
   We know that v = nλ. As the frequency (n) increases, its wavelength (λ) decreases but does not affect the velocity of the wave.
3. Amplitude: Velocity of sound does not depend on the amplitude of the vibrations.
4. Pressure: The velocity of sound in air or any gas in given by v = \(\sqrt{\frac{\gamma P}{d}}\); where ‘γ ’ is a constant, ‘P’ is the pressure and ‘d’ is the density.
   When the pressure of a gas is changed, its density also changes such that the ratio ‘P’/d is always a constant. Therefore, the variation of pressure of a gas does not affect the velocity of sound in it.

Comparison of Speed of Sound with Speed of Light

• The light can travel in vacuum but sound cannot.
• The speed of light in air is 3 × 10⁸ m/s which is about a million times greater than the speed of sound in air. (i.e. 330 m/s at 0 °C).
• The speed of light decreases in an optically denser medium while the speed of sound is more in solids, less in liquids and least in gases.
• In thunder, light is seen much earlier than sound of thunder is heard, even though they are produced simultaneously.
• In an athletic event, when the starter fires a gun, the sound of fire is heard a little later while the smoke is instantaneously seen. The reason is that v_light > V_sound .

Characteristics of Sound as we Hear It

1. Loudness: Loudness of sound is the measure of sound energy reaching the ear per second. Loudness or softness of a sound wave is the sensation that depends upon its amplitude.

E.g. When we strike the top of a table with more force, it vibrates and produces a loud sound. But, when we strike the top of table with lesser force, the vibrating table top produces soft sound waves. Louder sound has more amplitude.

2. Pitch: When a guitar n flute are played together then the sound emitted by the two musical instruments are different. The difference is due to one more characteristic of sound namely pitch.

Pitch is the sensation (brain’s interpretation) of the frequency of an emitter sound. The pitch of sound (Shrillness or flatness) depends on the frequency of vibration.

Faster the vibration of the source, higher is the frequency and higher the pitch and vice versa, Similarly, low pitch sound corresponds to low pitch.

3. Quality or Timbre: Quality or timbre of sound wave is that characteristic which helps us in distinguishing between two sounds of same loudness and same pitch.

Music and Noise

1. Music is the sound that is pleasant to hear/to the ears (e.g, sound coming out of musical instruments)
2. Noise is the sound that is unpleasant to hear/the ears (E.g., Sound produces by vehicles)

Tone and Note

1. A pure sound of single frequency is called tone.
2. An impure sound produced by mixture of many frequencies is called a note. It is pleasant to hear.

Sonic Boom: Speed of sound in air is 333 m/s, speed of faster runner = 12.5 m/s. Many objects such as aircrafts, bullets, and rocket planes travel at speed greater than the speed of sound in air. Such objects are said to be travelling with supersonic speed.

1. Objects moving with speed greater than speed of sound is said to have supersonic speed.
2. Supersonic aircraft produces shock waves in air due to its very high speed. The air pressure variation associated with shock waves produces a very sharp and loud sound called “Sonic boom”.

Chapter 5 Sound Reflection of Sound

Sound gets reflected at the surface of solid and liquid in the same way as light does. It follows the law of reflection. Bouncing back of sound wave from the surface of a solid or a liquid is called reflection of sound.

1. The laws of reflection are as follows:
2. The angle of incidence is equal to the angle of reflection.

The incident ray, reflected ray and normal at the point of incidence, they all lie in the same plane.

we have taken two identical cardboard tubes and arrange them as shown in the figure near the hard plywood. At one end of the cardboard, put in an alarm clock. Now try to listen to the clock tick-tock from the other by adjusting the position of the cardboard. Now measure the angles of incidence and reflection. Now lift the second pipe and try to hear the sound.

In this experiment you will notice that the angle of incidence is equal to the angle of reflection. Incident ray, reflected ray and normal all lie in the same plane.

Instead of hard plywood we can use any obstacle of big size which is either polished or rough. This is used for the reflection of sound waves.

Echo

The sound that comes to our ear after reflection is called echo or echo of sound.

Examples

1. Sound we hear after clap is an echo.
2. Sound heard after shouting in a big empty hall is also an echo.

The sensation of sound remains in human brain for about 0.1 s. It means the time interval between the original sound and echo must be at least 0.1 s. At times echo is heard multiple times. This happens when there is successive or multiple reflections. For example, thunderstorm.

As we know the speed of sound in air is 344 m/s. The distance travelled by sound in 0.1 s will be

D = v × T

= 344 × 0.1

= 34.4 m

It means to hear the echo properly, the minimum distance of the reflecting surface should be half of this distance, i.e. 34.4/2 = 17.2 m.

Temperature has impact on echo. So when temperature changes, distance will also change.

Reverberation

When echo is heard multiple times due to repeated and multiple reflections of sound from different reflecting surfaces, it causes persistence of sound. This is called reverberation.

Example: Sound echoes multiple times, when a man shouts at a top of the mountain cliff.

At places like auditorium, big halls sound echoes multiple times. To reduce ­reverberation, roofs and walls are covered by sound absorbing materials. These can be rough plaster, draperies or compressed fibre board. Reverberation can be prevented by stopping the reflection of sound. This could be done by:

• Using soft sound absorbent material, such as curtains, plant fibres, compressed fireboard, carpets are used in big halls.
• These materials absorb undesired reflected sound and reduce reverberation.

Uses of Multiple Reflection of Sound

On one hand reflection of sound can be irritating but on the other side it proves to be helpful. Let us see some examples where reflection of sound is useful.

1. Loudspeaker, Megaphone, Bulb horn: Devices like loudspeaker, megaphones are designed in such a way that they send the sound in a particular direction by multiple reflections without deviating it into various directions.

2. Soundboard: Soundboard is used in big halls or auditoriums where sound is sent towards auditorium. It works on the basis of law of reflection of sound waves.

Soundboard is a big concave board and it is placed behind the stage in such a manner that it focuses the speaker. Sound that is coming from the speaker falls on the soundboard which in turn gets reflected towards the audience. The result of this is that the audience who are sitting far from the speaker can ­easily hear the sound that is coming out from the speaker.

The walls of auditorium are also constructed in the form of curve. Hence, ceiling acts like a soundboard.

3. Stethoscope: Stethoscope is an instrument used by doctors for listening human breath. In stethoscope, human body’s sound is received by chest piece which is sent to ear by multiple reflections done inside the long tube.

Use of Multiple Reflection of Sound

• Measuring the depth of sea or ocean.
• Used for the detection of the position of objects. For example, sea rocks, hidden ice-berg in the sea and ocean shipwrecks.
• Investigating problem inside the human body.

 

 

Chapter 5 Sound Range of Hearing

The human ear is able to hear sound in a frequency range of about 20 Hz to 20,000 Hz. In other words the audible range of frequency is 20 Hz to 20 kHz. Any sound out of this range is called not audible.

The audible range of frequency varies from person to person and it also varies with age of the person. As the person grows older, his hearing power decreases.

Children can hear higher frequencies of up to 30 kHz while an old person can hear up to 12 kHz. Therefore, on an average the audible range is from 20 Hz to 20 kHz.

The human ear is most sensitive in the range 2 kHz to 3 kHz, where it can hear even a very feeble sound. Unlike human, animals can hear sounds of frequency below 20 Hz and above 20 kHz. Different animals have different range of frequency. Frequency ranges for hearing and speaking by humans and animals.

Let us see the frequency ranges for hearing and speaking of various animals and human.

 

Chapter 5 Sound Ultrasound and Its Applications

As studied in the last topic, the frequency of sound above 20 kHz is called ultrasound. Ultrasound can travel freely in solids and liquids. In case of gases, its intensity falls. In the medium, the speed of ultrasound is same as that of audible sound. In air, the speed of ultrasound is 330 m/s.

Properties of Ultrasound

Ultrasound has all the properties that an ordinary sound has. But because of its high frequency it has two additional properties.

• The energy carried by ultrasound is very high.
• The ultrasound can travel along a well-defined straight path. It does not bend appreciably at the edges of an obstacle because of its small wavelength.
  The above properties of ultrasound make it very useful for different purposes. It is extensively used in industries and for medical purposes.

Applications of Ultrasound

Some applications of ultrasound are given below:

• Bats avoid obstacles in their path by producing and hearing the ultrasound. They emit ultrasound which returns after striking an obstacle in their way. By hearing the reflected sound and from the time interval, they can judge the direction and the distance of the obstacle in their way.
• Ultrasound is used for drilling holes or making cuts of desired shape in materials like glass.
• It is used for cleaning very small objects like cleaning parts of watches or other electronic items. The objects are placed in the cleaning solution and ultrasonic waves are passed in the solution. This causes high frequency vibrations in the solution and makes the cleaning easier.
• Ultrasound is used for detection of defects in metals. Ultrasound will pass through the object if there is no defect but if there is some defect, ultrasound will get reflected back.
• Ultrasound is widely used for imaging the human organs. Ultrasonography is used to obtain the images of patient’s organs such as liver, gall bladder, uterus, etc. It helps to detect stone, tumour, etc. Echo cardiology is used to obtain the image of the heart.
• Ultrasound is used in surgery to remove cataract and in kidneys to break the stone in fine grains.
• In SONAR (abbreviated form of sound navigation and ranging) to detect and find the distance of objects under water, ultrasound is used.

Infrasonics or Infrasound

Infrasonic waves are the waves which has frequency less than 20 Hz. These waves are produced by large vibrating bodies. For example they are produced by the vibration of the earth’s surface during the earthquake.

These waves are also produced by some animals like elephants, rhinoceroses, whales, etc. These waves are not audible to human ears.

It has been observed that the animal’s behaviour becomes unusual just before tremor is felt. This is because the animals have the ability to detect infrasonic waves produced at the time of tremor.

Sound Navigation and Ranging (SONAR)

SONAR is a device which is used in the ships to locate rocks, icebergs, and submarines, old sank in seas etc.

Ultrasonic waves of high frequency are sent from a ship on the surface. The waves travel in straight line till they hit somebody like shipwreck or submarine. On hitting the body these waves are reflected back as shown in figure.

The transmitter sending the wave notes the time t between sending the signal and receiving it back. Let if d is the distance of the submarine from the ship, than the total distance travelled by the wave in time interval t is 2d.

\(\text { Using speed }=\frac{\text { Distance }}{\text { Time }}\)

v = \(\frac{2 d}{t}\)

Therefore d = \(\frac{1}{2}(v \times t)\)

The velocity v of the ultrasonic wave in water is same as that of audible range of water.

Chapter 5 Sound Master Your Test Question And Answers

Question 1. Describe various uses of multiple reflections of sound.
Answer:

Various uses of multiple reflections of sound:

Multiple reflection of sound is used to measure the depth of sea or ocean. It is also used to detect the position of objects for example sea rocks, hidden iceberg in the sea and ocean shipwrecks. It is also used to investigate problems inside the human body.

Question 2. What is the range of hearing of human being?
Answer:

Range of hearing of human being:

The range of hearing of human being is the sound of frequency between 20 Hz to 20000 Hz.

Question 3. Define infrasonic and ultrasonic sounds.
Answer:

Infrasonic and ultrasonic sounds:

The sound of frequency less than 20 Hz is called infrasonic sound while the sound of frequency greater than 20 kHz is called ultrasonic sound.

Question 4. What are the additional properties of ultrasound?
Answer:

The additional properties carried by ultrasound are:

• The energy carried is very high.
• It can travel along a well-defined straight path. It does not bend appreciably at the edges of the obstacle because of its small wavelength.

Question 5. What are the various applications of ultrasound?
Answer:

The various applications of ultrasound are:

• Bats avoid obstacles in their path by producing and hearing the ultrasound.
• It is used in drilling holes and making cuts of desired shape in the glass.
• It is used in cleaning very small objects like parts of watches or other electronic items.
• It is used for detection of defects in the metals.
• It is widely used for imaging of human organs.
• It is used in human surgery to remove cataract and in kidneys to break the stone in fine grains.
• It is used in SONAR (Sound navigation and ranging)

 

Chapter 5 Sound Structure of Human Ear

Ear acts as a receiver or detector of sound. The sound that is received in the form of pressure variations in air are converted into electrical signals within the ear. This sound on reaching our brain produces a sense of hearing. The human ear consists of three parts:

1. Outer ear: The outer ear looks complicated but it is functionally the simplest part of the ear. It consists of the ‘pinna’ or auricle (the visible projecting portion of the ear), the external acoustic meatus (the outside opening to the ear canal), and the external ear canal that leads to the ear drum. In sum, there is the pinna, the meatus and the canal. That’s all. The outer ear concentrates air vibrations on the ear drum and makes the drum vibrate. The outer ear is also called external ear.

2. Middle ear: The middle ear consists of an eardrum or tympanic membrane connected at the end of the auditory canal. The eardrum is a thin, tightly stretched membrane, also known as sheet (‘kaan ka parda’ in Hindi). The eardrum vibrates when compressions and rarefactions of sound wave hit it. A compression exerts an inward pressure on the outer surface of the eardrum. This forces the eardrum to move inward. However, a rarefaction does the opposite and moves the eardrum outwards.

Hence, the eardrum is made to vibrate by the successive compressions and rarefactions. The vibration of eardrum produces pressure variations within the middle ear. The three bones (hammer, anvil and stirrup) present in the middle ear amplify these pressure variations several times. The middle ear then transmits the sound wave’s amplified pressure variations to the inner ear.

3. Inner ear: The inner ear has a job to convert the sound wave’s amplified pressure variations into electrical signals. This work is done in the inner ear by cochlea, a snail-shaped organ. The cochlea is filled with water like fluid and its inner surface has large number of hair-like nerve cells.

The amplified pressure variations produce vibrations in the nerve cells and they in turn release electrical impulses. The electrical impulses are transmitted to the brain along the auditory nerve. The brain interprets the electrical impulses through a complex process, as sounds.

 

Chapter 5 Sound

Question 1. ______ acts as a receiver or detector of sound.
Answer. Ear

Question 2. The simplest part of ear is

1. Inner ear
2. Middle ear
3. Outer ear
4. Cerebellum

Answer. 3. Outer ear

Question 3. The ______ vibrates when compressions and rarefactions of sound wave hit it.
Answer. Ear drum

Question 4. The ______ has a job to convert the sound wave’s amplified pressure variations into electrical signals.
Answer. inner ear

Chapter 5 Sound Practice Exercises

Question 1. Note is a sound

1. of mixture of several frequencies
2. of mixture of two frequencies only
3. of a single frequency
4. always unpleasant to listen

Answer. 3. of a single frequency

Question 2. A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case

1. sound will be louder but pitch will not be different
2. sound will be louder and pitch will also be higher
3. sound will be louder but pitch will be lower
4. both loudness and pitch will remain unaffected

Answer. 1. sound will be louder but pitch will not be different

Question 3. In SONAR, we use

1. ultrasonic waves
2. infrasonic waves
3. radio waves
4. audible sound waves

Answer. 1. ultrasonic waves

Question 4. Sound travels in air if

1. particles of medium travel from one place to another
2. there is no moisture in the atmosphere
3. disturbance moves
4. both particles as well as disturbance travel from one place to another.

Answer. 3. disturbance moves

Question 5. When we change feeble sound to loud sound we increase its

1. frequency
2. amplitude
3. velocity
4. wavelength

Answer. 2. amplitude

Question 6. In the curve half the wavelength is

1. A B
2. B D
3. D E
4. A E

Answer. 2. B D

Question 7. Earthquake produces which kind of sound before the main shock wave begins

1. ultrasound
2. infrasound
3. audible sound
4. none of the above

Answer. 2. infrasound

Question 8. Infrasound can be heard by

1. dog
2. bat
3. rhinoceros
4. human beings

Answer. 3. rhinoceros

Question 9. Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting

1. intensity of sound only
2. amplitude of sound only
3. frequency of the sitar string with the frequency of other musical instruments
4. loudness of sound

Answer. 3. frequency of the sitar string with the frequency of other musical instruments

Chapter 5 Sound Fill in the Blanks

Question 1. Sound shows______variation with temperature.
Answer. Speed

Question 2. Infrasonic sound has less ______.
Answer. Frequency

Question 3. 1 kHz = ______Hz
Answer. 1000

Question 4. The maximum displacement of a wave from its mean position is known as ______.
Answer. Amplitude

Question 5. Wavelength is denoted by the symbol ______.
Answer. λ

Question 6. _____ is the unit of loudness of sound.
Answer. Decibel

Question 7. Echo is observed usually in ______ rooms.
Answer. Large

Question 8. In SONAR, ‘S’ stands for ______.
Answer. Sound

Question 9. In order to distinguish between two different sounds, they must be heard after an interval of ______.
Answer. 0.1 s

Question 10. If the time period is 0.05 second, then the frequency is ______ Hz.
Answer. 20

Chapter 5 Sound Match the Columns

Question 1. Choose the correct speed (m/s) of substance at temperature 25 °C.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-4, B-3, C-2, D-1
3. A-2, B-1, C-3, D-4
4. A-1, B-2, C-4, D-3

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

Question 2. Choose the correct S.I. unit.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-4, B-3, C-2, D-1
3. A-2, B-1, C-3, D-4
4. A-4, B-1, C-2, D-3

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

Question 3. Choose the correct symbol.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-2, B-4, C-1, D-3
3. A-2, B-1, C-3, D-4
4. A-4, B-1, C-2, D-3

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

Question 4. Choose the correct code.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-2, B-4, C-1, D-3
3. A-2, B-1, C-3, D-4
4. A-4, B-1, C-2, D-3

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

Chapter 5 Sound Assertion Reasoning

Direction: For the following questions the options will remain the following:

1. Both A and R are correct and R is correct explanation of A.
2. Both A and R are correct but R is not a logical explanation of A.
3. A is correct but R is incorrect.
4. R is correct but A is incorrect.

Question 1. Assertion: A boy shouts near mountain, he hears his sound again.
Reason: This is because of the reflection of sound.
Answer. 1. 1. Both A and R are correct and R is correct explanation of A.

Question 2. Assertion: Ultrasonic sound is used to clean electronic components.
Reason: Due to low frequency, dirt particles get detached from the components.
Answer. 2. Both A and R are correct but R is not a logical explanation of A.

Question 3.Assertion: The ceilings of concert halls and conference halls are made curved.
Reason: This way sound wave reaches all thepart after reflection from the curved surface.
Answer. 1. Both A and R are correct and R is correct explanation of A.

Chapter 5 Sound Comprehension Passage

Heinrich Rudolf Hertz was born on 22 February, 1857 in Germany. He was a German physicist. He was the first person who proved the existence of electromagnetic waves by his experiments. He laid the foundation for future development of radio, telephone, telegraph and television.

He also discovered the photoelectric effect when he noticed that the charged object loses its charge quickly when illuminated by ultraviolet radiation. It was later explained by Albert Einstein.

He also started experimenting on cathode rays. He discovered that cathode rays can penetrate in thin\ metal foil. He developed a version of cathode tube and worked on X-rays of ­different materials. But he never practically worked on actual 0020X rays.

The S.I. unit of frequency was named as Hertz in his honour.

Question 1. What is the S.I. unit of frequency?

1. Hertz
2. Metre
3. Second
4. m/s

Answer. 1. Hertz

Question 2. When was Hertz born?

1. 15 March, 1857
2. 22 February, 1857
3. 22 February, 1859
4. 19 November, 1882

Answer. 2. 22 February, 1857

Question 3. Which city was Hertz born?

1. London
2. Germany
3. Indonesia
4. America

Answer. 2. Germany

Question 4. Hertz was the first person to introduce ______.

1. Electromagnetic waves
2. Magnetic waves
3. Electricity
4. Gas

Answer. 1. Electromagnetic waves

Question 5. What was the full name of Hertz?

1. Henry Hertz
2. Rudolf Hertz
3. Heinrich Hertz
4. Heinrich Rudolf Hertz

Answer. 4. Heinrich Rudolf Hertz

Chapter 5 Sound Integer Type Question And Answers

Question 1. A sound source sends waves of 400 Hz. It produces waves of wavelength 2.5 m. What is the velocity of sound waves?
Answer. 1000 m/s

Question 2. The time period of a vibrating body is 0.05  s. What is the frequency of waves that it emits?
Answer. 20 Hz

Question 3. A bat can hear sound of frequencies up to 120 kHz. Determine the minimum wavelength of sound which it can hear. Speed of sound in air to be 344 m/s.
Answer. 2.867 x 10^-3 m

Question 4. A wave pulse of frequency 200 Hz, on a string moves a distance 8 m in 0.05 s. Calculate the wavelength of wave on string.
Answer. 0.8 m

Question 5. If velocity of sound in air is 340 m/s and frequency is 256 Hz. What will be the wavelength?
Answer. 1.25 m

Chapter 5 Sound Multiple Choice Question And Answers

Direction: Choose the correct option for each questions. There is only one correct response for each question.

Question 1. In SONAR, ______ waves are used.

1. Ultrasonic
2. Infrasonic
3. Radio
4. Audible sound

Answer. 1. Ultrasonic

Question 2. When we change feeble sound to loud sound we increase its ______.

1. Velocity
2. Wavelength
3. Frequency
4. Amplitude

Answer. 4. Amplitude

Question 3. Infrasound can be heard by which animal?

1. Human beings
2. Rhinoceros
3. Bat
4. Dog

Answer. 2. Rhinoceros

Question 4. Sound waves in air are ______.

1. Radio waves
2. Electromagnetic waves
3. Longitudinal waves
4. Transverse waves

Answer. 3. Longitudinal waves

Read and Learn More NEET Foundation Multiple Choice Questions

Question 5. The unit of quantity on which loudness of sound depends is ______.

1. Metre
2. Second
3. Hertz
4. m/s

Answer. 1. Metre

Question 6. Sound travels faster in ______.

1. Solid
2. Liquid
3. Gas
4. None of the above

Answer. 1. Solid

Read more

Chapter 5 Sound Short Answer Type Question And Answers

Question 1. If echo is heard 5 seconds later in a theatre, calculate the distance of the reflecting surface.
Answer.

Given:

If echo is heard 5 seconds later in a theatre

Total distance travelled by sound wave,

D = 5 × 340 = 1700 m

Distance of the reflecting surface

= D/2 = 1700/2 = 850 m

Question 2. Distinguish between intensity and loudness.
Answer.

Difference between intensity and loudness:

Question 3. What are the requirements of the medium for the sound to propagate?
Answer.

Requirements of the medium for the sound to propagate:

The medium required for propagation of sound must have the following properties:

The medium must be elastic so that its particles may come back to their initial positions after displacement on either side.

The medium must have inertia so that its particles may store mechanical energy.

The medium should be frictionless so that there is no loss of energy in propagation of sound through it.

Question 4. A tuning fork having a frequency of 250 Hz is made to vibrate by striking it against a rubber pod. What is the wavelength of the sound waves it produces in

1. air  
2. water
3. iron?

Velocity of sound in air, water and iron = 330 ms^-1, 1500 ms^-1 and 5 kms^-1 respectively.
Answer.

We know that

v = fλ

i.e., λ = \(\frac{v}{f}\)

In air, \(\lambda_a=\frac{330 \mathrm{~ms}^{-1}}{250 \mathrm{~s}^{-1}}\)

= 1.32m

Similarly,

\(\lambda_w=\frac{1500 \mathrm{~ms}^{-1}}{250 \mathrm{~s}^{-1}}\)

= 6.0m

In iron,

\(\lambda_{\text {iron }}=\frac{5000 \mathrm{~ms}^{-1}}{250 \mathrm{~s}^{-1}}\)

= 20.0 m

Question 5. If a tuning fork having a frequency of 400 Hz is made to vibrate continuously how many

1. Compression
2. Rarefactions

Pass a point P in air in 0.1 s?
Answer.

When a tuning fork vibrates once, it produces one sound wave, which consists of one compression and one rarefaction.

∴ Its frequency is 400 Hz, it vibrates 400 times in one second.

In one second, it vibrates 400 times.

So, in 0.1 second it vibrates 0.1 × 400 = 40 times.

So, it will produce 40 waves, i.e., 40 compressions and 40 rarefactions.

∴ Number of compressions passing a point P  in 0.1 second is 40.

Similarly, number of rarefaction passing a point P is 0.1 second is also 40.

Question 6. Displacement–Distance graph of particles vibrating in a gas under the influence of a tuning fork of frequency 512 Hz is as follows: Find the velocity of sound in the gas.

Answer.

Points A and B represent the consecutive compressions. Distance between them is called ‘wavelength’ (λ).

So, λ = (100 cm − 20 cm) = 80 cm

Distance between points C and D = (120 cm − 40 cm) = 80 cm is also the wavelength(λ).

∴ λ = 80 cm = 0.8 m

We know that

v = fλ

= 512 × 0.8

= 409.6 ms^-1

∴ Velocity of sound in the gas = 409.6 ms^-1

Question 7. A train is travelling in a railway track made of iron. A boy standing far away puts his ears to the track. He hears two sounds. Why? Calculate the time interval between these two sound waves.
Velocity of sound in air = v_a
Velocity of sound in iron = v_s (where v_s > v_a)
Answer.

It is because he receives the sound waves travelling through air as well as through iron. Since sound travels much faster through iron he will the hear sound travelling through iron first.

Time ‘t1’ taken by the sound to travel from the train to the boy through the iron rail is given by

\(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\)

\(t_1=\frac{d}{v_n}\)  (1)

Similarly, time ‘t2’ taken by sound to travel to him is,

\(t_2=\frac{d}{D_d}\)  (2)

∴ Time interval ∆t between two sound waves reaching the boy is

∆t = t₂ − t₁

= \(\left(\frac{d}{v_a}-\frac{d}{v_n}\right)\)

Question 8. Calculate the velocity of sound through iron given that the modulus of elasticity for iron = 2 × 10¹¹ Pa and density of iron = 8 × 10³ kgm^-3.
Answer.

Velocity of sound through a solid is given by

\(v=\sqrt{\frac{Y}{\rho}}\)

Where Y = young’s modulus

ρ = Density

V = \(\sqrt{\frac{\left(2 \times 10^{11}\right) \mathrm{Pa}}{\left(8 \times 10^3\right) \mathrm{kgm}^{-3}}}\)

= \(\sqrt{0.25 \times 10^8 \frac{\mathrm{Nm}^{-2}}{\mathrm{kgm}^{-3}}}\)

= \(\sqrt{25 \times 10^6 \frac{\mathrm{kgms}^{-2} \mathrm{~m}^{-2}}{\mathrm{kgm}^{-3}}}\)

= \(\sqrt{25 \times 10^6 \mathrm{~m}^2 \mathrm{~s}^{-2}}\)

= 5 x 10³ ms^-1

Question 9. How does a SONAR work?
Answer:

Working Of SONAR

SONAR is a device which is used in the ships to locate rocks, icebergs and submarines, old sank in the seas etc. Ultrasonic sound of high frequency are sent from a ship on the surface. The waves travel in the straight line till they hit something like shipwreck or submarine. On hitting the body these waves are reflected. The transmitter sending the waves note the time lag between sending the signal and receiving it back. This time lag is multiplied by speed of sound in water and the distance calculated is halved to get the actual distance of the object from the ship.

Question 10. Describe the structure and function of a human ear?
Answer:

Structure and function of a human ear:

The human ear consists of three parts:

1. Outer ear: It is functionally the simplest part of the ear. It consists of ‘pinna’ or auricle i.e. the visible portion of the ear, the external acoustic meatus i.e. the outside opening to the ear canal and the ear canal.
2. Middle ear: The middle ear consists of eardrum or tympanic membrane connected at the end of auditory canal. The eardrum vibrates when compression and rarefaction of the sound waves hit it. The three bones (hammer, anvil and stirrup) present in the middle ear the pressure variations several times. Thus the middle ear transmits the sound wave’s amplified pressure variation to the inner ear.
3. Inner ear: The inner ear converts the sound waves amplified pressure variation into electrical signals. This work is done by cochlea, a snail-shaped organ. The cochlea is filled with water like fluid and its inner surface has large number of hair like nerve cells. The amplified pressure variation produces vibrations in the nerve cells and they in turn releases electrical signals which are sent to the brain along the auditory nerve. The brain interprets the electrical signals through a complex process as sounds.

Question 11. What are the factors affecting the speed of sound in a gas?
Answer:

Speed of sound in a gas is affected by:

1. Density
   Temperature
2. Humidity
3. Direction of wind

Question 12. What are the factors which does not affect the speed of sound in air?
Answer:

The factors which does not affect the speed of sound in air are:

1. Wavelength
2. Frequency
3. Amplitude
4. Pressure

Question 13. What are the laws of reflection of sound?
Answer:

The laws of reflection of sound are:

• The angle of incidence is equal to angle of reflection.
• The incident ray, reflected ray and normal at the point of incidence, all lie in the same plane.

Question 14. What do you mean by reverberation?
Answer:

Reverberation:

When echo is heard multiple times, due to repeated and multiple reflections of sound from different reflecting surfaces, it causes persistence of sound. This phenomenon is called reverberation.

Question 15. Name the devices which use multiple reflections of sound.
Answer:

Loudspeaker, megaphone, soundboard and stethoscope are the devices which use multiple reflections of sound.

Question 16. A boy shouted at the top of the mountain. The echo is heard after 10 seconds. Speed of the sound is 450 m/s. How far is the mountain from the boy?
Answer:

v = 450 m/s

T = 10 s

Distance d = 450 × 10 = 4500 m

As sound has to travel twice the distance, hence distance between boy and mountain = 4500/2

= 2250 m

Question 17. Why sound is called a mechanical wave?
Answer:

Mechanical wave:

Sound is called a mechanical wave because it requires a medium to travel from one place to another.

Question 18. What properties should the medium have for the propagation of sound?
Answer:

The medium required for propagation of sound must have the following properties:

• The medium should be elastic so that its particles comes back to the mean position after displacement on the either side.
• The medium must have inertia so that its particles may store mechanical energy.
• The medium should be frictionless so that there is no loss of energy in propagation of sound through it.

Question 19. What do you mean by crest and trough?
Answer:

Crest And Trough:

The position of maximum upward displacement is called crest and position of maximum downward displacement is called trough. These two are created in the medium during the propagation of transverse wave.

Question 20. What is the difference between progressive wave and stationary wave?
Answer:

Progressive waves start at a point and moves indefinitely and infinitely to all parts of the medium. These waves transmit energy from one place to another. While standing waves appear to stand at a place and confined between two points in a medium. These waves store energy in them.

Question 21. What are the characteristics of a sound wave?
Answer:

The characteristics of the sound waves are:

• Amplitude
• Time period
• Frequency
• Wavelength
• Wave velocity

Question 22. On which factors does the speed of a sound in a medium depends?
Answer:

Speed of sound in a medium depends on elasticity of the medium as well as density of the medium.

Question 23. A sound wave has a frequency of 200 Hz. Wavelength is 15 m. How much time it will take to cover 2 km?
Answer:

ν = 200 Hz

λ = 15 m

Speed v = ν × λ = 200 × 15 = 3000 m/s

Time t = d/υ

= 2000/3000

= 0.67 seconds

Question 24. A sound has a frequency of 220 Hz. Its speed is 440 m/s. What will be the wavelength of the sound wave?
Answer:

ν = 220 Hz

v = 440 m/s

λ = v/ν = 440/220 = 2 m

Chapter 5 Sound Long Answer Type Question And Answers

Question 1. Differentiate between transverse and longitudinal waves.
Answer.

Difference between transverse and longitudinal waves:

Question 2. Explain how defects in a metal block can be detected using ultrasound.
Answer.

Defects in a metal block can be detected using ultrasound:

Ultrasounds are used to detect cracks in the metal blocks. The cracks inside the metal blocks, which are imperceptible from outside, decreases the strength of the structure.

Ultrasonic waves can pass through the metal block and detectors are conditioned to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected indicating the occurrence of the flaw or defect in that metal block.

Read and Learn More NEET Foundation Long Answer Questions

Question 3. Calculate the difference in Newton’s value of velocity of sound in air and that found by Pascal.
Answer.

The difference in Newton’s value of velocity of sound in air and that found by Pascal:

Velocity of sound in a gas is given by

v = \(\sqrt{\frac{E}{\rho}}\)

Where E is the modulus of elasticity of the gas.

Newton said ‘when sound travels through air through compression and rarefactions, the changes taking place in air are isothermal changes.’ For isothermal changes,

E = Pressure of gas

= Pressure of air

= 1.013 × 10⁵ Nm^-2

∴ According to Newton,

v_a = \(\sqrt{\frac{P_a}{\rho_a}}\)

v_a = \(\sqrt{\frac{1.013 \times 10^5 \mathrm{Nm}^{-2}}{1.29 \mathrm{kgm}^{-3}}}\)

Newton’s value of v_a = 280 ms^-1

Almost a century later, Laplace said ‘Sound travels very rapidly through air such rapid changes cannot be isothermal. They must be adiabatic in nature.’

For adiabatic changes,

E = YP

= 1.4 × 1.013 × 10⁵ Pa

So, according to Pascal,

va = \(\sqrt{\frac{Y P_a}{\rho_e}}\)

= \(\sqrt{\frac{1.4 \times 1.013 \times 10^5}{1.29}}\)

= 331.6 ms^-1

Velocity of sound in air at 0 °C is about 332 ms^-1.

So, Laplace’s value is very close to the actual value of velocity of sound in air. Thus Laplace corrected Newton. This is therefore, called

Laplace’s correction.

∆v = (331.6 − 280)ms^-1.

= 51.6 ms^-1.

Question 4. At what temperature the velocity of sound in air, will be double that of air at 0 °C ? (Given v0 = 332 ms^-1).
Answer.

We know that,

\(\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\) \(\frac{v_2}{v_0}=\sqrt{\frac{T_2}{T_1}}\)

\(\frac{2 v_0}{v_0}=\sqrt{\frac{T_2}{273}}\)   (∴ V₂ = 2V₀)

Squarring on both sides, we get

\(4=\frac{T_2}{273}\)

T₂ = 1092 K

= (1092 − 273) °C

= 819 °C

Question 5. Velocity of sound in a gas at STP is 400 ms^-1. If both the pressure and the absolute temperature of the gas are doubled, then how will the velocity of sound change?
Answer.

Given:

Velocity of sound in a gas at STP is 400 ms^-1. If both the pressure and the absolute temperature of the gas are doubled

Change in pressure has no effect on the velocity of sound. Only temperature change has,

Given: T₁ = 0 °C = 273 K, v₁ = 400 ms^-1, T₂ = 2 × 273 k,

We know that, T₂ = ?

\(\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\)

\frac{v_2}{400}=\sqrt{\frac{(2 \times 273)}{273}}

v₂ = 400 \sqrt{2}

= 565.7 ms^-1.

Question 6. A boy blows a whistle and its echo from a distant wall is heard after 1.8 seconds. How far is the wall from him? Velocity of sound in air at room temperature = 340 ms^-1.
Answer.

Given:

A boy blows a whistle and its echo from a distant wall is heard after 1.8 seconds.

If d be the distance of the wall from the boy, then sound travels a distance d to the wall and back to the boy.

So, total distance travelled by sound is 2d.

Time taken by sound to go to the wall(reflection of sound) and back.

\(t=\frac{\text { Distance }}{\text { Velocity }}\) \(t=\frac{2 d}{v}\)

=> \(1.6=\frac{2 d}{340}\)

2d = 340 x 1.6

∴ \(d=\frac{340 \times 1.6}{2}\)

= 272 m

Question 7. Two ships are 6 km apart in Indian Ocean. Signal sent by one ship is reflected by the ocean bed and received at the second ship after 6.6 s. Find the depth of the Indian Ocean. (Velocity of sound in water = 1500 ms^-1)
Answer.

Given:

Two ships are 6 km apart in Indian Ocean. Signal sent by one ship is reflected by the ocean bed and received at the second ship after 6.6 s.

The figure below shows two ships S₁ and S₂. Signal sent by ships S₁ travels to the ocean bed along S1B and gets reflected along BS₂ to reach the ship S₂. So, distance travelled by sound is S₁BS₂ = 2 S₁B.

In ∆ S₁BM,

\(S_1 B=\sqrt{d^2+\left(S_1 M\right)^2}\)

Time taken by (sound) signal to travel distance x = S₁BS₂ is given by

\(t=\frac{\text { Distance }}{\text { Velocity }}=\frac{S_1 B S_2}{v}\)

= \(\frac{S_1 B+B S_2}{v}\)

= \(\frac{2 S_1 B}{v}\)   (∴ Bs₂ = s₁B)

\(t=\frac{2 \sqrt{d^2+\left(S_1 M\right)^2}}{v}\)

∴ \(\sqrt{d^2+\left(S_1 M\right)^2}=\left(\frac{v t}{2}\right)\)

Squarring on both sides,

\(d^2+(S, M)^2=\left(\frac{v t}{2}\right)^2\) \(d^2+(3000)^2=\left(\frac{1500 \times 6.6}{2}\right)^2\)

d² + (3000)² = (4950)²

∴ d² = 4950² − 3000²

= (4950 + 3000)(4950 − 3000)

d2 = (7950)(1950)

\(d=\sqrt{(7950) \times(1950)}\)

⇒ d = 3937.4 m

⇒ d ≅ 3940 m

⇒ d ≅ 3.94 km

Question 8. A sound wave of wavelength 0.332 m has a time period of 10^-3 s. If the time period is decreased to 10^-4 s. Calculate the wavelength and frequency of the new wave. Name the subjective property of sound related to its frequency and the light related to its wavelength.
Answer.

Given:

A sound wave of wavelength 0.332 m has a time period of 10^-3 s. If the time period is decreased to 10^-4 s.

Here λ = 0.332 m

Time taken to travel,

t = 10^-3 s

Velocity = \(\frac{\lambda}{t}\)

v = \(\frac{0.33}{10^{-3}}\)

v = 0.33 × 10³

v = 330 m/s

Time period for second wave = 10^-4 s

Wavelength λ = vT

= 330 × 10^-4

= 0.033 m

Frequency = \(\frac{1}{T}\)

= \(\frac{1}{10^{-3}}\)

= 10³ Hz

Question 9. Derive a relation between Wave-Velocity, frequency and wavelength.
Answer.

A relation between Wave-Velocity, frequency and wavelength:

Let velocity of wave = v

Time period = T

Frequency = υ

Wavelength = λ

As per the definition of wavelength,

λ = Distance travelled by wave in one time period

= Wave Velocity × Time period

= v × T

vT = λ

t = \(\frac{1}{v}\)

\(v \times\left(\frac{1}{v}\right)=\lambda\)

Therefore, v = λυ.

Hence,

Wave Velocity = Frequency × Wavelength

Question 10. Meera is standing between two hills. She shouted loudly and hears first echo after 0.5 sec and second echo after 1 sec. What is the distance between two hills?
Answer.

Given:

Meera is standing between two hills. She shouted loudly and hears first echo after 0.5 sec and second echo after 1 sec.

Let the distance between nearest cliff and

Meera = x m and the distance between distant cliff and Meera = y m

Distance between two cliff = (x + y) m

Total distance covered by sound to produce first echo = 2x m and time = 0.5 sec.

2x = 340 × 0.5

\(x=340 \times \frac{0.5}{2}=85 \mathrm{~m}\)

Total distance covered by sound to produce second echo = 2y m and time = 1 sec

2y = 340 × 1

y = \(340 \times \frac{1}{2}\)

y = 170m

So distance between the two cliffs = (85 + 170) = 255 m

Question 11. A man fires a gun and hears its echo after 5 seconds. The man then moves 310 m towards the hill and fires his gun again. This time he hears the echo after 3 seconds. Calculate the speed of sound.
Answer.

Given:

A man fires a gun and hears its echo after 5 seconds. The man then moves 310 m towards the hill and fires his gun again. This time he hears the echo after 3 seconds.

Let d be the distance between the man and the hill in the beginning

v = \(\frac{2 d}{t}\)

v = \(\frac{2 d}{5}\)   (1)

He moves 310 m towards the hill. Therefore distance will be (d − 310) m.

v = \(2 \frac{(d-310)}{3}\)   (2)

Since velocity of sound is same, equating (1) and (2).

\(\frac{2 d}{5}=2 \frac{(d-310)}{3}\)

3d = 5d − 1550

2d = 1550

d = 775 m

Velocity of sound v = \(2 \times \frac{775}{5}\)

= 310 ms^-1

Question 12. An engine is approaching a hill at a constant speed. When it is at a distance 0.9 km, it blows whistles, whose echo is heard by the driver after 5 s. If the speed of sound is 340 ms^-1,calculate the speed of the engine.
Answer.

Given:

An engine is approaching a hill at a constant speed. When it is at a distance 0.9 km, it blows whistles, whose echo is heard by the driver after 5 s.

Let v_e be the speed of the engine.

Distance covered by the engine in 5 s = 5 v_e

Distance covered by sound in reaching the hill and coming back to moving driver

= 900 × 2 − 5v_e

= 1800 − 5v_e

According to given condition,

t = \(\frac{\left(1800-5 v_e\right)}{v}\)

As t = 5 s and v (speed of sound) = 340 ms^-1

5 = \(\frac{(1800-5 v)}{340}\)

1700 = 1800 − 5v_e

Or v_e = 20 ms^-1

Physics Chapter 4 Work And Energy

In previous chapter we have discussed about fundamental concepts of physics, which deal with the laws of nature like motion, forces, and Newtons laws of motion and gravitation, etc. Work and energy are the other two basic principles of nature which we come across knowingly or unknowingly in our day to day life.

In this chapter, we will discuss about some of the basic concepts on which work and energy are co-related. Also, energy and power are closely related to work.

All living beings need food to survive. Complex food molecules are broken down by our body and converted into simple soluble form, which in turn is absorbed by the body to get ‘Energy’. Thus, energy can be defined as the element which allows us to perform any function.

For our day to day activities like singing, dancing, running, etc., we need energy. Without energy we cannot perform any work.

For machines, cars, etc., the energy is provided by electricity or the fuel which we provide. Thus, any object whether living or non-living requires energy to do any work.

Chapter 4 Work And Energy

In the language of science, work is said to be done only if the force applied in a specific direction produces motion in the body. In other words, if the body displaces from its original position, work is said to be done otherwise it is not done irrespective of all your efforts. In our everyday life, we consider any useful mental or physical labour as work.

Certain activities like playing in a field, chit chatting with friends, singing a song, watching television, attending a party are sometimes not counted as work. What defines ‘work’ depends on how we use it in science.

From the above mentioned examples, it is clear that how work we usually do is different from the scientific definition of work. To understand the concept better, we would study the concept of work done by a constant force.

We will consider a constant force ’F’ acting on the body and body being displaced in the direction of the force by a distance ’S’ as shown.

Let ’W’ be the work done.

According to the definition,

Work done = force × displacement

Mathematically,

W = F × S

Elaborating the above equation, the work done by a body is equal to the magnitude of the force applied on the body multiplied by the displacement carried in the direction of force. Work has only magnitude and no direction. If either force or displacement in the above equation is zero then the work becomes zero.

If in the above equation, 1 newton of force is applied on the body producing a ­displacement of 1 metre in the body in the direction of motion, then the work done is said to be 1 Nm or 1 Joule. (1 Nm = 1 Joule).

If force F acts in the direction of displacement at an angle θ, then in the direction of displacement the rectangular component will be represented by F cos θ.

Work done = Force(rectangular component) × displacement

W = F cos θ × S

Therefore, depending on the value of θ (i.e., cos θ) work done can be positive, negative or zero.

Concept of Positive and Negative Work

Work can be positive or negative depending upon the direction of force applied. If the force applied is along the direction of the displacement, work done is positive. Example of such situation is pulling of a toy by a kid (parallel to the ground), pushing the door, etc. In these examples, work done is equal to the product of force and displacement.

On the other hand, when the force applied and displacement act in the opposite directions, the work done is said to be negative. Example of this type of work is, when a body is moving with a uniform velocity and suddenly a retarding force is exerted on the body to halt its motion.

The retarding force is opposite to the direction of motion. The angle between the force applied and the displacement is 180°. The work done in this case is said to be negative. Work done is represented by F × (–S) or (–F) × S. Negative work done is indicated by the minus sign.

 

Chapter 4 Work And Energy Track Your Learning Question And Answers

Question 1. The angle between the force & displacement is_________ in case of negative work.

1. 45°
2. 90°
3. 0°
4. 180°

Answer. 3. 0°

Question 2. Which of the above quantity does not depend upon work done on an object?

1. displacement
2. angle between force and displacement
3. force applied
4. initial velocity of the object

Answer. 4. initial velocity of the object

Question 3. If the force applied is along the direction of the displacement, work done is said to be ______.

1. Negative
2. Positive
3. Neutral
4. Cannot be determined

Answer. 2. Positive

Question 4. Nm is a SI unit of

1. Work
2. Power
3. Acceleration
4. Force

Answer. 1. Work

Question 5. kWh is the unit of

1. Work
2. Power
3. Acceleration
4. energy

Answer. 4. energy

Chapter 4 Work And Energy

Every process on this earth requires energy. Our earth has evolved because of energy. The life processes that keep us alive require energy. Every single process needs energy. Just like work, let us understand what energy is, from where do we get energy in the language of science.

In simple language, energy is defined as the capacity of doing work. It is the property possessed by objects which allows them to do work or can exert force on another object. Which means energy is transferred from one object to the other.

The latter may move in the direction of force experienced and hence receives the energy and therefore do some work. Thus, the first object has a capacity to do work. This means that any object that can do work, possesses energy.

Commercial Unit of Energy

When we have to express large quantities of energy, joule is not used. We use a bigger unit of energy called kilowatt hour (kWh). Let us understand with the help of an example, suppose we have a machine that uses 1000  J of energy every second. If this machine is used continuously for one hour, it will consume 1 kWh of energy. Thus, 1 kWh is the energy used in one hour at the rate of 1000 J s^-1 (or 1 kW).

1 kW h = 1 kW × 1 h

1000 W × 3600 s = 3600000 J

1 kW h = 3.6 × 10⁶ J

The energy used in households, industries and commercial establishments is usually expressed in kilowatthour. For example, electrical energy used during a month is expressed in terms of ‘units’. Here, 1 ‘unit’ means 1 kilowatt hour.

Classification of Resources

Natural resources are valuable and easily available to us. They are classified on the basis of their ability to replenish or recover. They are broadly classified into:

1. Renewable resources
2. Non Renewable resources

1. Renewable resources: When the energy source used is easily replenished in a short period and there are practically limitless reserves (inexhaustible). An example is the solar energy that is the source of energy from the sun, or the wind used as an energy resource. Other renewable energies are original solar, natural wind (atmospheric flows), natural ­geothermal, oceanic tidal, natural waterfall (hydraulic flows), etc.
2. Non Renewable resources: They are limited sources of energy on earth (exhaustible) in quantity and therefore are ­non-replenishable. The non-renewable energy sources include, ­non-exclusively, fossil source like petroleum, natural gas, coal, etc.

Different Sources of Energy

There are various sources of energy that are used across the world by human beings to generate power and do the work. Over the years, there are other sources being discovered however, none of them has reached the stage where they can fulfill the power requirements of the modern era.

Human beings rely majorly on natural resources, unfortunately some of them are combustible. For this reason, we should look for artificial and alternate sources.

The reason why we are looking for alternate sources of energy is to produce electricity and run machines on a massive scale.

Solar Energy

Solar energy is the chief source of energy generated from the sun in the form of electric or thermal energy. Solar energy is captured in large solar panels made up of silicon and arsenic which convert the sun’s rays into usable electricity. Besides generating electricity, solar energy is also used in thermal applications.

Wind Energy

In wind energy, wind is used to generate electricity with the help of kinetic energy created by moving air. Wind energy is transformed into electrical energy with the help of wind turbines or wind energy conversion systems. The kinetic energy is then changed to rotational energy, by moving a shaft which is further connected to a ­generator which later produces electrical energy.

Geothermal Energy

Geo means ‘earth’ and thermal means ‘energy’;It is the produced from within the earth. Geothermal energy can be used for heating and cooling purposes or to generate clean electricity. However, for the generation of electricity, high or medium temperature resources are required.

Hydrogen Energy

Hydrogen has the ability to power fuel cells in zero-emission electric vehicles and the fuel cell’s potential for high efficiency. Hydrogen is a tremendous source of energy and can be used as a source of fuel to power ships, vehicles, homes, industries and rockets. It is renewable source of energy. It is environment friendly.

Tidal Energy

Tidal energy converts the energy obtained from tides into other forms of energy mainly electricity. Although because of its complications, it is not yet widely used. Tidal energy has potential for future electricity generation. Tidal power is a eco-friendly energy source.

Hydroelectric Energy

Hydroelectric energy involves water flowing through a pipe before pushing against and turning turbine blades connected to an electric generator. Hydropower provides 16 percent of the world’s electricity.

Biomass Energy

Biomass energy is produced from organic material and is commonly used throughout the world. Chlorophyll present in plants captures the sun’s energy by converting ­carbon dioxide present in the air and water from the ground into carbohydrates through the process of photosynthesis. When the plants are burned, the water and carbon dioxide is again released back into the atmosphere.

Biomass energy is used for heating and cooking in homes and also as a fuel in industrial production. This type of energy produces large amount of carbon dioxide into the atmosphere which is a greenhouse gas, thus it is not very efficient.

Nuclear Power

It is the most novel way of getting energy and quantitatively the most important renewable source of energy. Nuclear energy originates from the fission or fusion of uranium atoms. This produces massive heat to generate steam, which is then used by a turbine generator to generate electricity. Since, nuclear power plants do not burn fuel, they are environment friendly.

Fossil Fuels (Coal, Oil and Natural Gas)

Fossil fuels are the world’s dominant source of energy. Oil is converted into many products, the most used of which is gasoline. Natural gas is starting to become more common, vehicles are seen running on it. To get to the fossil fuel and convert it to use there has to be a heavy destruction and pollution of the ­environment.

The fossil fuel reserves are also limited, expecting to last only another 100 years given are the basic rate of consumption. It is estimated by 2040 that maximum of the fossil fuel deposits would get exhausted.

Understanding Energy Better

We come across different forms of energy in our daily lives. The biggest source of energy is the sun and also the tiniest source comes from nuclei of the atoms. Every action that we do involves energy from hitting a cricket ball, cycling, hitting a nail with the hammer, etc. The body which does work loses energy and the one on which work is done gains energy.

The demand for energy is increasing day by day and the pressure of extracting energy possesses a threat to our planet. We will discuss about energy demands and other threats later in the chapter.

A body which has energy has the capability of doing work. A body which has energy can exert a force on other bodies. Due to this, energy is transferred from one body to the other. The body which received energy may move/does work. The units of work are same as that of energy (Joule). Thus, 1 joule of energy is required to do 1 J of work.

Forms of Energy

The energy present around us is available in different forms like heat energy, sound energy, light energy, electricity, chemical, mechanical (kinetic + potential), etc. They can be interchanged from one form to another. When work done on an object is known, upon which energy is acquired then it is called mechanical energy. They are of two types:

Forms of energy

Kinetic Energy

The energy possessed by a body by virtue of its motion is called as kinetic energy. A body moving can do more work than a stable one. A rotating wheel, moving windmill, bullet fired from a gun, speeding car, etc. Greater the speed of the object, greater is the kinetic energy. Thus, objects in motion possess energy called as kinetic energy.

Now, student may ask how much energy (kinetic) does a moving body possess? We will try to devise a ­formula for that.

Consider the above figure:

Consider a body of mass ’m’ starts moving from rest, with uniform velocity ’u’. After a time interval ’t’ its velocity becomes v.

If initial velocity of the body is u or vi = 0, final velocity vf = v and the displacement of body is ’S’. Then

First of all we will find the acceleration of body.

Using the equation of motion

2aS = v_f² – v_i²

Putting the above mentioned values

2aS = v² – 0

a = \(\frac{v^2}{2 S}\)

Now force is given by

F = ma

Putting the value of acceleration

F = \(m\left(\frac{v^2}{2 S}\right)\)

As we know that

Work done = F⋅S

Putting the value of F

Work done = \(\left(\frac{m v^2}{2 S}\right)(S)\)

Work done = \(\frac{m v^2}{2}\)

or

Work done = \(\frac{1}{2} m V^2\)

Since the work done in motion is called ‘kinetic energy’

i.e. K.E. = Work done

or

K.E. = \(\frac{1}{2} m V^2\)

Potential Energy

The energy possessed by the body by virtue of its height or position is called as potential energy. The energy gets stored in the body because of the work done on the object. The energy while doing work is stored in the body as potential energy.

It does not cause any alterations in its speed or velocity. Some of the examples where we come across with potential energy are that of pulling of rubber band, winding the key of a toy car, water in the water tank at the top of the house, etc.

Potential Energy in a Body at Some Height

An object has energy known as potential energy when raised at a certain height. Greater the potential energy, greater is the height. Work is done on the object, against the gravity in bringing an object to a height. The energy gets stored in the object.

Gravitational potential energy of an object above the ground at a certain point is defined as raising the body to that point against the gravity.

Let us try to devise a formula to calculate the potential energy.

Consider an object of mass, m. Let it be raised through a height h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object mg. The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done,

W = force × displacement

= mg × h

= mgh

Since work done on the object is equal to mgh, energy equal to mgh units is gained by the object. This is the potential energy (EP) of the object.

E_p = mgh

The potential energy of an object at a height depends on the ground level or the zero level. An object in a given position can have a certain potential energy with respect to one level and a different value of potential energy with respect to another level. At ground potential energy is zero and at some height it is equal to mgh.

The potential attained by the body is independent of the path followed. The potential energy depends only on the initial and final positions. Energy is independent of the path followed.

Conservative and Non-Conservative Forces

Work done by gravity in moving object from one place to another depends only on the initial and final positions and doesn’t depend on the path taken. The work done by gravity from A to B is same by path 1, 2 and 3.

Hence the work done by such forces depends only on initial and final position and not on the path taken. Hence such forces are called conservative forces. Examples of conservative forces are Gravitational forces, spring force, etc.

On the other hand non conservative forces are those in which work done depends on path taken. For example fiction force is non conservative force.

Law of Conservation of Energy

This law states that energy can neither be created nor destroyed. It can only be changed from one form to the other. The total energy of the system remains constant.

According to the law of conservation of energy, energy can only be converted from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation remains the same.

The law of conservation of energy is valid in all situations and for all kinds of transformations.

Consider a simple example. Let an object of mass m be made to fall freely from a height h. At the start, the potential energy is mgh and kinetic energy is zero. Why is the kinetic energy zero? It is zero because its velocity is zero. The total energy of the object is thus mgh.

As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, the kinetic energy would be \(\frac{1}{2} m V^2\). As the fall of the object continues, the potential energy would decrease while the kinetic energy would increase.

When the object is about to reach the ground, h = 0 and v will be the highest. Therefore, the kinetic energy would be the largest and potential energy will be the least. However, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is,

Potential energy + kinetic energy = constant

\(m g h+\frac{1}{2} m v^2=\mathrm{constant}\)

The sum of kinetic and potential energies gives the total mechanical energy of the system.

Efficient use of Energy

Energy is very important to us. We need energy in every form or the other to sustain or to carry out any work. From the naturally gifted resources, we get energy in one form or the other, for example, coal, petroleum, sunlight, etc. It is our duty to preserve and protect these resources. Judicious use of energy is very important to keep the resources available for future generations.

 

Chapter 4 Work And Energy Track Your Learning Question And Answers

Question 1. ______ is defined as the capacity of doing work.

1. Work
2. Energy
3. Power
4. None

Answer. 2. Energy

Question 2. The main source of energy is ______, which gives energy to all the entities of the universe either directly or indirectly.

1. Sun
2. Sea
3. Ocean
4. Coal

Answer. 1. Sun

Question 3. The energy used in households, industries and commercial establishments is usually expressed in:

1. Joules
2. Watt
3. Kilowatt hour
4. None of the options

Answer. 3. Kilowatt hour

Question 4. The largest group of geothermal power plants in the world are located in ______.

1. Germany
2. India
3. United States
4. UK

Answer. 3. United States

Question 5. Tidal energy uses rise and fall of tides to convert ______ energy of incoming and outgoing tides into electrical energy.

1. Potential
2. Mechanical
3. Electrical
4. Kinetic

Answer. 4. Kinetic

Question 6. When the speed of a bike increases by 120%, then its kinetic energy increases by:

1. 4.84 times
2. 3 times
3. 1.44 times
4. By 240 %

Answer. 1. 4.84 times

Question 7. How fast should a man of 50 kg run so that his kinetic energy reaches 500 J?

1. 5 m/s
2. 10 m/s
3. 20 m/s
4. 2(5)1/2 m/s

Answer. 4. 2(5)1/2 m/s

Chapter 4 Work And Energy Rate of Doing Work

We all consume energy at some rate. Therefore, rate of consuming energy is called power. Power measures the speed of work done, that is, how fast or slow the work is done. Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by:

\(\text { Power }=\frac{\text { work }}{\text { time }} \text { or } P=\frac{W}{t}\)

The unit of power is watt having the symbol W. 1 watt is the power of an agent, which does work at the rate of 1 joule per second. We can also say that power is 1 W when the rate of consumption of energy is 1 J s^-1. 1 watt = 1 joule/second or 1 W = 1 J s^-1. We express larger rates of energy transfer in kilowatts (kW).

1 kilowatt = 1000 watts

1 kW = 1000 W

1 kW = 1000 J s^-1

Key Points to Remember

1. Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force. The unit of work is joule: 1 joule = 1 newton × 1 metre.
2. Work done on an object by a force would be zero if the displacement of the object is zero.
3. An object having capability to do work is said to possess energy. Energy has the same unit as that of work.
4. An object in motion possesses what is known as the kinetic energy of the object. An object of mass, m moving with velocity v has a kinetic energy of \(\frac{1}{2} m V^2\).
5. The energy possessed by a body due to its position called the potential energy. The gravitational potential energy of an object of mass, m raised through a height, h from the earth’s surface is given by mgh.
6. According to the law of conservation of energy, energy can only be transformed from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation always remains constant.
7. Energy exists in nature in several forms such as kinetic energy, potential energy, heat energy, chemical energy etc. The sum of the kinetic and potential energies of an object is called its mechanical energy.
8. Power is defined as the rate of doing work. The SI unit of power is watt. 1 W = 1 J/s.
9. The energy used in one hour at the rate of 1kW is called 1 kW h.

Chapter 4 Work And Energy Very Short Question and Answers

Question 1. Define power. What is the SI unit of power?
Answer:

Power:

Power is defined as total work done divided by total time taken. The SI unit of power is Watt.

Question 2. What is the commercial unit of energy? Convert it into Js^-1
Answer:

Commercial unit of energy:

The commercial unit of energy is kilowatt-hour (kWh).

Question 3. A boy of mass 60 kg runs up a staircase of 50 steps in 10 s. If the height of each step is 10 cm, find his power. Take g = 10 m s^-2.
Answer:

Weight of the boy = mg = 60 × 10 ms^-2

= 600 N

Height of the staircase,

h = 50 × 10/100 m

= 5 m

Time taken to climb = 10 s

P = work done/time

P = mgh/t

P = 600 × 5/10 s

P = 300 W

Thus, power is 300 W.

Question 4. An electric bulb of 60 W is used for 6 h per day. Calculate the ‘units’ of energy consumed in one day by the bulb.
Answer:

Power of electric bulb = 60 W

= 0.06 kW.

Time used,

t = 6 h

Energy = power × time taken

= 0.06 kW × 6 h

= 0.36 kWh

= 0.36 ‘units’.

The energy consumed by the bulb is 0.36 ‘units’.

Chapter 4 Work And Energy

Question 1. A pump transfers 500 L of water to the overhead tank of a building of height 12 m in 15 minutes. Calculate the power of motor pump.
(Take g = 10 ms^-2) (Mass of 1 L of water = 1 kg)

1. P = 66.7 W
2. P = 68 W
3. P = 72 W
4. P = 83 W

Answer. 1. P = 66.7 W

Question 2. A machine does 192 J of work in 12 Sec. What is the power of the machine?

1. 8 W
2. 2 W
3. 16 W
4. 10 W

Answer. 3. 16 W

Question 3. A weighting 500 kg runs up a hill rising himself vertically 10 m in 40 Sec. Calculate power. given g = 9.8 m^-1

1. 1220 W
2. 2000 W
3. 1623 W
4. 1225 W

Answer. 2. 2000 W

Question 4. A rickshaw puller pulls the rickshaw by applying a force of 200 N. If the rickshaw moves with constant velocity of 10 ms^-1. Find the power of rickshaw puller.

1. 2220 W
2. 2000 W
3. 2623 W
4. 1005 W

Answer. 2. 2000 W

Question 5. Calculate the time taken 60 W bulb to consume 3600 J of energy.

1. 3 min
2. 60 min
3. 6 min
4. 1 min

Answer. 4. 1 min

Chapter 4 Work And Energy Practice Exercises

Question 1. When a body falls freely towards the earth, then its total energy

1. increases
2. decreases
3. remains constant
4. first increases and then decreases

Answer. 3. remains constant

Question 2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car

1. does not change
2. becomes twice to that of initial
3. becomes 4 times that of initial
4. becomes 16 times that of initial

Answer. 1. does not change

Question 3. In case of negative work the angle between the force and displacement is

1. 0°
2. 45°
3. 90°
4. 180°

Answer. 4. 180°

Question 4. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5  kg. Both spheres are dropped simultaneously from a tower. When they are 10m above the ground, they have the same

1. acceleration
2. momenta
3. potential energy
4. kinetic energy

Answer. 1. acceleration

Question 5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g = 10 m s^-2)

1. 6 × 103 J
2. 6 J
3. 0.6 J
4. zero

Answer. 4. zero

Question 6. Which one of the following is not the unit of energy?

1. joule
2. newton metre
3. kilowatt
4. kilowatt hour

Answer. 3. kilowatt

Question 7. The work done on an object does not depend upon the

1. displacement
2. force applied
3. angle between force and displacement
4. initial velocity of the object

Answer. 4. initial velocity of the object

Question 8. Water stored in a dam possesses

1. no energy
2. electrical energy
3. kinetic energy
4. potential energy

Answer. 4. potential energy

Question 9. A body is falling from a height h. After it has fallen a height h/2 it will possess

1. only potential energy
2. only kinetic energy
3. half potential and half kinetic energy
4. more kinetic and less potential energy

Answer. 3. half potential and half kinetic energy

Chapter 4 Work And Energy Fill in the Blanks

Question 1. If force and displacement are in the same direction, the work would be ______.
Answer. Positive

Question 2. The work done on a 5 kg body to displace it by ______ is 5 J, given that force applied is 1 N.
Answer. 5 m

Question 3. Work done on a body get stored as ______.
Answer. Potential energy

Question 4. A bird sitting at a height has only ______ energy.
Answer. Potential

Question 5. Kinetic energy is a ______ quantity.
Answer. Scalar

Question 6. If the speed of a body is doubled, its kinetic energy must become ______.
Answer. Four times

Question 7. The sum of potential energy and kinetic energy is called ______ energy.
Answer. Mechanical

Question 8. When a ball is thrown up, ______ energy is converted into potential energy.
Answer. Kinetic

Question 9. Electricity is measured by electric meters installed in our homes that measure electric energy in units of ______.
Answer. Kilowatt-hour (kWh)

Question 10. Energy is a ______ quantity.
Answer. Scalar

 

Chapter 4 Work And Energy Match the Columns

Question 1. Match the column 1 with 2.

Select the correct option:

1. A-3, B-1, C-4, D-2
2. A-4, B-3, C-2, D-1
3. A-3, B-4, C-1, D-2
4. A-1, B-3, C-2, D-4

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

Question 2. Match the column 1 with 2.

Select the correct option:

1. A-2, B-3, C-4, D-1
2. A-4, B-3, C-2, D-1
3. A-3, B-4, C-1, D-2
4. A-1, B-3, C-2, D-4

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

Chapter 4 Work And Energy Assertion Reasoning

Direction: Choose the correct answer from the following choices:

1. Assertion and reason are both correct statements and reason is explanation for assertion.
2. Assertion and reason are both correct statements but reason is not the correct explanation for assertion.
3. Assertion is correct statement but reason is incorrect statement.
4. Assertion is incorrect statement but reason is correct.

Question 1. Assertion (A): Humans are machines, they can do lot of work.
Reason (R): People are able to do work because of energy.
Answer. 1. Assertion and reason are both correct statements and reason is explanation for assertion.

Question 2. Assertion (A): Thus, work done by force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force.
Reason (R): Work has both magnitude and direction.
Answer. 3. Assertion is correct statement but reason is incorrect statement.

Question 3. Assertion (A): Work done is negative when the force acts opposite to the direction of displacement.
Reason (R): Work done is positive when the force is in the direction of displacement.
Answer. 2. Assertion and reason are both correct statements but reason is not the correct explanation for assertion.

Question 4. Assertion (A): The energy possessed by an object is measured in terms of its capacity of doing work.
Reason (R): The unit of energy is, therefore, the same as that of work, that is, joule (J).
Answer. 1. Assertion and reason are both correct statements and reason is explanation for assertion.

Question 5. Assertion (A): Flowing water, ­blowing wind, a running athlete, etc., possess kinetic energy.
Reason (R): Kinetic energy is the energy possessed by an object due to its motion.
Answer. 1. Assertion and reason are both correct statements and reason is explanation for assertion.

Chapter 4 Work And Energy Comprehension Passage

An object at a certain height possess potential energy. Now, work is done on it, against the gravity by bringing the object to some other point.

Gravitational potential energy of an object above the ground at a certain point is defined as raising the body to that point against the gravity.

Let us try to devise a formula to calculate the potential energy.

Consider an object of mass m. Let it be raised through a height h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object mg.

The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done.

W = force × displacement

= mg × h

= mgh

Since work done on the object is equal to mgh, energy equal to mgh units is gained by the object. This is the potential energy (EP) of the object.

E_p = mgh

Question 1. When a body is raised above the ground against gravity, which energy is used?

1. Potential energy
2. Kinetic energy
3. Gravitational potential energy
4. Gravitational kinetic energy

Answer. 3. Gravitational potential energy

Question 2. ‘Mgh’ is the formula for

1. Kinetic energy
2. Potential energy
3. Both (a) and (b)
4. None of the above

Answer. 2. Potential energy

Chapter 4 Work And Energy Integer Type Question And Answers

Question 1. Calculate the kinetic energy of the body of mass 4 kg moving with the velocity of 0.1 metre per second.
Answer. 0.02 J

Question 2. How much work is done by a force 20 N in moving an object through a distance of 2 m in direction of the force?
Answer. 40 J

Question 3. What is the work to be done to increase the velocity of a car from 20 km h-1 to 40 km h^-1 if the mass of the car is 1200 kg?
Answer. 2.22 x 105 J

Question 4. 10 tube lights each of 50 W are operated for 15 hours. Calculate electrical energy consumed in ‘units’.
Answer. 7.5 kWh

Question 5. Rahul, having her own mass 50 kg, climbs 20 m height along with 30 kg mass in 40 s. Calculate her power and work done. (take g = 10 m/s²)
Answer. W = 16 kJ P = 400 W

Chapter 4 Work And Energy Very Short Question and Answers

Question 1. Define energy and its commercial unit. Express it in terms of Joules.
Answer:

Energy and its commercial unit.:

Energy is defined as the capacity of doing work. The commercial unit of energy is kilowatt-hour.

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J

Question 2. How are resources classified?
Answer:

Classification Of resources:

When energy source used is easily replenished in a short period and the reserve is limitless is called renewable source. The sources of energy which are limited in quantity and are exhaustible are called non-renewable sources of energy.

Question 3. Name the different sources of energy.
Answer:

The different sources of energy

The five sources of energy are solar energy, wind energy, geothermal energy, hydroelectric energy and biomass energy

Question 4. What are the different forms of energy?
Answer:

Different forms of energy:

The various forms include mechanical energy (potential energy + kinetic energy), chemical energy, heat energy, light energy and electrical energy.

Question 5. Define mechanical energy and its types.
Answer:

Mechanical energy and its types:

Mechanical energy is of two types i.e., the energy possessed by a body due to its motion is called kinetic energy. The energy possessed by a body due to its position or height is called potential energy.

Question 6. Define conservative and non-conservative sources of energy.
Answer:

Conservative and non-conservative sources of energy:

If the work done by a force depends only on initial and final positions but not on the path taken, then the force is said to be conservative. If the work done by a force depend on the path taken then the force is said to be non-conservative.

Question 7. Define law of conservation of energy.
Answer:

Law of conservation of energy:

Energy can neither be created nor destroyed. It can only be changed from one form to another.

Question 8. A car is moving with the uniform velocity of 50 km/h. What is the kinetic energy of the object of mass 40 kg kept in the car?
Answer:

Given:

A car is moving with the uniform velocity of 50 km/h.

We know that both objects and car are moving with the same velocity

i.e., v = 50 km/h = 50 × 103/(60 × 60)

= 13.88 m/s²

Mass of the object,

m = 40 kg

K.E. = \(\frac{1}{2} m V^2\)

K.E. = \(\frac{1}{2} \times 40 \times(13.88)^2\)

K.E. = 3853.08 J

Question 9. The acceleration due to gravity is 20 m/s², what will be the potential energy of a body of mass 1 kg kept at the height of 10 m?
Answer:

Given:

The acceleration due to gravity is 20 m/s²,

Potential Energy = mgh

Mass, m = 1 kg

Acceleration due to gravity,

g = 20 m/s²

Height, h = 10 m

Potential Energy = 1 × 20 × 10

= 200 J

Hence, potential energy is 200 Joules.

Question 10. Calculate potential energy of water in a tank having l = 4 m, b = 3 m and h = 2, situated on top of a house whose height is 9 m.
Take rw = 1 x 10³ kg/m³ and g = 10 ms^-2
Answer:

Volume of water = Volume of tank

= lbh

= 4 m × 3 m × 2 m

= 24 m³

Mass of water = Volume of density

= 24 × 10³ kg

Mean height of water above the ground

H = h₁ + h₂

= 9 m + 1 m

= 10 m

Therefore, Potential energy = MgH

= 24 × 10³ × 10 × 10

= 24 × 10⁵ J

Question 11. A body of mass 0.5 kg is in rest at a height of 20 m above the ground.

1. Find its total energy.
2. What is the total energy, if it starts falling freely and has fallen through a height of 5 m? (Take g = 10 ms^-2)
3. What conclusion can be drawn from these calculations?

Answer:

(1) P.E. of the body = mgh

= 0.5 × 10 × 20

= 100 J

K.E. of the body = \(\frac{1}{2} m V^2\)

= \(\frac{1}{2} \times 0.5 \times 0\)

= 0

Total energy = P.E. + K.E.

= 100 + 0

= 100 J

(2)

Velocity of the body at point B is given by

v² = u² + 2aS

= 0 + 2 × 10 × 5

= 100

∴ v = 10 ms^-1

∴ Its K.E. at point B = \(\frac{1}{2} m V^2\)

= \(\frac{1}{2} \times 0.5 \times(10)^2\)

= \(\frac{1}{2} \times 0.5 \times 100\)

= 25 J

Its P.E. at point B = mgh2

= 0.5 × 10 × 15

= 75 J

Its total energy at point B = K.E. + P.E.

= 25 J + 75 J

= 100 J

Thus, we find that the total energy of a freely falling body remains constant, though its potential energy gradually change into kinetic energy.

Question 12. Calculate the kinetic energy of a body of mass 4 kg moving with the velocity of 0.2 meter per second.
Answer:

Kinetic Energy = \(\frac{1}{2} m V^2\)

Here mass = 4 kg

Velocity = 0.2 m/s

So, by putting the values in formula:

Kinetic energy = \(\frac{1}{2} \times 4 \times(0.2)^2\)

= \(\frac{1}{2} \times 4 \times 0.2 \times 0.2\)

= 0.08

Question 13. Define work and its types.
Answer:

Work and its types:

If a force is applied on a body which led to displacement in the body then work is done. Work is the product of magnitude of force and displacement.

Positive work: If a force is applied in the same direction to displacement, then work done is said to be positive.

Negative work: If a force is applied in the opposite direction of displacement, then work done is said to be negative.

Question 14. A swimmer makes a swing jump between two points, by holding one end of a rope, other end of which is tied to some higher point. What type of work is done by rope in jumping of the swimmer from one point to another?
Answer:

Zero work done.

Question 15. Calculate the work done in operating the crane, if it lifts a mass of 1600 kg through a vertical height of 40 m. 
Answer:

g = 9.8 m/s²

F = mg = 1600 kg × 9.8 m/s²

= 15680 N

W = F × S = 15680 N × 40 m = 627,200 J

Question 16. How much work is done by a force of 15 N in moving an object through a distance of 2 m in the direction of force?
Answer:

W = F × S

Hence force, F = 15 N

Distance, S = 2 m

Work done, W = 15 × 2

W = 30 Joules

Chapter 4 Work And Energy Multiple Choice Question And Answers

Direction: Choose the correct option for each questions. There is only one correct response for each question.

Question 1. We need ______ for other activities like playing, singing, reading, writing, thinking, jumping, cycling and running.

1. Energy
2. Work
3. Motion
4. None of the above

Answer. 1. Energy

Question 2. What is the unit of work?

1. Newton metre (Nm)
2. Coulomb
3. Ampere
4. None of the above

Answer. 1. Newton metre (Nm)

Question 3. Which is the biggest natural source of energy to us?

1. Moon
2. Stars
3. Sea
4. Sun

Answer. 1. Moon

Question 4. On which factor the work done on an object does not depend?

1. Displacement
2. Force applied
3. Angle between force and displacement
4. Initial velocity of the object

Answer. 4. Initial velocity of the object

Question 5. Water stored in a dam possesses

1. No energy
2. Electrical energy
3. Potential energy
4. Kinetic energy

Answer. 4. Kinetic energy

Question 6. A body is falling from a height h. After it has fallen a height h/2 , it will possess

1. Only potential energy
2. Only kinetic energy
3. Half potential and half kinetic energy
4. More kinetic and less potential energy

Answer. 3. Half potential and half kinetic energy

Question 7. How are Joule (J) and ergs (erg) related?

1. J = 10⁷ erg
2. 1 erg = 10^-7 J
3. 1 J = 10^-7 erg
4. None of the above

Answer. 3. 1 J = 10^-7 erg

Question 8. Work done = Force × ______.

1. Displacement
2. Acceleration
3. Velocity
4. Speed

Answer. 2. Acceleration

Question 9. 1 Joule = 1 ______.

1. Nm²
2. kg m/s²
3. Nm
4. N²m²

Answer. 1. Nm²

Question 10. Which form of energy does the flowing water possess?

1. Gravitational
2. Potential
3. Kinetic
4. electricity

Answer. 3. Kinetic

Question 11. 3730 watts = ______ h.p.

1. 5
2. 2
3. 746
4. 6

Answer. 3. 746

Question 12. The P.E. of a body at a certain height is 200  J. The kinetic energy possessed by it when it just touches the surface of the earth is

1. > P.E.
2. < P.E.
3. = P.E.
4. None of the above

Answer. 1. > P.E.

Question 13. Power is a measure of the

1. Rate of change of momentum
2. Force which produces motion
3. Change of energy
4. Rate of change of energy

Answer. 4. Rate of change of energy

Question 14. 1.5 kW = ______ watts.

1. 150
2. 15000
3. 1500
4. 15

Answer. 2. 15000

Question 15. What is the energy of the simple pendulum when it is at its mean position?

1. Potential energy
2. Kinetic energy
3. Both (a) and (b)
4. Sound energy

Answer. 3. Both (a) and (b)

Question 16. Name the physical quantity which is equal to the product of force and velocity.

1. Work
2. Power
3. Energy
4. Current

Answer. 2. Power

Question 17. What is the example of kinetic energy in the following options?

1. A moving bus
2. A moving particle in electric field
3. A stretched rubber band just released
4. All the above

Answer. 2. A moving particle in electric field

Question 18. A light and the heavy body have equal momenta, which one has greater kinetic energy?

1. A light body
2. A heavy body
3. Both have same K.E
4. None of the above

Answer. 4. Both have same K.E

Question 19. Sum of kinetic and potential energy is called

1. Mechanical energy
2. Chemical energy
3. Electrical energy
4. Magnetic energy

Answer. 1. Mechanical energy

Question 20. Kinetic energy is given by

1. F × S
2. \(\frac{1}{2} m V^2\)
3. Mgh
4. None of the above

Answer. 1. F × S

Question 21. 1 kJ equals to

1. 100 J
2. 10 J
3. 1000 J
4. None of the above

Answer. 2. 10 J

Question 22. Work done in raising an object from the ground to that point against gravity is called

1. gravitational potential energy
2. gravitational kinetic energy
3. gravitational energy
4. none of the above

Answer. 3. gravitational energy

Question 23. The energy used in households, industries and commercial establishments are usually expressed in ______.

1. kilowatt hour
2. Joules
3. kilo joule
4. None of the above

Answer. 1. kilowatt hour

Question 24. Which is the formula of potential energy?

1. Mgh
2. \(\frac{1}{2} m V^2\)
3. F × S
4. None of the above

Answer. 1. Mgh

Question 25. What are the conditions needed for work to be done?

1. a force should act on an object and
2. the object must be displaced
3. there should be a chemical energy applied

1. 1 and 2
2. 2 and 3
3. 1 and 3
4. All the above

Answer. 1. 1 and 2

Question 26. Which of the following statements are true?

1. Work done by force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force.
2. Work has only magnitude and no direction.
3. Here the unit of work is newton metre (N m) or joule (J)
4. There are two conditions need to be satisfied for work to be done

1. 1 and 2
2. 2 and 4
3. 1, 2 and 3
4. All the above

Answer. 1. 1 and 2

Question 27. What is the unit of work?

1. Newton metre
2. Joule
3. Ampere
4. Coulomb

1. 1 and 3
2. 3 and 4
3. 1 and 2
4. None of the above

Answer. 4. None of the above

Question 28. Which statement is correct regarding energy?

1. The energy possessed by an object is measured in terms of its capacity of doing work.

2. The unit of energy is Joule.

3. 1 kJ equals 100 J.

1. 1 and 2
2. 2 and 3
3. 1 and 3
4. All the above

Answer. 3. 1 and 3

Question 29. Which of the following are the examples of kinetic energy?

1. A falling coconut
2. a speeding car
3. a rolling stone
4. a flying aircraft

1. 1 and 2
2. 2 and 3
3. 3 and 4
4. All the above

Answer. 1. 1 and 2

Question 30. Which of the following are examples of potential energy?

1. Flowing water
2. Blowing wind
3. A coiled spring
4. Wheels on roller skates before someone skates

1. 1 and 2
2. 3 and 2
3. 4 and 1
4. 3 and 4

Answer. 1. 1 and 2

Question 31. Which statement is correct?

1. Power is defined as the rate of doing work or the rate of transfer of energy.
2. Power = work/time
3. The unit of power is watt
4. 1 W = 1 J s^-1

1. 1 and 2
2. 2 and 4
3. 3 and 1
4. All the above

Answer. 4. All the above

Question 32. Work done in raising a box on a platform depends on

1. how fast it is raised
2. strength of the man
3.  negative work
4. zero work

Answer. 4. zero work

Question 33. Work done upon a body is

1. a vector quantity
2. a scalar quantity
3. always positive
4. always negative

Answer. 3. always positive

Question 34. Kilowatt hour (kWh) represents the unit of

1. power
2. impulse
3. momentum
4. none of these

Answer. 2. impulse

Question 35. When two unequal masses possess the same momentum, then kinetic energy of the heavier mass is _______ kinetic energy of the lighter mass.

1. same as
2. greater than
3. smaller than
4. much greater than

Answer. 4. much greater than

Question 36. The number of joules contained in 1 kWh is

1. 36 × 10²
2. 36 × 10³
3. 36 × 10⁴
4. 3.6 × 10⁶

Answer. 3. 36 × 10⁴

Question 37. A completely inelastic collision is one in which the two colliding particles

1. are separated after the collision
2. remain together after the collision
3. split into small fragments flying in all directions
4. none of the above

Answer. 4. none of the above

Question 38. A body moves through a distance of 3 m in the following different ways. In which case is the maximum work done?

1. When pushed over an inclined plane
2. When lifted vertically upward
3. When pushed over smooth rollers
4. When pushed on a plane horizontal surface

Answer. 2. When lifted vertically upward

Question 39. A truck and a car are moving on a smooth, level road such that the K.E. associated with them is same. Breakes are applied to both of them simultaneously. Which one will cover a greater distance before it stops?

1. Car
2. Truck
3. Both will cover the same distance
4. kinetic and potential energies

Answer. 2. Truck

Question 40. Two bullets P and Q, masses 10 and 20 g, are moving in the same direction towards a target with velocities of 20 and 10 m/s respectively. Which one of the bullets will pierce a greater distance through the target?

1. P
2. Q
3. Both will cover the same distance
4. Nothing can be decided

Answer. 3. Both will cover the same distance

Question 41. When the force applied and the displacement of the body are inclined at 90° with each other, then work done is

1. infinite
2. maximum
3. zero
4. unity

Answer. 1. infinite

Question 42. kg m² s^-2 represents the unit of

1. kinetic energy
2. work done
3. potential energy
4. all of these

Answer. 3. potential energy

Question 43. If sand drops vertically at the rate of 2 kg/sec on a conveyor belt moving horizontally with the velocity of 0.2 m/sec, then the extra force required to keep the belt moving is

1. 0.04 N
2. 0.08 N
3. 0.4 N
4. 0.2 N

Answer. 4. 0.2 N

Question 44. A body is dropped from a certain height to the ground. When it is halfway down, it possesses,

1. only K.E.
2. both K.E. and P.E.
3. only P.E.
4. zero energy

Answer. 3. only P.E.

Question 45. The energy required to raise a given volume of water from a well can be

1. mega watts
2. mega newton
3. mega joules
4. kilo watts

Answer. 2. mega newton

Question 46. If a force F is applied on a body and it moves with velocity v, then power will be

1. F ×v
2. \(\frac{F}{v^2}\)
3. \(\frac{F}{v}\)
4. F × v²

Answer. 3. \(\frac{F}{v}\)

Question 47. Which of the following graphs closely represents the P.E. (U) of a freely falling body and its height (h) above the ground?

1. 
2. 
3. 
4. 

Answer.

1. 

Question 48. The displacement x of a particle moving in one dimension, under the action of a constant force, is related to the time t by the equation, t x = + 3 where x is in metres and t in seconds. The displacement of the particle when its velocity is zero, is

1. 0
2. 6 m
3. 12 m
4. 18 m

Answer. 1. 0

Question 49. Asha lifts a doll from the floor and places it on a table. If the weight of the doll is known, what else does one need to know in order to calculate the work Asha has done on the doll?

1. Time required
2. Height of the table
3. Mass of the ball
4. Cost of the doll or the table

Answer. 1. Time required

Question 50. The work done in lifting a mass of 1 kg to a height of 9.8 m is

1. 1 J
2. (9.8)² J
3. 9.8 J
4. None of these

Answer. 2. (9.8)² J

Question 51. In which of the following cases, will the work done be maximum? The body is moved through a distance s on the ground

Answer.

Question 52. Work done by a centripetal force

1. increases by decreasing the radius of the circle
2. decreases by increasing the radius of the circle
3. increases by increasing the mass of the body
4. is always zero

Answer. 2. decreases by increasing the radius of the circle

Question 53. Certain weight is attached with a spring. It is pulled down and then released. It oscillates up and down. Its K.E. will be

1. maximum in the middle of the movement
2. maximum at the bottom
3. maximum just before it is released
4. constant

Answer. 4. constant

Question 54. A 1  kg mass falls from a height of 10 m into a sand box. What is the speed of the mass just before hitting the sand box? If it travels a distance of 2 cm into the sand before coming to rest, what is the average retarding force?

1. 12 m/sec and 3600 N
2. 14 m/sec and 4900 N
3. 16 m/sec and 6400 N
4. 18 m/sec and 8100 N

Answer. 1. 12 m/sec and 3600 N

Question 55. If L, M denote the angular momentum and mass of a particle and p its linear momentum, which of the following can represent the kinetic energy of the particle moving in a circle of radius R?

1. \(\frac{L^2}{2 M}\)
2. \(\frac{p^2}{M}\)
3. \(\frac{L^2}{2 M R^2}\)
4. \(\frac{1}{2} M p\)

Answer. 2. \(\frac{p^2}{M}\)

Question 56. A particle of mass 4 m which is at rest and explodes into three fragments. Two of the fragments each of mass mare found to move with a speed v in mutually perpendicular directions. The total energy released in the explosion is

1. 2mv²
2. \(\frac{1}{2} m v^2\)
3. mv²
4. \(\frac{3}{2} m v^2\)

Answer. 3. mv²

Question 57. kWh represents the unit for

1. force
2.  power
3. time
4. energy

Answer. 4. energy

Question 58. Energy cannot be measured in

1. Js
2. Ws
3. kWh
4. erg

Answer. 4. erg

Question 59. A steam engine converts

1. heat energy into sound energy
2. heat energy into mechanical energy
3. mechanical energy into heat energy
4. electrical energy into sound energy

Answer. 1. heat energy into sound energy

Question 60. A man throws bricks to the height of 12 m where they reach with a speed of 12 m/s. If he throws the bricks such that they just reach this height, then what percentage of energy will he save?

1. 19%
2. 38%
3. 57%
4. 76%

Answer. 2. 38%

Question 61. Two bodies with masses MA and MB are moving with equal kinetic energy. Their linear momenta are numerically in a ratio |PA| : |PB|, the ratio of their masses will be

1. MB : MA
2. MA : MB
3. \(\sqrt{M_A}: \sqrt{M_B}\)
4. \(M_{\mathrm{A}}^2: M_{\mathrm{B}}^2\)

Answer. 2. MA : MB

Question 62. Mechanical work done is equal to (symbols have their usual meanings)

1. W = F/S
2. W = FS
3. W = F + S
4. W = F – S

Answer. 3. W = F + S

Question 63. Which of the following graphs best represents the graphical relation between momentum (P) and kinetic energy (K) for a body in motion?

1. 
2. 
3. 
4. 

Answer.

2.

Question 64. An elevator is designed to lift a load of 1000 kg through 6 floors of a building on an average of 3.5 m per floor in 6 sec. Power of the elevator, neglecting other losses, will be

1. 3.43 × 10⁴ watt
2. 4.33 × 10⁴ watt
3. 2.21 × 10⁴ watt
4. 5.65 × 10⁴ watt

Answer. 4. 5.65 × 10⁴ watt

Question 65. When the momentum of a body increases by 100 %, then its K.E. increases by

1. 20 %
2. 40 %
3. 100 %
4. 300 %

Answer. 1. 20 %

Question 66. No work is said to have been done when an object moves at an angle of _____ with the direction of the force.

1. 0°
2. 90°
3. 180°
4. between 90° and 180°

Answer. 4. between 90° and 180°

Question 67. When a body is whirled in a circle, then work done on it is

1. positive
2. negative
3. zero
4. infinite

Answer. 2. negative

Question 68. A crane is used to lift 1000 kg of coal from a mine 100 m deep. If the time taken by the crane is 1 hr, then find the power of the crane, assuming the efficiency of the crane to be 80 %. (g = 9.8 m/s²)

1. 2567 watts
2. 2403 watts
3. 3403 watts
4. 3761 watts

Answer. 3. 3403 watts

Question 69. The flowing water of a river possesses

1. gravitational energy
2. potential energy
3. electrical energy
4. kinetic energy

Answer. 3. electrical energy

Question 70. The mass of an object P is double the mass of object Q. If both move with the same velocity, then the ratio of K.E. of P to Q is

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

Answer. 4. 4 : 1

Question 71. A truck can move up on a road having the gradient of 1 m rise for every 50 m with a speed of 15 km/hr. The resisting force is equal to 1/25 th the weight of the truck. How fast will the same truck move down the hill with the same horse power?

1. 30 km/hr
2. 45 km/hr
3. 60 km/hr
4. 75 km/hr

Answer. 2. 45 km/hr

Question 72. A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one fourth the initial velocity. The mass of the other body is

1. 3 kg
2. 0.6 kg
3. 2.4 kg
4. 4 kg

Answer. 2. 0.6 kg

Question 73. A body rolling down a hill has

1. K.E. only
2. P.E. only
3. Neither K.E. nor P.E.
4. Both (a) and (b) above

Answer. 2. P.E. only

Question 74. A total of 4900 joules was expended in lifting a 50 kg mass. The mass was raised to the height of

1. 98 m
2. 960 m
3. 245 m
4. 10 m

Answer. 4. 10 m

Question 75. A ball of mass 200 g falls from a height of 5 m. What is its K.E. when it just reaches the ground?

1. 9.8 J
2. 98
3. 980 J
4. None of these

Answer. 4. None of these

Question 76. A stretched spring possesses

1. kinetic energy
2. elastic potential energy
3. electric energy
4. magnetic energy

Answer. 1. kinetic energy

Question 77. When a person climbs a hill, he possesses

1. only K.E
2. only P.E.
3. both K.E. and P.E.
4. none of these

Answer. 2. only P.E.

Question 78. An iron sphere of mass 30 kg has the same diameter as an aluminium sphere whose mass is 10.5 kg. The spheres are dropped simultaneously from a cliff. When they are 10m from the ground, they have the same

1. acceleration
2. momentum
3. potential energy
4. K.E.

Answer. 3. potential energy

Question 79. A stone of mass m kg is whirled in a vertical circle of radius 20 cm. The difference in the kinetic energies at the lowest and the topmost positions is

1. 4 mg joules
2. 0.4 mg joules
3. 40 mg joules
4. None of these

Answer. 1. 4 mg joules

 Chapter 4 Work And Energy Long Answer Type Question And Answers

Question 1. A bullet of mass 50 g travelling horizontally with a speed of 200 ms^-1 strikes a glass pave 4 mm thick and falls dead, on emerging from it. How much work is done by the bullet against the opposing force of the glass pave?
Answer.

Given:

A bullet of mass 50 g travelling horizontally with a speed of 200 ms^-1 strikes a glass pave 4 mm thick and falls dead, on emerging from it.

Initial K.E. of the bullet = \(\frac{1}{2} m u^2\)

Final K.E. of the bullet = \(\frac{1}{2} m v^2\)

Loss in K.E. of the bullet = Initial K.E. – Final K.E.

= \(\frac{1}{2} m u^2-\frac{1}{2} m v^2\)

= \(\frac{1}{2} m\left(u^2-v^2\right)\)

By Work – Energy theorem,

Work done by a body = Loss in its K.E.

W = \(\frac{1}{2} m\left(u^2-v^2\right)\)

= \(\frac{1}{2} \times \frac{50}{1000}\left(200^2-0^2\right)\)

= \(\frac{1}{2} \times \frac{5}{100} \times 200 \times 200\)

= 1000J

= 1 x 10³ J

Question 2. If a heavy truck (mass = M) and a light car (mass m which is less than M) possess same amount of momentum, then which one possess greater kinetic energy? Explain.
Answer.

We know that momentum ‘p’ of a body of mass ‘m’ moving with velocity v is given by

p = mv

Squaring on both sides,

p² = (mv)² = m²v²

p² = m⋅mv

= \(2 m\left(\frac{1}{2} m v^2\right)\)

p² = 2mE_k

So, K.E., \(E_\kappa=\frac{p^2}{2 m}\)

If p is same, i.e., constant, then

\(E_K=\frac{C}{m}\)

i.e., E_k is inversely proportional to mass.

It  means, a body with smaller mass possess more K.E.

∴ Car will possess more K.E. than the truck.

Question 3. A body fall from height H. If t₁ is time taken for covering the first half height and t₂ be the time taken for second half. Which of these relations is true for t₁ and t₂.

1. t₁ > t₂
2. t₁ < t₂
3. t₁= t₂
4. Depends on the mass of the body

Answer.

Given:

A body fall from height H. If t₁ is time taken for covering the first half height and t₂ be the time taken for second half.

Let H be the height, then

First Half

\(\frac{H}{2}=\left(\frac{1}{2}\right) g t_1^2\)   (1)

Or  \(\left(\frac{1}{2}\right) g t_1^2=\frac{H}{2}\)

Also v = gt₁

Second Half

\(\frac{H}{2}=v t_2+\left(\frac{1}{2}\right) g t_2^2\)

or \(\left(\frac{H}{2}\right)=g t_1 t_2+\left(\frac{1}{2}\right) g t_2^2\)

or \(\left(\frac{1}{2}\right) g t_2^2=\left(\frac{H}{2}\right)-g t_1 t_2\)   (2)

t₂² + 2t₁t₂ − t₁² = 0

or \(t_2=\frac{\left[-2 t_1+\sqrt{\left(4 t_1^2+4 t_1^2\right)}\right]}{2}\)

or \(t_2=\frac{\left[-2 t_1+2 t_1 \sqrt{2}\right]}{2}\)

t₂ = 0.4 t₁

so, t₁ > t₂

Hence a is correct.

Question 4. A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline. The plane is 3 m long and is inclined at 20. the coefficient of friction between the package and the inclined plane is .40.What minimum initial K.E. must the boy supply to the package given as sin20 = .342 cos20 = .940?
Answer.

Given:

A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline.

The plane is 3 m long and is inclined at 20. the coefficient of friction between the package and the inclined plane is .40.

If the package travels the entire length S of the incline, the frictional force will perform work − μNS where μ is the coefficient of friction and N is normal reaction.

Let ‘h’ be the height of the incline plane then the gravitational potential energy of the package will increase by mgh.

Now let us assume ‘v’ be the speed given to the package so as to reach the top

Then kinetic energy at the initial point = \(\frac{1}{2} m v^2\)

Now applying work energy theorem

K.E_f – K.E_i = Work done by the gravitational force + Work done by the frictional force

Now since K.E_f = 0

Also Work done by the gravitational force

= –(change in gravitational potential energy)

= −mgh

Therefore

\(\frac{1}{2} m v^2\) = mgh – μNS

or \(\frac{1}{2} m v^2\) = mgh + μN

Now S = 3

N = mgcosθ

h = S sinθ

Substituting all the values

\(\frac{1}{2} m v^2\) = 42.2 J

Question 5. Deduce a formula for K.E of a body.
Answer.

A formula for K.E of a body:

Consider a body of mass ‘m’ starts moving from rest, with uniform velocity ‘u’. After a time interval ‘t’ its velocity becomes v.

If initial velocity of the body is u or vi = 0, final velocity vf = v and the displacement of body is ‘S’. Then

First of all we will find the acceleration of the body.

Using equation of motion

2aS = v_f² − v_i²

Substituting the above mentioned values

2aS = v² − 0

a = \(\frac{v^2}{2 S}\)

Now force is given by

F = ma

Substituting the value of acceleration

F = \(m\left(\frac{v^2}{2 S}\right)\)

As we know that,

Work done = F.S

Substituting the value of F

Work done = \(\left(\frac{m v^2}{2 S}\right)(S)\)

Work done = \(\frac{m v^2}{2}\)

or Work done = \(\frac{1}{2} m v^2\)

Since the work done in motion is called ‘Kinetic Energy’

i.e., K.E. = Work done

or \(\frac{1}{2} m v^2\)

Question 6. A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table. Calculate the work done by the force in 10 s and show that this is equal to the change in K.E. of the body.
Answer.

Given:

A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table.

Here, m = 1.0 kg, u = 0, F = 0.5 N, t = 10s

a = \(\frac{F}{m}=\frac{0.5}{1.0}=0.5 \mathrm{~ms}^{-2}\)

From S = \(u t+\frac{1}{2} a t^2\)

S = \(0+\frac{1}{2} \times 0.5(10)^2=2.5 \mathrm{~m}\)

Work done = F × S

= 0.5 × 25

= 12.5 J

From v = u + at = 0 + 0.5 × 10

= 5 ms–1

Change in K.E. = \(\frac{1}{2} m\left(v^2-u^2\right)\)

= \(\frac{1}{2} \times 1.0\left(5^2-0\right)\)

= 12.5 J

Chapter 4 Work And Energy Short Answer Type Questions

Question 1. A force of 20 N acts on a body to displace it through 10 m on a level road. Calculate the work done if force and displacement make an angle 60 degree with each other.
Answer. 

Given:

A force of 20 N acts on a body to displace it through 10 m on a level road.

Here force F = 20 N, Displacement, S = 10 m,

θ = 60 degrees

Now Work done, W = FS cos θ

= 20 × 10 × cos 60

= 20 × 10 × 0.5 = 100 J

Question 2. Define work. Write the formula of work when force and displacement do not act in same direction.
Answer.

Work:

Work is the product of force and displacement.

When force and displacement do not act in same direction, the formula for work is given by,

W = FS cos θ

Where θ is the angle between force and displacement.

Question 3. Mention the difference between potential energy and kinetic energy. A sliding box of mass 1 kg slows down from 10 m/s to 4 m/s. Calculate the change in kinetic energy.
Answer.

Given:

The difference between potential energy and kinetic energy. A sliding box of mass 1 kg slows down from 10 m/s to 4 m/s.

Potential energy of a body is the energy possessed by a body due to its position while kinetic energy of a body is the energy possessed due to velocity of the body.

Change in kinetic energy,

ΔK = 1/2m(v² – u²)

ΔK = (42 – 10²)/2

= -84/2 = -42 J

Question 4. Which physical quantity has Nm as its SI unit? Two different bodies of masses in the ratio of 1 : 2 are moving with same speed. What is the ratio of their kinetic energy?
Answer.

Given:

Two different bodies of masses in the ratio of 1 : 2 are moving with same speed

Work has the SI unit Nm.

Ratio of Kinetic energies

\(\frac{K_1}{K_2}=\frac{\left(\frac{1}{2} m_1 v_1^2\right)}{\left(\frac{1}{2} m_2 v_2^2\right)}\) \(\frac{K_1}{K_2}=\left(\frac{m_1}{m_2}\right)\left(\frac{v_1^2}{v_2^2}\right)\) \(\frac{K_1}{K_2}=\frac{1}{2} \times 1 \quad\left(\text { since } v_1=v_2 \& \frac{m_1}{m_2}=\frac{1}{2}\right)\)

So \(\frac{K_1}{K_2}\) = 0.5

Question 5. In lifting a body to 10 m, 800 J of work was done. Calculate its weight.
Answer.

Given:

In lifting a body to 10 m, 800 J of work was done.

Work done W = mgh

800 = mg × 10

mg = 800/10 = 80 N

Thus, weight of the body is 80 N.

Question 6. How much work is done in lifting a book weighing 800 g through a height of 1.5 m? (Take g = 10 ms^-2)
Answer.

We know that,

W = Force × displacement

⇒ W = mg × S

⇒ W = mgh (∵ S = h)

⇒ W = 0.8 kg × 10 ms^-2 × 1.5 m

⇒ W = 12 J

Question 7. When can work done by a force be zero? Give examples.
Answer.

We know that, in general work done is given by

W = FS cos θ

Where θ is the angle between direction of force and displacement.

W = 0 = FS cos θ

⇒ either S = 0

or cos θ = 0

i.e., θ = 90°

So work done by a force is zero.

When either displacement is zero or angle between directions of force and displacement is zero.

Examples

1. When a force is applied to a body and it doesn’t move

Example: a wall.

2. Displacement is in horizontal direction and force in vertical direction. So, work done by gravity on a bucket, which a man is carrying along horizontal ground.

Question 8. How much work is done in pulling a trolley through 6 m, as shown in the below figure?

Answer.

We know that,

W = FS cos θ

= 100 N × 6 m × cos(60°)

= 100 × 6 × 0.5 Nm

= 300 J

Question 9. You are standing on roof top of a tower of height 25 m. What is the potential energy of a ball of mass 150 g kept on the ground? (Take g = 10 ms^-2)
Answer.

In this case, potential energy will be negative since the body is below the reference level which is on the roof top.

P.E. = mgh

= 0.150 kg × 10 ms^-2 × (−25 m)

= −37.5 J

Note: This numerical tells us that potential energy of a body can be negative.

Example: Potential energy of coal in a coalmine is negative with respect to the observer on ground.

Question 10. If the velocity of a body is doubled, how will its kinetic energy change? Compare new kinetic energy with the old one.
Answer.

Consider a body of mass ‘m’ moving with a velocity ‘v1’.

Then, its

\(\text { K.E. }=\frac{1}{2} m v_1^2\)

∴ \(E_1=\frac{1}{2} m v_1^2\)    (1)

Now, its velocity is doubled.

So, v₂ = 2v₁

∴ Its new kinetic energy

\(E_2=\frac{1}{2} m v_2^2\)

= \(\frac{1}{2} m\left(2 v_1^2\right)\)

= \(4 \cdot \frac{1}{2} m v_1^2\)

= 4 E₁ form (1)

Thus, its kinetic energy becomes four times.

Question 11. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in m/sec) of a particle is zero, than what is the speed (in m/sec) after 5 s?
Answer.

Given:

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle.

We know that

\(\text { Power }=\frac{d W}{d t}\)

W = 0.5 × 5 = 2.5 = K.E._f − K.E._i

Since the initial speed of the particle is zero, the initial kinetic energy will also be zero.

⇒ \(2.5=\frac{M}{2}\left(v_f^2-v_i^2\right)\)

⇒ \(2.5=\frac{M}{2}\left(v_f^2-0\right)\)

⇒ \(v_f^2=2.5 \times \frac{2}{0.2}=25\)

⇒ v_f = 5

Question 12. Deduce the formula of P.E. of a body.
Answer.

P.E. of a body:

Consider an object of mass, m. Let it be raised through a height h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object mg.

The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done,

W = Force × Displacement

= mg × h

= mgh

Since work done on the object is equal to mgh, the energy equal to mgh units is gained by the object. This is the potential energy (EP) of the object.

E_p = mgh

Question 13. When a rubber is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx², where a and b are constants. What is the work done in stretching the un-stretched rubber b and by L?
Answer.

Given:

When a rubber is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx², where a and b are constants

We have

F = ax + bx²

W = Fdx

The work done by stretching the un-stretched rubber b and by L is

\(W=\int_0^L\left(a x+b x^2\right) d x\)

= \(\left[a\left(\frac{x^2}{2}\right)+b\left(\frac{x^3}{3}\right)\right]_0^L\)

= \(a\left(\frac{L^2}{2}\right)+b\left(\frac{L^3}{3}\right)\)

Chapter 3 Gravitation Long Answer Type Question And Answers

Question 1. Interconnected vessel as shown in figure is filled with an ideal liquid P₁ and P₂ are airtight pistons having areas A₁ = 2 cm² and A₂ = 5 cm² respectively. A weight of 3 kg is kept on piston P₁. Calculate

1. pressure acting on piston P₁
2. pressure on piston P₂
3. force with which P₂ moves up.

Answer.

Given:

Interconnected vessel as shown in figure is filled with an ideal liquid P₁ and P₂ are airtight pistons having areas A₁ = 2 cm² and A₂ = 5 cm² respectively.

A weight of 3 kg is kept on piston P₁.

Pressure P₁ exerted by the piston P₁ on the confined liquid is given by

P₁ = \(\frac{\text { Force }}{\text { Area }}=\frac{\text { Weight }}{\text { Area }}=\frac{M_g}{A_1}\)

= \(\frac{3 \times 10 \mathrm{~N}}{2 \times 10^{-1} \mathrm{~m}^2}\)

P₁ = 15 × 10⁴ Pa, (downwards)

By Pascal’s law, this pressure is communicated to the piston P₂.

∴ Upward pressure P₂ acting on piston P₂

= P₁

= 15 × 10⁴ Pa, (upwards)

We know that,

P = \(\frac{F}{A}\)

∴ F = PA

= P₂A₂

= \(15 \times 10^4 \frac{\mathrm{N}}{\mathrm{m}^2} \times 5 \times 10^{-4} \mathrm{~m}^2\)

= 75 N

Read and Learn More NEET Foundation Long Answer Questions

Question 2. Nine tenth of an iceberg having a total volume of 500 m3 is submerged in ocean. Calculate the buoyant force acting on the iceberg.

1. Take ρoceanwater = 1.02 × 10³ kgm^-3
2. g = 10 ms^-2

Answer.

Given:

Nine tenth of an iceberg having a total volume of 500 m3 is submerged in ocean

Buoyant force or upthrust is given by

U = Vρ g

where

V = Volume of immersed part of solid.

= \(\left(\frac{9}{10} \times 500\right) \mathrm{m}^3\)

= 450 m³

So U = V ρ g

= \(450 \mathrm{~m}^3 \times 1.02 \times 10^3 \frac{\mathrm{kg}}{\mathrm{m}^3} \times 10 \mathrm{~m}\)

= 45 × 1.02 × 10⁵ N

= 4.59 × 10⁶ N

Question 3. A piece of metal weighs 200 gf in air and 180 gf in water. Finds its relative density.
Answer.

Given:

A piece of metal weighs 200 gf in air and 180 gf in water.

\(\text { R. D. of a solid }=\frac{\text { Weight in air }}{\text { Loss of weight in water }}\)

= \(\frac{W_1}{W_1-W_2}\)

= \(\frac{200 \mathrm{gf}}{(200-180) g f}\)

= \(\frac{200}{20}\)

∴ RD of metal = 10 (No unit).

Question 4. Imagine a planet whose both diameter and mass are one half of the Earth. The day’s temperature of this planet’s surface reaches up to 800 K. Find whether oxygen molecules are possible in the atmosphere of this planet.
Answer.

Given:

Imagine a planet whose both diameter and mass are one half of the Earth. The day’s temperature of this planet’s surface reaches up to 800 K.

Escape velocity,

v_e = \(\sqrt{2} G M / R\)

Let v_p = escape velocity on the planet

v_e = escape velocity on the earth

\(\frac{v_p}{v_e}=\sqrt{\left(\frac{M_p}{R_p} \times \frac{R_e}{M_r}\right)}=\sqrt{\left(\frac{1}{2} \times \frac{2}{1}\right)}=1\)

v_p = v_e = 11.2 km/s

From kinetic theory of gases

v_rms = \(\sqrt{3 R T / M})=\sqrt{3 N K T / M}=\sqrt{3 N K T / N m}\)

where N = Avogadro’s number

m = mass of oxygen molecule

K = Boltzmann constant

v_rms = \(\sqrt{3 R T / m}\)

= \(\sqrt{\left(\frac{\left(3 \times 1.38 \times 10^{-23} \times 800\right)}{5.3 \times 10^{-2 h}}\right)}\)

(m = 5.3 × 10−26 kg)

v_rms = 0.79 km/s

As v_rms is very small compared to the escape velocity on the planet, molecules cannot escape from the surface of the planet’s atmosphere.

Question 5. Deduce the relationship between g and G.
Answer.

Relationship between g and G:

Let g = acceleration due to gravity at a planet

M = mass

R = radius

m = mass of object

By Newton’s law of motion, force on a body due to gravity on its surface

F = mass × acceleration due to gravity

= m g

By Newton’s gravitational law, force is

F = GMm/R²

Therefore,

GMm/R² = mg

Acceleration due to gravity g = GM/R²

Question 6. How much below the surface of earth does the acceleration due to gravity

1. reduce to 36%
2. reduce by 36%, of its value on the surface of earth, (radius of earth = 6400 km).

Answer. Case (1)

We have,

g_h = g (1 – d/R)

Or d = (g – g_h) R/g (1)

Here

g_h = 36/100 g (2)

Using (1) and (2)

g_h = 36/100 g

d = (1 – 36/100) R

d = (100 – 36)/100 × 6400

d = 4096 km

Case (2)

Here

g_h = g – 36/100 g

i.e., g_h = 64/100 g (3)

Using equation (1) and (3)

d = (g – 64/100 g) 6400/g

d = 100 – 64/100 × 6400

d = 2304 km

Question 7. Deduce gravitational force between

1. gravitational force between earth and the  sun and
2. gravitational force between the moon and  the earth.

Answer. (1) Gravitational force between earth and the sun

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of sun, m₂ = 2 × 10³⁰kg

Distance between sun and earth, R = 1.5 × 10¹¹m

Gravitational force between the sun and the earth,

F = \(G \frac{m_1 m_2}{r^2}\)

F = (6.67 × 10^-11 Nm² kg^-2 × 6 × 10²⁴ kg × 2 × 10³⁰kg)/(1.5 × 10¹¹m) ²

F = 3.6 × 10²² N

(2) Gravitational force between the moon and the earth

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of moon, m₂ = 7.4 × 10²² kg

Gravitational force between earth and the moon is

F = \(G \frac{m_1 m_2}{r^2}\)

F = (6.67 × 10^-11Nm²kg^-2× 6 × 10²⁴kg × 7.4 × 10²² kg)/(3.8 × 10⁸m)²

F = 2.05 × 10²⁰ N

Question 8. Explain why a sheet of paper falls slower than one that is crumpled into a ball?
Answer.

The sheet of paper falls slower than the one that is crumpled into a ball because the air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the sheet of paper is more than the resistance offered by air to the paper ball because the sheet has larger area.

Question 9. The escape velocity of a body on the surface of the earth is 11.2 km/s. A body is projected away with twice this speed. What is the speed of the body at infinity? Ignore the presence of other heavenly bodies.
Answer.

If v is the velocity of projection and v’ is the velocity at infinity, then we have by energy conservation principle.

\(\frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m v^{\prime 2}+0\)

Here v = 2v_e

Thus,

\(\left(\frac{1}{2}\right) \cdot 4 v_e^2-\frac{G M}{R}=\frac{1}{2} v^{\prime 2}\) \(2 v_e^2-\frac{G M}{R}=\frac{1}{2} v^{\prime 2}\)

Now, v_e = \(\frac{\sqrt{2 G M}}{R}\)

\(2 v_e^2-\frac{v_e^2}{2}=\frac{1}{2} v^{\prime 2}\)

Or, v’² = 3 v_e²

Or, v’ = \(\sqrt{3} v_e=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}=19.4 \mathrm{~km} / \mathrm{s}\)

Question 10. Deduce the equation showing the variation in the value of g with depth below the surface of the earth.
Answer.

Let us consider a body of mass m at a depth h below the surface of earth. Then, radius of the inner solid sphere of the earth = R – h.

Volume of the inner solid sphere of the earth = \(\frac{4}{3} \pi(R-h)^3 d\)

If d is the average density of the earth, then

Mass of inner solid sphere of earth = \(\frac{4}{3} \pi(R-h)^3 d\).

According to the law of gravitation,

mg_d = \(G \times \frac{4}{3} \pi(R-h)^3 d \times m /(R-h)^2\)

This gives

g_d = \(G \times \frac{4}{3} \pi(R-h) d (1)\)

On the surface of earth,

g = \(\frac{G M}{R^2}\)

= \(G \times \frac{4}{3} \pi R^3 \frac{d}{R^2}\)

= \(G \times \frac{4}{3} \pi R d (2)\)

From equation (1) and (2)

g_d/g = \(G \times \frac{4}{3} \pi(R-h) \times d / G \times \frac{4}{3} \pi R d\)

= (R – h)/R

Or g_d = g (1 – h/R) or (R – h)/R < 1

So, g_d < g

Thus, the value of g at a depth inside the earth is less than that on the surface of the earth.

The value (R – h)/R decreases with the value of h, i.e. depth below the surface of the earth. So the value of g decreases as we go down below the surface of earth.

Question 11. A stone is dropped from the edge of the roof.

1. How long does it take to fall 4.9 m?
2. How fast does it move at the end of the fall?
3. How fast does it move at the end of 7.9 m?
4. What is its acceleration after 1 s and after 2 s?

Answer. (1) As the stone is dropped, its initial velocity,

u = 0, h = 4.9 m

a = g = 9.8 ms^-2, time, t =?

From

h = \(u t+\frac{1}{2} g \mathrm{t}^2\)

4.9 = \(0+\frac{1}{2} \times 9.8 t^2=4.9 t^2\)

Or \(f^2=\frac{4.9}{4.9}=1, t=\sqrt{1}=1 \mathrm{~s}\)

(2) Final velocity, v = ? at t = 1 s

From

v = u + gt

v = 0 + 9.8 × 1 = 9.8 m s^-1

(3) Let v be the final velocity when h = 7.9 m

From

v² – u² = 2ah

v² – 0 = 2 (9.8) 7.9

Or v = \(\sqrt{2 \times 9.8 \times 7.9}=\sqrt{154.84}\)

v = 12.4 m s^–1

(4) The acceleration of a freely falling body remains the same at all times, i.e., a = g = 9.8 ms^-1 after 1 s and 2 s.