Sound Waves in Class 11 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

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## Mind Maps

*Class 11*

*Class 11*

**Kinematics 1D****Kinematics 2D****Gravitation****Simple Harmonic Motion (SHM)****Mechanical Properties of Fluids: Fluid Dynamics****Mechanical Properties of Fluids:Fluid Statics****Mechanical Properties of Fluids:Viscosity****Mechanical Properties of Fluids:Surface Tension****Sound Waves**

**Class 12**

**Class 12**

**Electric Potential****Electric Field and Charges****Capacitor****Electromagnetic Induction (EMI)****Ray Optics: Reflection of Light****Ray Optics : Refraction****Ray Optics : Optical Instruments****Alternating Current****Magnetic Effects of Current****Magnetism and Matters****Current Electricity****Wave Optics****Wave on String****Modern Physics: Atomic Structure****Modern Physics: Photoelectric Effect****Modern Physics: Nuclei****Modern Physics: Radioactivity***Modern Physics: X Rays*

## Physics Revision Videos for Class 11, 12, JEE & NEET

### Class 11 Chapters for Revision

**Kinematics 1D****Kinematics 2D****Gravitation****Simple Harmonic Motion (SHM)****Sound waves****Mechanical Properties of Fluids: Fluid Statics & Dynamics****Mechanical Properties of Fluids: Viscosity****Mechanical Properties of Fluids: Surface Tension**

### Class 12 Chapters for Revision

**Electrostatics****Capacitor****Current Electricity****Magnetic Effect of Current****Magnetism and Matters****Alternating Current****Electromagnetic Induction (EMI)****Ray Optics****Wave Optics****Wave on String****Modern Physics: Atomic Structure****Modern Physics: Photoelectric Effect****Modern Physics: Nuclei****Modern Physics: Radioactivity****Modern Physics: X Rays**

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Lets study about Moment of Inertia of a Rigid Body and Physical Significance of Moment of Inertia here.

**MOMENT OF INERTIA OF RIGID BODY**

It is the property of a body due to which it opposes any change in its state of rest or uniform rotation. Analytically for a particle of mass m rotating in a circle of radius R, moment of inertia of a particle of mass m about the axis of rotation is given by:

$\mathrm{I}=\mathrm{mR}^{2}$ . . . . .(1)

So for a body made up of number of particles (discrete distribution) of masses $m_{1}, m_{2}, \ldots \ldots \ldots$. etc. at

distance $r_{1}, r_{2}, \ldots$ etc. respectively from the axis of rotation, then moment of inertia of all these particles are

$m_{1} r_{1}^{2}, m_{2} r_{2}^{2}, \ldots$ and moment of intertia of body ( $I$ ) is the algebric sum of moment of inertia of all these

particles-

i.e. $\quad \mathrm{I}=\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2} \mathrm{r}_{2}^{2}+\ldots \ldots+\mathrm{m}_{\mathrm{r}_{1} \mathrm{i}}^{2}+\mathrm{m}_{\mathrm{n}} \mathrm{r}_{\mathrm{n}}^{2}=\sum_{i=1}^{\mathrm{n}} \mathrm{m}_{\mathrm{r}_{1}} \mathrm{r}_{1}^{2} \ldots \ldots(2)$

$\quad \quad \mathrm{m}_{\mathrm{i}}=\mathrm{mass}$ of ith particle

$r_{i}=$ distance of ith particle from axis of rotation

While for a continuous distribution of mass, treating the element of mass dm as particle at position r from

axis of Rotation.

$$

\begin{array}{ll}{\text { dI }=\mathrm{dm} \mathrm{r}^{2},} & {\text { i.e., } \quad \mathrm{I}=\int \mathrm{r}^{2} \mathrm{d} \mathrm{m}}\end{array}

$$ . . . . (3)

It is a scalar quantity

Unit : M.K.S.: $\mathrm{kg}-\mathrm{m}^{2},$ C. G.S. : $\mathrm{gm}-\mathrm{cm}^{2}$

Dimension : $\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{0}$

Moment of inertia depends on the following factors.

i) Mass of body

ii) Mass distribution of body or shape, size and density of body.

iii) On the position of the axis of rotation.

The more is the distribution of mass with respect to axis of rotation the more will be moment of

inertia.

**Moment of Inertia does not depend on the following factors.**

(i) Angular velocity

(ii) Angular Acceleration

(iii) Torque

(iv) Angular Momentum

**PHYSICAL SIGNIFICANCE OF MOMENT OF INERTIA:**Comparison of the expression for rotational motion with corresponding relations for translatory motion as

given in table I, we get moment of inertia plays same role in rotatory motion as mass in translatory

motion, i.e., if a body has large moment of inertia, it is difficult to start rotation or to stop it if rotating. Large

moment of inertia also helps in keeping the motion uniform. Due to this reason stationary engines are

provided with fly-wheels having large moment of inertia.

**Inertia for rotational motion is moment of inertia.**

Introduction to Rotational Dynamics

Moment of Inertia: Perpendicular and Parallel axis theorem

Law of Conservation of Angular Momentum

Conservation of Angular Momentum Examples

Kinetic Energy of a Rotating Body

Combine Translational and Rotational Motion

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Get important questions of Wave for Board exams. Download or View 11th Physics important questions for exam point of view. These important questions will play significant role in clearing concepts of Physics. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 Physics chapter wise CBSE.

Click Here for Detailed Notes of any chapter.

**Q.1. What is the angle between particle velocity and wave velocity in ( 1) Transverse wave ( 2) Longitudinal wave?**

**Ans.** Angle between particle velocity and wave velocity

for transverse wave is $\pi / 2$ and that in case of

longitudinal wave is either zero or $\pi$.

**Q.2 Explain why it is possible to detect the approaching of a distant train by placing the ear very close to railway line?**

**Ans.** This is so because sound travels 16 times faster in iron than in air.

**Q.3 What will be the nature of graph between the pressure of a gas and the speed of sound waves passing through the gas?**

**Ans**. The graph will be a straight line parallel to pressure axis, as speed of sound is independent of the pressure of gas.

**Q.4 When sound waves travel from air to water, does the frequency of wave changes? The wavelength ? The speed?**

**Ans. **The frequency does not change. The speed and wavelength change.

**Q.5 What are those two important characteristics which a material medium must have for the propagation of mechanical waves?**

**Ans.** Elasticity and Inertia.

**Q.6 Why explosions on the other planets can not be heard on the earth?**

**Ans.** Because there is no material medium in the space between the planets, through which sound may

**Q.7 the violin is taken to a hot room, how will the pitch of the note produced be affected?**

**Ans**. The frequency of the note emitted gets increased.

**Q.8 What do you mean by overtones ? **

**Ans.** The notes produced by an instrument of sound other than the fundamental note are called overtones.

**Q.9 What physical quantity changes when a source of sound moves and the listener is stationary ?**

**Ans.** Wavelength of sound waves changes.

**Q.10 Will be Doppler effect, when the directions of motion of the source or observer is perpendicular to the direction of propagation of sound ? **

**Ans.** No, there is no Doppler’s effect, when the source or observer moves perpendicular to the direction of propagation of sound.

**Q.11 Given below are some examples of wave motion. State in each case, if the wave motion is transverse, Longitudinal or a combination of both **

**(i) Motion of a kink in a long coil spring produced by displacing one end of spring sideways.**

**(ii) Waves produced in a cylinder containing a liquid by moving its piston back and forth.**

**(iii) Waves produced by motor boat sailing in water.**

**(iv) Light waves travelling from sun to the earth.**

**(v) Ultrasonic waves in air produced by a vibrating quartz crystal.**

**Ans.** (i) Wave motion is transverse, because the vibration of particles (Kinks) of the spring are at right angles to the direction of wave propagation.

(ii) The wave motion is longitudinal, because the molecules of the liquid vibrate to and fro about their mean position along the direction of propagation of the wave.

(iii) The given wave motion is a combination of longitudinal and transverse waves.

(iv) The wave motion is transverse, because the light wave is an electromagnetic waves in which electric and magnetic fields vibrate in the direction perpendicular to each other and also to the direction of propagation of the wave.

(v) Ultrasonic waves produced by a quartz crystal in air are longitudinal, because the molecules of air vibrate to & fro about their mean positions along the direction of propagation of wave due to vibration of quartz crystal.

**Q.12 **Why the shape of a pulse gets deformed during propagation in a dispersive medium ?

**Ans.** When a pulse passes through a dispersive medium, the wavelength of wave changes. Hence, the shape of pulse changes i.e. it gets deformed.

**Q.13** When a stone is thrown on the surface of water. Waves travel out, from where does the energy come ? [

**Ans.** The energy carried by the water waves comes from Kinetic energy of stone hitting the water surface.

**Q.14 Why is sound heard in $\mathrm{CO}_{2}$ more intense in comparison to sound heard in air ? **

**Ans.** This is because intensity of sound increases with increase in density of the medium.

**Q.15 Which type of hearing aids are required by persons on the surface of moon and why?**

**Ans.** We require those hearing aids, which may transmit and detect electromagnetic waves, because there is no atmosphere on the moon.

**Q.16 Explain why frequency is the fundamental property of a wave ? **

**Ans **When a wave travels from one medium to other, its length as well as velocity may change but frequency does not change. This is the reason, we say that frequency is the fundamental property of a wave.

**Q.17 Explain, why does sound travels faster on a cloudy day as compared to that on a dry day ? [**

**Ans. **The amount of water vapour present in the atmosphere is much greater on a cloudy day than on a dry day. As the water vapours are lighter than dry air, hence density of wet air, becomes, less than that of dry air. Now, because the speed of sound is inversely proportional to square root of the density, hence sound travels faster on a cloudy day than on a dry day.

**Q.18 If a balloon is filled with $\mathbf{c} o_{2}$ gas then how will it behave for sound as a lens ? On filling of hydrogen gas ? **

**Ans.** In $\mathrm{CO}_{2}$, the velocity of sound is less than air, when, the balloon filled with will behave as convex lens. On the other hand the balloon filled with gas will behave as concave lens because velocity of sound is greater in hydrogen than in air.

**Q.19 Why is the sound produced in air not heard by a person deep inside the water ? **

**Ans.** There is a great difference in speed of sound in air and in water. Hence, there is little refraction of sound from air into water, most of the sound is reflected from surface of water. It is due to this reason that where there is a glass wall between two rooms, the conversation in one room can not be heard in other room.

**Q.20 Explain why sound of very high velocity can not be transmitted through a material of high density but of low elasticity ? **

**Ans**. The speed of sound wave is given by,

$v=\sqrt{\frac{E}{\rho}}$

**Q.21 Explain why interference is not possible in the sounds produced by two violins ?**

**Ans.** For interference between two waves the phase difference between the waves must remainconstant. Each violin produces wave-train of sound discontinuously. Hence, the phase of a wave-train arriving at a point varies with time. Clearly, the phase difference between waves coming at a point from two separate violins will vary with time. Hence, an average intensity of sound will be heard everywhere.

**Q.22 As in sound can beats be observed by two light sources ? Explain. **

**Ans.** No, to observe beats by two light sources the phase difference between the sources should change regularly. In light sources however this change occur at random, because the light source consists of a large number of atoms and each atom emits wave independently

**Q.23 Explain why there are so many holes in a flute ?**

**Ans.** The flute is basically an open organ pipe. The location of the open end can be changed by keeping the one hole open and closing the other holes. Thus, the frequency of the note produced by the flute can be changed.

**Q.24 A bucket is placed below a water top. We can estimate the height of water level reached in the bucket from a distance simply by hearing the sound. Explain, why it is so ?**

**Ans.** The frequency of note emitted by an air column is inversely proportional to its length $\left(n \propto \frac{1}{l}\right)$. Therefore, as the length of air column decreases, frequency increases. i.e. the note becomes shrill. As the water level in bucket rises, the length of air column of the bucket goes on decreasing and the sound produced goes on becoming more and more shrill. On this basis we can estimate from a distance the height up to which the bucket has been filled with water.

**Q.25 Why does the pitch of a note produced by a wooden open end pipe becomes sharper when the temperature rises ? [2]**

**Ans.** With rise in temperature, the velocity of sound increases. The fundamental frequency of an open

prgan pipe is given by $n=\frac{v}{2 l}$, hence with increase

in value of v, n increases and the pitch of note becomes sharper.

**Q.26 What is the difference between an echo and a reverbration ? **

**Ans.** An echo is produced when sound reflected from a distant obstacle comes back after an interval of

$\frac{1}{10}$ second or more. In an echo, the original and reflected sounds are heard separately.

Reverbration, on the other hand, consists of successive reflections which follow each other so **quickly that they can not produce separate echos. **

**Q.27. A violin note and a sitar note may have same frequency and yet we can distinguish between two notes. Explain, why it is ? **

**Ans.** This is due to the fact that overtones (integral multiples of fundamental frequency) produced by two sources may be different, in other words the quality of sound produced by two instruments of same fundamental frequency is different.

**Q.28 Of what material is the running fork made and why ? **

**Ans**. A tunning fork is made of an alloy of steel, Nickel and chromium, called elinvar i.e. the material for which the elasticity does not change.

**Q.29 Why a given sound is louder in a hall than in the open ? **

**Ans.** In a hall, repeated reflections of sound takes place from the walls and the ceiling. The reflected sounds mix up with the direct sound, resulting in increase in the intensity of sound in the hall. But in open, no such reflections of sound are possible and as such faint sound is heard in open.

**Q.30 Why all the stringed instruments are provided with hollow boxes ? **

**Ans.** The stringed instruments are provided with hollow boxes in order to increase the surface area of vibration, which increases the loudness or intensity of sound produced, $(\text { As } I \propto A)$ . Moreover the air inside the hollow box is set into forced vibrations which also increases the loudness of sound produced.

**Q.31 Explain how the bats ascertain distances, directions, nature and size of the obstacles without any eyes ? **

**Ans**. The bats can emit ultrasonic waves (sound waves of frequency above . The waves emitted by bats are reflected by the obstacles and is received by it. Bats can also detect the ultrasonic waves. The time interval between the emission of ultrasonic wave and its reception back after reflection give the information of distance, nature of obstacle and its direction. Then bats can ascertion distances, direction, nature and size of the object inspite of their poor eyesight.

**Q.32 In summer, sound of a siren is heard louder in the night than in the day to a person on earth. Explain why it is so ? **

**Ans.** During the day time in summer the surface of the earth becomes hot due to sun rays. Hence the air near the earth is hotter than the air above. i.e. as we move above the earth, the air becomes denser. Hence, the wave starting from the sound source situated on the earth bends continuously towards the normal and goes away from the earth.

Therefore, the sound of siren is heard feeble on earth.

On the other hand, in night time the air near the earth becomes rapidly cold and so now in going from the earth the air becomes rare or less dense. Hence, the sound waves ultimately, after total internal reflection, return to earth. Hence, the sound of siren is heard loud on the earth.

**Q.33 Explain the factors on which the pitch of a tunning fork depends. **

**Ans.** The factors on which the pitch of a tunning fork depends are –

Length of the prongs (l) : The pitch is inversely proportional to square of the length$(l)$

of the prongs i.e. $v \propto \frac{1}{l^{2}}$

(ii) Thickness of the prongs (b): The pitch is

directly proportional to the thickness (b) of the

prongs i.e.

(iii) Elasticity of material (v): The pitch is directly

proportional to the square root of the elasticity

(Y) of the material i.e..

(iv) Density of material : The pitch is inversely

proportional to the square root of the density of

the material ie..

**Q.34 State Newton’s formula for velocity of sound in air. Point out the error and discuss Laplaces correction. **

**Ans.** According to Newton the velocity of longitudinal waves through any medium (solid, liquid or gas) depends upon the elasticity and density of medium as

$v=\sqrt{\frac{E}{\rho}}, \mathrm{E}=$ coeff. of elasticity of medium

$\rho=$ density of medium.

For air or gas (since gas has only one type of

elasticity) has bulk modulus only.

$v=\sqrt{\frac{K}{\rho}}$ ……………….(i)

Newton assumed that sound travels through a gas in the form of compression and rarefactions and this phenomenon takes place so slowly that temperature of gas (or medium) remains constant. i.e. the process is isothermal.

Under isothermal conditions, constant.

Differentiating on both sides, we get,

$P d V+V d P=0$

$\Rightarrow P=-\frac{V d P}{d V}=-\frac{d P}{\frac{d V}{V}}=K$

Hence, from equation ( 1), velocity of sound,

$v=\sqrt{\frac{P}{\rho}}$

Where $P=$ atmospheric pressure at N.T.P.

$=76 \mathrm{cm}$. of $\mathrm{Hg} \quad=0.76 \times 13.6 \times 10^{3} \times 9.8$

$=1.013 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}$

and density of air of N.T.P.

But experimental value of velocity of sound in air at N.T.P. is i.e. there is a large difference between theoretical and experimental values. This shows that there is an error in Newton’s formula. Later on, this error was solved by french mathematician Laplace and is called Laplace’s correction

Laplace’s Correction

According to Laplace the change in pressure and volume of a gas, when sound wave propagates through it are not isothermal but adiabatic.

For adiabatic process, gas equation is, constant.

Where

Differentiating above equation, we get,

(by definition)

From equation (i), we have,

Velocity of sound in air,

(Since air is a mixture of diatomic gas;

The above value agrees fairly well with the experimental values.

**Q.35 Establish relation between particle velocity and wave velocity. **

**Ans. **Wave Velocity : It is the velocity with which a progressive wave travels through the medium. It is a constant quantity and is given by, .

Where, n = frequency of wave & is wavelength.

Particle Velocity

A wave propagates through a medium due to repeated periodic motion of the particle of the medium i.e. velocity of a particle continuously changes with time.

The instantaneous particle velocity of a particle is defined as the time rate of change of its displacement. It is denoted by ‘u’.

The displacement of a particle in a progressive wave at any time ‘t’ is given by,

Hence, the instantaneous particle velocity is given by,

Now,

i.e.

maximum particle velocity

The slope of the displacement curve at position x is given by

Now,

i.e.

Hence, particle velocity at a given position at a given time is equal to product of wave velocity

and negative slope of the wave curve at the given position and time.

**Q.36 What are standing (or stationary) waves ? Give analytical treatment of formation of standing waves. **

**Ans.** When two sets of progressive waves of same type (i.e. both longitudinal or both transverse) having the same amplitude and same time period (or frequency) travelling with same speed along the same straight line in opposite directions superimpose, a new set of waves are formed. These are called standing or stationary waves.

In stationary waves, there are certain points of the medium, which are permanently at rest i.e. their displacement is zero all throughout. These points are called nodes. Similarly, there are other points which vibrate with largest amplitude. These points are called antinodes.

Analytical method

Let a plane progressive wave of amplitude ‘a’ is travelling with speed along positive direction of axis. The equation of the wave is,

Let this wave is reflected from a free boundary and the reflected wave advances in negative direction of x–axis. The equation. of reflected wave.

If, however, the waves were reflected from a rigid boundary, then the equation. of reflected wave will be,

sign shows that a phase change of takes place on reflection).

According to principle of superposition, the resultant displacement,

i.e.

Variation of y w.r.t. x

Substituting in above equation., the value of becomes alternately +1 and –1. Thus, at these points the displacement y is always max. These points are called ‘antinodes’ and are separated from one another by a distance of . Similarly, substituting in above equations, the value of becomes zero. Thus at these points, the displacement y is always zero & are called nodes and these are also separated from one another by a distance of .

Variation of y w.r.t. time (t)

Substituting is above equation. the value of becomes zero i.e. at these points, all particles of the medium pass through their mean position simultaneously twice in each vibration.

(2) Substituting the value of becomes alternately +1 and –1. At these points the displacements y is max. Hence, all the particles of the medium (except nodes) are in position to maximum displacement twice in each vibration.

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Get Rotational Motion important problems with solutions for Board exams. View 11th Physics important questions developed by top IITian faculties for exam point of view. These important problems with solutions will play significant role in clearing concepts related to rotational motion chapter. This question bank is designed by keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 & 12 Physics chapters.In this article you will learn the concept of rotational motion with important problems and having their solutions too.

**Q. Which component of linear momentum does not contribute to angular momentum ?**

**Sol.**Radial component of linear momentum.

**Q. What is the cause and outcome in a rotational motion ?**

**Sol.**Cause of rotational motion – Torque Out come in rotational motion – Angular momentum

**Q. Is it possible to open a pen up with one finger ? Why ?**

**Sol.**Is it possible to open a pen up with one finger ? Why ?

**Q. A rifle barrel has a spiral groove which imparts spin to the bullet. Why ?**

**Sol.**Angular momentum gained by the bullet provides better accuracy.

**Q. For uniform circular motion, does the direction of the centripetal force depend on the sense of rotation (**

*i.e.*, clock wise or anti clockwise) ?

**Sol.**No, direction of centripetal force remains constant either it is clockwise and anticlockwise rotation.

**Q. If a body is rotating, is it necessarily being acted upon by an external torque ?**

**Sol.**No, torque is required only for producing angular acceleration. For uniform rotation, no torque is needed.

**Q. A projectile acquires angular momentum about the point of projection during flight. Does it violate the conservation of angular momentum ?**

**Sol.**A projectile acquires angular momentum about the point of projection during flight. Does it violate the conservation of angular momentum ?

**Q. Give the physical significance of moment of inertia ?**

**Sol.**The moment of inertia plays the same role in rotational motion as the mass does in linear motion.

**Q. The moment of inertia plays the same role in rotational motion as the mass does in linear motion.**

**Sol.**Yes, moment of inertia of a body changes with the change of the axis of rotation.

**Q. Is there any difference between moment of inertia and rotational inertia ?**

**or**

**Why is moment of inertia called rotational inertia ?**

**Sol.**The moment of inertia is another name for rotational inertia.

The moment of inertia is called rotational interia for the reason that it gives the measure of inertia in rotational motion.

**Q. Why are the spokes fitted in a cycle wheel ?**

**Sol.**Why are the spokes fitted in a cycle wheel ?

**Q. Two solid spheres of same mass are made of metals of different densities. Which of them has a larger moment of inertia about a diameter ?**

**Sol.**Two solid spheres of same mass are made of metals of different densities. Which of them has a larger moment of inertia about a diameter ?

**Q. A ring and a circular disc of different material have equal masses and equal radii. Which one will have a larger moment of inertia about an axis passing through its centre of mass and perpendicular to its plane ?**

**Sol.**A ring has a larger moment of inertia because its entire mass is at a larger distance from the axis of rotation than the average distance of the mass particles of the disc.

**Q. A disc of metal is melted and recast in the form of a solid sphere. What will happen to the moment of inertia about a vertical axis passing through the centre ?**

**Sol.**The moment of inertia will decrease, because the material particles will, on the average come closer to the axis.

**Q. Is friction necessary for a body to roll ? Why ?**

**Sol.**If no other external force acts on a body, friction is necessary for a body to roll. Frictional force being tangential provides torque. Any force at the centre of mass will provide no torque and so cannot roll the body.

*Short Answer Type [2 & 3 Marks]*

**Q. Can a body in translatory motion have angular momentum ? Explain.**

**Sol.**Yes, a body in translatory motion shall have angular momentum unless the fixed point about which angular momentum is taken lies on the line of motion of the body. This follows from

$|L|=r p \sin \phi \cdot L=0,$ only when $\phi=0$ or $\phi=180^{\circ}$

**Q. A planet revolves around a massive star in a highly elliptical orbit. Is its angular momentum constant over the entire orbit ?**

**Sol.**Yes, angular momentum of the planet is constant over the entire orbit. This is because revolution of planet around the star is under the effect of gravitational force between the star and the planet. This is a radial force whose torque is zero. Therefore, angular momentum of the planet is constant (vector), whatever be the nature of the orbit.

**Q. A star of mass**

*M*and radius $10^{6} \mathrm{km}$ rotates about its axis with an angular speed of $10^{-6} \mathrm{s}^{-1}$. What is the angular speed of the star when it collapses (due to inward gravitational forces) to a radius of $10^{4} k m$. Assume that there is no mass loss by the star during collapse.

**Sol.**Let $I_{1}$ and $\omega_{1}$ be the initial moment of inertia and the angular speed of the star about its own axis of rotation, and $I_{2}$ and $\omega_{2}$ be the corresponding quantities after the star is collapsed. since the collapse is due to the internal forces (gravitational attraction of the outer layers by the inner layers of the star), the angular momentum of the star remains conserved during collapse. Thus,

The collapsed star rotates with a speed $10^{4}$ times the initial angular speed.

**Q. A child stands at the centre of a turn table with his two arms outstretched. The turn table is set rotating with an angular speed of 40**

*r.p.m.*(a) How much is the angular speed of the child, if he folds his hands back reducing the moment of inertia to (2/5) time the initial value ? Assume that the turn table rotates without friction.

(b) Show that the childs’s new *K.E.* of rotation is more than the initial *K.E.* of rotation. How do you account for this increase in *K*.*E*. ?

** [NCERT]**

**Sol.**Here, initial angular speed,

$\omega_{1}=40 \mathrm{rev} / \mathrm{min} ; \omega_{2}=?$

final moment of inertia, $I_{2}=\frac{2}{5} I_{1}$

As no external torque acts in the process, therefore, *L*= constant

i.e. K.E. of rotation increases. This is because the child spends internal energy in folding back his hands.

**Q. What is the angular momentum of the projectile when it is at the highest point about an axis through the point of projection ?**

**Sol.**At the highest point of projectile motion the velocity is $u \cos \theta$ acting horizontally. The perpendicular distance between the point of projection and the highest point

**Q. A planet moves around the sun under the effect of gravitational force exerted by the sun. Why the torque on the planet due to the gravitational force is zero ?**

**Sol.**The torque on the planet due the sun, $\vec{\tau}=\vec{r} \times \vec{F}$ Where $\vec{r}$ is position vector of the planet $w \cdot t .$ the sun and $\vec{F}$ is the gravitational force on the planet since the gravitational force on the planet acts along the line joining the planet to the sun, the vectors and are always parallel and hence.

**Q. A tube of length**

*l*is filled with a liquid density . Find the force experienced at one of its edge if rotated about the other edge with an angular velocity .

**Sol.**Angular velocity will be same for all points. Consider a small portion of length

*dx*at a distance

*x*from the axis.

Mass of $d x=\rho . A d x,$ where $A$ is cross sectional area.

Force experienced by $d x=m r \omega^{2}=\rho A d x . x \omega^{2}$

Force at the edge $=\int_{0}^{l} \rho A \omega^{2} x d x \Rightarrow F=\rho A \omega^{2} \frac{l^{2}}{2}$

**Q. A grindstone has a moment of inertia . A constant torque is applied and the grindstone is found to have a speed of 150**

*r.p.m.*in 10*sec*after starting from rest. Calculate the torque.

**Sol.**Here, Moment of inertia of grindstone, $I=6 \mathrm{kg} m^{2}$

Initial angular velocity, $\omega_{1}=0$

Final angular velocity,

$\omega_{2}=2 \pi n=2 \pi \times \frac{150}{60}=5 \pi \mathrm{rad} / \mathrm{sec}$

Time for which torque acts, $t=10 \mathrm{sec} .$

$\therefore$ Angular acceleration

$(\alpha)=\frac{\omega_{2}-\omega_{1}}{t}=\frac{5 \pi-0}{10}=\frac{\pi}{2} r a d / \sec ^{2}$

As $\tau=I \alpha \quad \therefore \tau=6 \times \frac{\pi}{2}=3 \pi \mathrm{Ns}$

**Q. A disc rotating about its axis with angular speed $\omega_{0}$ is placed lightly (without any transnational pull) on a perfectly frictionless table. The radius of the disc is**

*R*. What are the linear velocities of the points*A*,*B*and*C*on the disc shown in fig. ? Will the disc roll in the direction indicated ?

**[NCERT]**

**Sol.**Using the relation $v=r \omega,$ we get

For point $A, v_{A}=R \omega_{0},$ along $A X$

For point $B, v_{B}=R \omega_{0},$ along $B X^{\prime}$

For point $C, v_{C}=\left(\frac{R}{2}\right) \omega_{0}$ parallel to $A X,$

The disc will not rotate, because it is placed on a perfectly frictionless table. Without friction, rolling is not possible.

**Q. To maintain a rotor at uniform angular speed of $200 \mathrm{rads}^{-2}$$200 \mathrm{rads}^{-2}$, an engine needs to transmit a torque of 180**

*N m*. What is the power of the engine required ? **[NCERT]**

**Sol.**For a ‘uniform’ angular speed of rotation, no torque is needed, provided there is no friction. In practice, however, a torque has to be applied to the rotor to counterbalance the frictional torque.

If $\tau$ be the frictional torque, then the required power of the engine transmitting the torque is given by

$P=\tau \omega$

when $\omega$ is the angular speed. Here $\tau=180 N m$ and

$\omega=200 r a d s^{-1},$ therefore

$P=180 N m \times 200 s^{-1}=36000 W=36 k W$

**Q. Prove that the torque acting due to a force $\mathrm{F}$ in the**

**$x y$ plane is, $\tau=\mathrm{xF}_{y}-\mathrm{yF}_{\mathrm{x}}$**

**Sol.**We know $\vec{\tau}=\vec{r} \times \vec{F}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ and

$\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}+F_{z} \hat{k},$ we have,

$\tau=\tau_{x} \hat{i}+\tau_{y} \hat{j}+\tau_{z} \hat{k}=\left|\begin{array}{ccc}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {x} & {y} & {z} \\ {F_{x}} & {F_{y}} & {F_{z}}\end{array}\right|$

Comparing the coefficients of $\hat{i}, \hat{j}$ and $\hat{k},$ we have

$\tau_{z}=x F_{y}-y F_{x}$

**Q. Explain why friction is necessary to make the disc roll in the direction indicated. (a) Give the direction of frictional force at**

*B*, and the sense of frictional torque, before perfect rolling begins. (b) What is the force of friction after prefect rolling begins ?

**Sol.**To roll a disc, we require a torque, which can be provided only by a tangential force. As force of friction is the only tangential force in this case, it is necessary.

(a) As frictional force at *B* opposes the velocity of point *B*, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards.

(b) As frictional force at *B* decreases the velocity of the point of contact *B* with the surface, the perfect rollling beings only when velocity of point *B* becomes zero. Also, force of friction would become zero at this stage.

**Q. State the theorems of Parallel axes.**

**Sol.**Parallel AxesTheorem

The moment of inertia about an axis passing parallel

to the axis through the centre of mass of a rigid

body is the sum of the moment of inertia ( $l_{m}$ ) of

the body about the axis through centre of mass and

the product of its mass $M$ and square of the

separation $\left(a^{2}\right)$ between the parallel axis.

**Q. State the perpendicular axes theorem.**

**Sol.**Perpendicular Axes Theorem

According to this theorem the sum of moment of inertia of a plane lamina about two mutually perpendicular lying in its plane is equal to its moment of inertia about an axis perpendicular axes to the plane of lamina and passing through the point of intersection of first two axes.

**Q. In a fly wheel, most of the mass is concentrated at the rim. Explain, why.**

**Sol.**In a fly wheel, most of the mass is concentrated at the rim. Due to this peculiar shape, the flywheel possesses a large value of moment of inertia for the given mass and the radius.

If such a wheel gains or loses some rotational energy, it brings about only a very small change in its angular speed. In this way, a flywheel helps to maintain uniform motion.

**Q. A rod of length**

*l*and mass M held vertically is let go down, without slipping at the point of contact. What is the velocity of the top end at the time of touching the ground ?

**Sol.**Loss in

*P.E.*= Gain in rotational Kinetic energy

**Q. Two identical cylinders ‘run a race’ starting from rest at the top of an inclined plane, one slides without rolling and other rolls without slipping. Assuming that no mechanical energy is dissipated in heat, which one will win ?**

**Sol.**When a cylinder slides without rolling,

$E=\frac{1}{2} m v_{1}^{2} \Rightarrow v_{1}=\sqrt{\frac{2 E}{m}}$

When the cylinder rolls without slipping

$E=\frac{1}{2} m v_{2}^{2}+\frac{1}{2} I w^{2} \Rightarrow E=\frac{1}{2} m v_{2}^{2}+\frac{1}{2}\left(\frac{1}{2} m r^{2}\right) w^{2}$

$=\frac{1}{2} m v_{2}^{2}+\frac{1}{4} m v_{2}^{2}=\frac{3}{4} m v_{2}^{2} \quad[\because v=r w]$

$\therefore v_{2}=\sqrt{\frac{4 E}{3 m}}$

As $v_{1}>v_{2},$ therefore sliding cylinder will win the

race,

**Q. Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string ?**

**Sol.**Consider that a stone of mass $M$ is revolved along a circular path by tying it to a thread of length $l .$

Moment of inertia of stone, $I=M l^{2}$

Suppose that a torque $\tau$ is applied to produce an angular acceleration $\alpha .$ Then, $\tau=I \alpha=\left(M l^{2}\right) \alpha$ Or $\alpha=\frac{\tau}{M l^{2}}$

It follows that to produced the same amount of angular acceleration; the larger torque is required, when longer string is used to revolve a stone of given mass. Hence, it is more difficult to revolve a stone by tying it to a longer string.

**Q. Three point mass are located at the vertices of an equilateral triangle of length a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through ?**

**[NCERT]**

**Sol.**In fig. $A B C$ is an equilateral triangle. Let $A O \perp B C$, so $A O$ is also bisector of $B C$ i.e. $A O$ is median of $\triangle A B C .$

We have to calculate moment of inertia of the system about *AO*.

**Q. A solid cylinder of mass 20**

*kg*rotates about its axis with angular speed $100 \mathrm{s}^{-1}$. The radius of the cylinder is 0.25*m*. What is the kinetic energy associated with the rotation of the cylinder ? What is the magnitude of angular momentum of the cylinder about its axis ?

**Sol.**

**Q. What is the moment of inertia of a uniform circular disc of radius**

*R*and mass*M*about an axis passing through its centre and normal to the disc. The moment of inertia of the disc about any of its diameter is given to be $\frac{1}{4} \mathrm{MR}^{2}$. **[NCERT]**

**Sol.**We are given, moment of inertia of the disc about any of its diameter $=\frac{1}{4} M R^{2}$

Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through a point on its edge and normal to the disc

$=2 \times \frac{1}{4} M R^{2}=\frac{1}{2} M R^{2}$

**Q. A solid cylinder of mass M and radius R is held at rest in a horizontal position. Two strings are wound around the cylinder. As the string gets un-wound, find the tension in the string and acceleration in the cylinder.**

**Sol.**$M g-2 T=M a \quad ; \quad 2 T R=I \alpha=\frac{M R^{2}}{2} \frac{a}{R}$

**Q. Explain, why friction is necessary to make the disc roll in the direction indicated. (a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins. (b) What is the force of friction after perfect rolling begins ?**

**Sol.**The make the disc roll, a torque is required which in turn requires tangential force on it. Here, in this problem, only frictional force can provide the tangential force.

(a) Frictional force at point B opposes the velocity of point B. Therefore, frictional force at the point of contact is in the same direction as the arrow. The diretion of frictional torque is such that it opposes the angular motion. since is normal to the plane of the paper in inward direction, the direction of frictional torque is normal to paper and in outward direction.

(b) Frictional force at point B decreases the velocity of the point of contact B with the surface. The perfect rolling takes place, when the velocity of point B becomes zero and once this is so, the force of friction also becomes zero.

**Q. A symmetric lamina of mass M consists of a square shape with a semicircular section over each of the edge of the square as shown in fig. The side of the square is 2**

*a*. The moment of inertia of the lamina about of an axis through its centre of mass and perpendicular to the plane is $1.6 \mathrm{Ma}^{2}$. Find the moment of inertia of the lamina about the tangent AB in the plane of the lamina. **[I.I.T 1997]**

**Sol.**Here, M.I. of plane lamina about an axis through its C.M. and perpendicular to its plane,

$I=1.6 \mathrm{M} a^{2}$

If $I_{x}$ and $I_{y}$ are moments of inertia of the plane lamina about the axes $X X^{\prime}$ and $Y Y^{\prime}$ respectively, then according to the theorem of perpendicular axes,

$I_{x}+I_{y}=I$

since the lamina is symmetrical about the axes and

$I_{x}=I_{y}$

$\therefore \quad I_{x}+I_{x}=I$

or $\quad I_{x}=\frac{1}{2} I=\frac{1}{2} \times 1.6 \mathrm{Ma}^{2}=0.8 \mathrm{Ma}^{2}$

According to the theorem of parallel axes, we have

$I_{A B}=I_{x}+M(O Y)^{2}=0.8 M a^{2}+M(2 a)^{2}=4.8 \mathrm{Ma}^{2}$

**Q. A ring, a disc and a sphere all of the same radius and same mass roll down an inclined plane from the same height**

*h*. Which of the three reaches the bottom (i) earliest (ii) latest ? **[NCERT]**

**Sol.**We know that acceleration of an object down an inclined plane is given by

$a=\frac{g \sin \theta}{1+I / m r^{2}}$ … (i)

For a ring, $I=m r^{2}$

$a_{\sin g}=\frac{g \sin \theta}{1+1}=0.5 g \sin \theta$

For a disc, $I=\frac{1}{2} m r^{2}$

$\therefore \quad \frac{g \sin \theta}{1+\frac{1}{2}}=\frac{2}{3} g \sin \theta \quad=0.67 g \sin \theta$

For a sphere, $I=\frac{2}{5} m r^{2}$

$\therefore \quad a_{\text {sphere }}=\frac{g \sin \theta}{1+\frac{2}{5}}=\frac{5}{7} g \sin \theta=0.71 g \sin \theta$

As is maximum it will reach the bottom at the earliest. Again, as is minimum, it will reach the bottom at the end.

The result does not depend upon the mass or the radices of the rolling body.

*Long Answer Type [5 Marks]*

**Q. A uniform bar of length 6**

*a*and mass 8*m*lies on a smooth horizontal surface. Two point masses*m*and 2*m*moving in the same plane with speed 2*v*and*v*respectively strike at*a*and 2*a*and stick to the rod. Find**(i) Moment of Inertia of the system after the collision.**

**(ii) Angular momentum associated**

**(iii) Angular velocity, just after the collision**

**Sol.**Angular momentum with the two masses before collision

$L_{i}=m 2 v \cdot a+2 a 2 m v=6 m v a$

moment of the bar about an axis through its central axis,

$I_{b}=\frac{8 m(6 a)^{2}}{12}=\frac{8 \times 36 \times m a^{2}}{12}=24 m a^{2}$

(i) Moment of inertia after collision $=24 m a^{2}+m a^{2}+2 m \cdot 4 a^{2}$

$I_{f}=33 m a^{2}$

(ii) Angular momentum associated $=L_{i}=6 \mathrm{mva}$

(iii) Angular velocity after collision $=\omega=\frac{L_{i}}{I_{f}}$

i.e., $\quad \omega=\frac{6 m v a}{33 m a^{2}}=\frac{6 v}{33 a}=\frac{2 v}{11 a}$

**Q. What will be the duration of the day, if the earth suddenly shrinks to $\frac{1}{64}$ of its original volume, mass remaining unchanged ?**

**Sol.**Let $T_{1}$ and $T_{2}$ be the periods of revolution of theearth before and after its contraction ; and $R_{1}$ and $R_{2}$ be its radii before and after the contraction.

Now, according to the principle of conservation of angular momentum, $I_{1} \omega_{1}=I_{2} \omega_{2}$

where $I_{1}=\frac{2}{5} M R_{1}^{2}$ and $I_{2}=\frac{2}{5} M R_{2}^{2}$ are moments of

inertia of the earth before and after its contraction. it may be noted that the mass of the earth $(M)$ remains unchanged.

$\therefore \frac{2}{5} M R_{1}^{2}\left(\frac{2 \pi}{T_{1}}\right)=\frac{2}{5} M R_{2}^{2}\left(\frac{2 \pi}{T_{2}}\right)$

$\Rightarrow T_{2}=\left(\frac{R_{2}}{R_{1}}\right)^{2} \times T_{1}$ …(i)

As the earth shrinks to $\frac{1}{64}$ of its volume, we have final volume $=\frac{1}{64}$ (initial volume)

substituting the value of $\frac{R_{2}}{R_{1}}$ in equation (i) we get $T_{2}=\left(\frac{1}{4}\right)^{2} T_{1}=\frac{T_{1}}{16}$ i.e. duration of the day will be of

1.5 hour.

**Q. Two discs of moments of inertia $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ and about their respective axis (normal to the disc and passing through the centre), and angular speed $\omega_{1}$ and $\omega_{2}$ and are brought into contact face to face with their axes of rotation coincident.**

**(a) What is the angular speed of the two-disc system ?**

**(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take $\omega_{1}^{1} \omega_{2}$.**

**Sol.**(a) Total initial angular momentum of the two discs $=I_{1} \omega_{1}+I_{2} \omega_{2}$

since the two discs are brought into contact face to face, with their axes of rotation coincident, the M.I. of the combined system will be $I_{1}+I_{2} .$ If $\omega$ is the

angular speed of the combined system, then final angular momentum of the system $=\left(I_{1}+I_{2}\right) \omega$

since no external torque acts on the system, we have $I_{1} \omega_{1}+I_{2} \omega_{2}=\left(I_{1}+I_{2}\right) \omega \Rightarrow \omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$

(b) Initial K.E. of rotation of the two discs $=\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2}$

Final K.E. of the combined system $=\frac{1}{2}\left(I_{1}+I_{2}\right) \omega^{2}=\frac{1}{2}\left(I_{1}+I_{2}\right)\left(\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}\right)^{2}=\frac{1}{2} \frac{\left(I_{1} \omega_{1}+I_{2} \omega_{2}\right)^{2}}{I_{1}+I_{2}}$

Now, (initial K.E.) – (final K.E.)

$=\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2}-\frac{1}{2} \frac{\left(I_{1} \omega_{1}+I_{2} \omega_{2}\right)^{2}}{I_{1}+I_{2}}$

$=\frac{1}{2} \times \frac{I_{1}^{2} \omega_{1}^{2}+I_{1} I_{2} \omega_{1}^{2}+I_{1} I_{2} \omega_{2}^{2}+I_{2}^{2} \omega_{2}^{2}-I_{1}^{2} \omega_{1}^{2}-I_{2}^{2} \omega_{2}^{2}-2 I_{1} I_{2} \omega_{1} \omega_{2}}{I_{1}+I_{2}}$

$=\frac{I_{1} I_{2}}{2\left(I_{1}+I_{2}\right)}\left(\omega_{1}^{2}+\omega_{2}^{2}-2 \omega_{1} \omega_{2}\right)=\frac{I_{1} I_{2}\left(\omega_{1}-\omega_{2}\right)^{2}}{2 I_{1}+I_{2}}$

= a positive quantity

Hence the rotational kinetic energy of the combined system is less than the sum of the initial energies of the two discs.

The loss of energy is due to dissipation of energy in frictional contact of the two discs. It may be pointed out that the two discs rotate as a single combined system with a common angular speed, because of frictional contact only. However, the angular momentum will be conserved as the frictional torque is an internal torque.

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So in this article we have provided solved important questions of Gravitation chapter that will help class 11 and 12 students.

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**Do the force of friction and other contact force arise because of gravitational force? If not, what is the origin of these forces?**

**Sol.**The force of friction or any other contact forces, may be quite large and so it does not arise because of weak gravitational attraction. Such forces have electrical origin.

**Charge can be shielded from electrical forces by putting it inside a hollow conductor. Can a body be shielded from gravitational influence of nearby matter by putting it inside hollow sphere or by some other means?**

**Sol.**No, a body can not be shielded from the gravitational field of nearby matter by any method.

**The value of g on moon is $1 / 6$ th of that of earth. If a body is taken from earth to moon, then what will the change in its weight, Inertial mass and Gravitational mass.**

**Sol.**The weight of body on moon will be $1 / 6$ th of that on the earth, but there will be no change in gravitational mass and inertial mass.

**If the earth stops rotating about its axis, what will be the effect on the value of $^{\circ} \mathrm{g}^{\prime} ?$ Will this effect be same at all places?**

**Sol.**The value of will increase at all places except at the poles. The increase will be different at different places, maximum at equator.

**Can we determine the gravitational mass of a body inside on artificial satellite?**

**Sol.**No, artificial satellite is like a freely falling body so that $g=0 .$

**The astronauts in a satellite orbiting the earth feel weightlessness. Does the weightlessness depend upon the distance of satellite from earth?**

**Sol.**No, the gravitational acceleration of the astronaut w.r.t. satellite is zero, whatever be the distance of the satellite from earth.

**If the kinetic energy of a satellite revolving in an orbit close to earth happens to be doubled, will the satellite escape?**

**
**

**
**

**Sol.**The velocity of satellite will become equal to the escape velocity, hence it will escape.**When an object falls to the ground, the earth moves up to it. Why is earth’s motion not noticeable?**

**Sol.**Because action and reaction are equal and opposite, hence the force with which earth is attracted towards the object is equal to the force with which earth attracts the object. $(-m g, m$ is the mass of the object). If $a^{\prime}$ be the acceleration produced in earth and $M$ is the mass of earth, then $a=\frac{m g}{M}$ As $M>>m,$ hence $a<<g$ i.e. acceleration produced in earth is so small that it is not noticeable.

**An artificial satellite revolves in its orbit around the earth without using any fuel. But aeroplane requires fuel to fly at a certain height. Why so?**

**Sol.**Air is required for aeroplane to fly hence the aeroplane requires fuel against the frictional force due to air. On the other hand, the satellite orbits at a much more height where almost no air is present and its friction in negligible.

**An elephant and an ant are to be projected out of earth into space. Do we need different velocities to do so?**

**Sol.**The escape velocity $v_{e}=\sqrt{2 g R}$ does not depend upon the mass of the projectile. Thus we need the same velocity to project an elephant and an ant into space.

**Why are space rockets usually launched from west to east in the equatorial plane?**

**Sol.**since earth rotates from west to east and as such all points on the earth have velocity from west to east. Moreover, this velocity is maximum in the equatorial plane as . This maximum velocity is added to the launching velocity of the rocket and hence launching becomes easier.

**Does a rocket really need the escape velocity of**

**$11.2 \mathrm{km} / \mathrm{s}$ initially to escape from the earth?**

**Sol.**No, rocket can have any velocity at the start. The rocket can continue to increase the velocity due to thrust provided by the escaping gases that will carry it to a desired position.

**Does a planet revolving around sun in an elliptical orbit have a constant**

(i) Linear speed

(ii) Angular momentum

(iii) Kinetic energy

(iv) Potential energy

(v) Total energy throughout its orbit

**Sol.**(i) According to the law of conservation of angular momentum, the planet moves faster when it is close to the sun and moves slower when it is farther away from the sun. Hence linear speed of the planet does not remain constant.

(ii) According to the law of conservation of angular momentum, the angular momentum of planet remains constant.

(iii) Since linear speed of the planet around the sun changes, therefore, its kinetic energy also changes continuously.

(iv) Potential energy depends upon the distance between sun and planet, hence potential energy changes as the distance between the sun and

planet changes continuously.

(v) Total energy of planet remains constant.

**In a spaceship moving in gravity free region, the astronaut will not be able to distinguish between up and down. Explain.**

**Sol.**The upward and downward sense is due to gravitational force of attraction between the body and the earth. In spaceship gravitational force is counter balanced by the centripetal force needed by satellite to move around the earth in circular orbit. Hence in absence of force, the astronaut will not be able to distinguish between up and down.

**What do you understand by geo-stationary and polar satellites. Discuss their important uses.**

**Sol.**Geo-stationary or Geo-synchronous satellite : A satellite which appears stationary to an observer on earth is called geo-stationary satellite. Obviously, the velocity of such satellite relative to earth is zero. Clearly time of revolution of satellite hrs. This satellite is also called geo-synchronous satellite as its angular velocity is synchronised with angular speed of earth about its axis. i.e. this satellite revolves around the earth with same angular speed in same direction as is done by earth around its axis.

Essential conditions for geo-stationary satellite: (a) A geo-stationary satellite should be at a height of nearly $36,000 \mathrm{km}$ above the equator of earth.

(b) The period of revolution around the earth should be the same as that of earth about its axis i.e. exactly 24 hours.

(c) It should revolve in an orbit concentric and co- planer with the equatorial plane, so the plane of orbit of the satellite in normal to the axis of rotation of the earth.

(d) Its sense of rotation should be same as that of earth about its own axis i.e. from west to east. Its orbital velocity is nearly 3.1 $k m /$ sec.

Uses : Geo-stationary satellites are used for purposes of communications as they act as reflectors of suitable waves carrying the messages. Hence they are also called telecommunication satellites. India’s INSAT- $2 \mathrm{B}, 2 \mathrm{C}$ are communication satellites.

Polar satellites : Such satellite which revolves in polar orbit around the earth is called polar satellites. A polar orbit is one whose angle of inclination with equatorial plane of earth is $90^{\circ}$ and a satellite in polar orbit passes over the both north and south geographic poles respectively once in orbit.

(i) For ground water survey.

(ii) For detection the areas under forest.

(iii) For preparing wasteland maps.

(iv) For assessment of drought.

(v) For the assessment of crop diseases.

(vi) For detecting the potential fishing zones.

(vii) For identifying the sources of pollution.

(viii) To locate the position and movement of the troops of enemy and to locate the place and time for any nuclear explosion etc. i.e. for spying purposes.

**State Kepler’s laws of planetary motion and explain the deduction of Kepler’s second law and third law of planetary motion.**

**Sol.**On the basis of his investigations, Kepler formulated the following three laws of planetary motion:

(i) First law (law of orbit): Every planet revolves around the sun in an elliptical orbit, keeping sun at one focus of ellipse.

(ii) Second law (Law of area): The radius vector drawn from sun to a planet sweeps out equal areas in equal intervals of time. i.e. areal velocity of the planet around the sun is constant.

(iii) Third law (Law of period): The square of time period of revolution of a planet around the sun is directly proportional to the cube of semi major axis of its elliptical orbit i.e. $T^{2} \propto R^{3}$ Deduction of second law: Let a particle rotates in $X-Y$ plane about $z$ -axis. At any time $t, \overrightarrow{O P}=\vec{r},$ be the position vector of particle. The area swept by the position vector in small time $d t,$ $|d \vec{A}|=$ area of $\Delta O P Q=\frac{1}{2}|\vec{r} \times d \vec{r}|$

since, a planet revolves around the sun, in an elliptical orbit under the influence of gravitational pull of sun on planets and this pull (or force) acts along the line joining the centres of the sun and theplanet and is directed towards the sun. Hence, the torque acting on planet is zero.

i.e. areal velocity of the planet around the sun is constant, which is second law of planetary motion. Deduction of third law : Let a planet of mass $^{\circ} \mathrm{m}^{\prime}$ is orbiting around the sun of mass $M$ in a circular orbit of radius $^{\circ} r^{\prime}$ with constant angular velocity $\omega,$ then,

$\frac{G M m}{r^{2}}=m \omega^{2} r=m\left(\frac{2 \pi}{T}\right)^{2} r \Rightarrow T^{2}=\left(\frac{4 \pi^{2}}{G M}\right) r^{3}$

i.e. $T^{2} \propto r^{3} \quad\left(\because \frac{4 \pi^{2}}{G M}=\text { constant }\right)$

Above result holds equally good for elliptical orbit provided we replace $r$ with $^{*} a^{\prime},$ the semi major axis of ellipse.

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Get (SHM) simple harmonic motion questions and answers for physics class 11 exams.View 11th Physics important questions for exam point of view. These important questions will play significant role in clearing concepts of Physics chapters. This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 Physics chapter wise CBSE.

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**Q. A man with wrist watch on his hands fall from the top of a tower. Does the watch give correct time?**

**Sol.**

**Q. A vibrating simple pendulum of period is placed in a lift which is accelerating downwards. What will be the effect on the time period?**

**Sol.**

**Q. Where is tension maximum in the string of a simple pendulum?**

**Sol.**

**Q. Can a motion be oscillatory but not simple harmonic? If your answer is yes give an example and if not explain why?**

**Sol.**

**Q. Which of the following examples represents nearly) S.H.M. and which represents periodic but not S.H.M.?**

**(i) The rotation of earth about its own axis.**

**(ii) Motion of an oscillating mercury column in a U-tube.**

**(iii) Motion of a ball bearing inside a smooth curved bowl when released from a point slightly above the lower most position.**

**(iv) General vibration of polyatomic molecule about its equilibrium configuration.**

**Sol.**

**Q. A girl is sitting on a swing. Another girl sits by her side. What will be the effect on the periodic time of the swing?**

**Sol.**No effect, as periodic time is independent of mass.

**Q. Given below are $(x-t)$ plots for linear motion of a particle. Which of the plots represent periodic motion ? What is the period of motion (in case of periodic motion)?**

**Sol.**

**Q. The bob of a vibrating simple pendulum is made of ice. How will the period of swing will charge when ice starts melting?**

**Sol.**The period of swing of simple pendulum will remain unchanged till the location of centre of gravity of the bob left after melting the ice remains at a fixed distance from the point of suspension. But, if the centre of gravity of ice bob after melting is raised upwards, then the effective length of pendulum decreases and hence the time period of swing decreases. If the centre of gravity of ice shifts on lower side, the time period of swing increases.

**Q. A ball of mass ‘m’ is dropped in a tunnel along the diameter of earth from a height ‘h’ (h< <R) above the surface of earth. Is the motion simple harmonic?**

**Sol.**

**Q. Which of the following functions of time represent (a) Periodic (b) Non-periodic motion ? Give the period of for each periodic motion $(\omega$ is any positive constant).**

**(ii) sin $\omega t+\cos \omega t$**

**(ii) $\sin \omega t+\cos 2 \omega t+\sin 4 \omega t$**

**(iii) Se $^{\circ t}$**

**(iv) $\log \omega t$**

**Sol.**

**Q. If a pendulum clock is taken on the mountain top, does it lose or gain time, assuming it is correct at a lower elevation?**

**Sol.**

**Q. The soldiers marching on a suspended bridge are advised to go out of steps. Why?**

**Sol.**

**Q. Glass window-panes may be broken by a far away explosion. Explain why?**

**Sol.**

**Q. Find expression for time period and frequency for the oscillations of a liquid in a U-tube.**

**Sol.**

**Q. Explain the oscillations of a loaded spring and find the expression for time period and frequency in case of vertical spring.**

**Sol.**

Learn the concepts related to simple harmonic motion with these important questions and answers. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc.

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Get circular motion practice problems with answers for class 11 physics. View 11th Physics important questions for exam point of view. These important questions will play significant role in clearing concepts of Physics. This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 Physics chapters.

Improve your concepts of circular motion with these practice problems with answers and increase your marks in exams.

Click Here for Detailed Notes of Circular Motion along with other chapters.

**Q. Can a particle accelerate when its speed is constant? Explain.**

**Sol.**A particle can be accelerated if its velocitychanges. A particle having uniform circular moton has constant speed but its dirction of motion changes continuously. Due to this, there is change in velocity and hence the particle is accelerated. Is it possible for a body to have constant speed but variable velocity? If so, give example.

**Q. Is it possible for a body to have constant speed but variable velocity $?$ If so, give example.**

**Sol.**Hes. When a body moves in a circle with constant speed, its velocity is variable. This because velocity is a vector quantity whose magnitude is speed but its direction changes continuously due to the change of the direction of motion.

**Q. Can a body have a constant velocity and still a variable speed?**

**Sol.**No. Velocity is a vector quantity and speed is its magnitude. A vector quantity changes if either its magnitude or direction changes. Since the magnitude of velocity (i.e., speed) changes and hence the body can not have a constant velocity.

**Q. Is circular motion possible at constant speed or at constant velocity? Explain.**

**Sol.**Circular motion is possible at constant speed because in circular motion, the magnitude of the velocity i.e., speed remains constant while the direction of motion changes continuously.

**Q.**

**Is uniform circular motion an example of uniform acceleration. Why?**

**Sol.**No, since acceleration is always towards the centre.

**Q. Name the physical quantity which remains same in an uniform circular motion.**

**Sol.**Kinetic energy and speed.

**Q. Two particles are moving with common speed v such that they are always at a constant distance ‘ d ‘ apart and their velocities are always equal and opposite. After what time, they turn to their initial positions?**

**Sol.**The two particles should be moving in circular path of radius d / 2 with same speed. Since tangent at any point gives the direction of velocity, they will always be opposite. They will turn to their initial position after completing one circular path.

**Q. Why do the passengers of a car rounding a curve are thrown outward?**

**Sol.**When the car turns round a curve, the passengers sitting in the car experience an outward force, i.e. centrifugal force due to the absence of the necessary centripetal force.

**Q. Is it possible to be accelerate a particle if it is travelling at constant speed?**

**Sol.**Yes, when a particle moves in a circles with constant speed, its velocity is variable because the direction of motion is continuously changing. accelerating.

**Q. What do you mean by centrifuge? Explain its**

**principle.**

**Sol.**It is a device used for separating mixtures of substances with different densities. It works on the principle that the centrifugal force, $F=m r w^{2}$ acting on a body in circular motion is,

(i) Directly proportional to mass ( $m$ ) of the body.

(ii) Directly proportional to the distance of the body from its axis of rotation.

The heavier particles experience large amounts of centrifugal force, i.e., they will be push outward, and get separated from the lighter particles.

**Q. Moon is continuously revolving round the earth without falling towards it. Justify, why it doses so?**

**Sol.**For revolving round the earth, the moon provides itself the necessary centripetal force at the expense of its weight. We know that a body falls under the effect of its weight. Since the weight of the moon gets used in providing it the necessary centripetal force, it revolves round the earth, without falling towards it.

**Q. For uniform circular motion, does the direction of the centripetal force depend upon the sense of rotation?**

**Sol.**The direction of the centripetal force does not depend, whether the body is moving in clockwise or anticlockwise direction. It is always directed along the radius and towards the centre of the circle.

**Q. Why does a cyclist lean to one side, while going along a curve? In which direction does he lean?**

** Or**

**Why does a cyclist lean inward while negotiating a curve? Explain with a diagram.**

**Sol.**While going along a curve, the cyclist leans to one side, so as to provide himself the necessary centripetal force. He leans in the inward direction i.e., towards the centre of the circular curve.

**Q. A body is suspended by a string of length and is projected horizontally with velocity . Calculate the tangential and radial acceleration when the string rises by from its initial position.Also find the difference in velocity.**

**Sol.**$l=1 m,$ applying conservation of energy, we have

$m g(1-\cos 60)=\frac{1}{2} m v_{B}^{2}$

$V_{B}=2 g\left(1-\frac{1}{2}\right)=g$

Centripetal acceleration $=g \sin 60^{\circ}=\frac{\sqrt{3}}{2} g$

Velocity difference $=v_{B}-v_{A}=10-4 m s^{-1}$

**Q. The acceleration associated with a mass moving in a circular path is to be found. It is given that the velocity at any instant is where is a constant. Classify the motion and find acceleration.**

**Sol.**since velocity changes with time, the motion in circular path involves tangential acceleration. So it is a non-uniform circular motion.

Net acceleration

$=a_{n} \sqrt{a_{r}^{2}+a_{t}^{2}}=\sqrt{\left(K^{2} r t^{2}\right)^{2}+(K r)^{2}}=K r \sqrt{1+K^{2} t^{4}}$

**Q. What is the need for banking the tracks?**

**Sol.**On a plane road, the centripetal force required to negotiate the curve is provided by the force of friction between the tyres of the vehicle and the surface of the road. Thus, the force of friction depends upon the nature of the surfaces. The force of friction may provide the necessary centripetal force if the speed of the vehicle is small enough. On the other hand, if the speed of the vehicle at the curve is not small enough, then theforce of friction will be unable to provide the necessary centripetal force to the vehicle and it may skid away. Therefore, to provide the necessary centripetal force to the vehicle at higher speeds, roads are banked.

**Q.**A train round an unbanked circular track of radius at a speed of . The mass of the train is. What provides the centripetal force required for this purpose ? The engine or the rails ? The outer or the inner rail? Which rail will wear out faster, the outer or the inner rail? What is the angle of banking required to prevent wearing out of the rails?

**Sol.**Given that and The centripetal force required for negotiating the curve on the unbanked road is provided by the lateral thrust by the outer rail on the flanges of the wheels. According to Newton’s third law of motion the train exerts an equal an opposite force on the outer rail. Hence, the outer rail will wear out faster than the inner rail.Further, given that,Speed of train, Radius of the bend, .

If is the angle of banking required to prevent

wearing out of the rails,

$\theta=37^{\circ} 25^{\prime}$

**Q. A large mass and a small mass hung at the two ends of a string that passes over a smooth tube. The mass moves around in a circular path which lies in the horizontal plane. The length of the string from the mass to the top of the tube is and is the angle this length makes with the vertical. What should be the frequency of rotation of the mass so that mass remains stationary?**

**Sol.**

**The various forces acting on $M$ and $m$ are**

$T=M g$

$T \cos \theta=m g$

$T=M g$

$T \cos \theta=m g$

$\begin{array}{l}{T \sin \theta=\frac{m v^{2}}{r}=m r \omega^{2}} \\ {\text { Where } r \text { is the radius of the circular path and } \omega} \\ {\text { is the angular velocity. }} \\ {r=l \sin \theta} & {\text { [From(iii)] }}\end{array}$

**Q. A conical pendulum consist of a string P Q whose upper end is fixed and lower and is attached with a bob. The bob is drawn aside, and given a horizontal push so that it describes a horizontal circle with angular speed in such a way that P Q makes a constant angle with the vertical. As the string traces the surface of a cone, it is known as a conical pendulum. Deduce an expression for the time period for the pendulum.**

**Sol.**

**Q. A cyclist moving with a velocity of approaches a -turn of radius his speed at a rate of . Calculate the acceleration of the cyclist on the turn.**

**Sol.**

$v=7.5 \mathrm{ms}^{-1}, r=80 \mathrm{m}$ centripetal acceleration is,

When the cyclist applies brakes at $P$ of the circular

turn, the cyclist applies brakes at $P$ of the circular

Acceleration will act opposite to the velocity i.e.,

$a_{t}=0.5 \mathrm{ms}^{-2}$

$a=\sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{(0.7)^{2}+(0.5)^{2}}=0.86 \mathrm{ms}^{-2}$

Let $\theta$ be the angle made by net acceleration with

the velocity of the cyclist, then

$\tan \theta=\frac{a_{c}}{a_{t}}=\frac{0.7}{0.5}=1.4$

**Q. A car of mass moves with a constant speed over a (a) horizontal flat surface (b) a convex bridge (c) a concave bridge. What force is exerted by the car on the bridge in each of these cases, as it passes the middle point of the bridge ? Take radius of curvature of the bridge in the last cases as r.**

**Sol.**The car is acted upon in the Vertical direction by two forces: its weight $M g$ and the normal reaction

$R$.

(a) When the car runs over a horizontal bridge, the normal reaction, say $R_{1}$ is just equal and opposite to its weight $M g$ i.e., $R_{1}=M g$

Thus, force on the bridge is equal to the weight of

the car.

(b) When the car runs over a convex bridge, it requires a centripetal force vertically downwards.

If $R_{2}$ is normal reaction, then

Thus, force on the bridge is less then the weight of the car.

**Q. A pendulum consisting of a bob of mass and string of length is moved upto the horizontal position and released. What should the minimum strength of the string be to withstand the tension as the pendulum passes through the position of equilibrium?**

**Sol.**When the bob is released from the horizontal position. A, it will reach the equilibrium position by moving along circular path as shown figure. At point , the bob of the pendulum is acted upon by two forces: its weight m g and tension in the string. The resultant of these forces provides the necessary centripetal force i.e.,

Kinetic energy at of the bob point $O=$ Potential

energy at point $A$

$\frac{1}{2} m v^{2}=m g l \Rightarrow v=\sqrt{2 g l}$

Substituting for $v$ in equation (i), we have

The string should have minimum strength to stand a tension three times the weight of the bob.

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thGet important laws of motion class 11 questions with answers developed by expert faculties to provide affordable and quality education to students. View class 11 Physics important questions with answers for exam point of view from the topic Nwton’s Laws of Motion fully solved. These important questions will play significant role in clearing concepts of the chapter. This question bank is designed by keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions with their answers for Class 11& 12 chapters.

In this article student’s will get laws of motion questions with answers which is an important topic of Class 11 Physics.

###### Click Here for Detailed Notes of Newton’s Laws of Motion along with other chapters and subjects.

**Q. Name one factor on which the inertia of a body depends.**

**Sol.**

**Q. If you jerk a piece of paper from under a book quick enough, the book will not move. Why?**

**Sol.**

**Q. Is a bus moving along a circular track, an inertial frame of reference?**

**Sol.**

**Q. A body is moving with uniform velocity. Can it said to be in equilibrium?**

**Sol.**

**Q. Why do we beat dusty blanket with a stick to remove dust particle?**

**Sol.**

**Q. An astronaut accidentally gets separated out of his small spaceship accelerating in inter-stellar space at a constant rate of . What is the acceleration of the astronaut the instant after the is outside the spaceship?**

**Sol.**

**Q. Why do we use shock absorbers in automobiles?**

**Sol.**

**Q. Why bogies of a train are provided with the buffers?**

**Sol.**

**Q. If a body is not at rest, the net external force acting on it cannot be zero. Is it true or false?**

**Sol.**

**Q. What is the impulsive force?**

**Sol.**

**Q. A thief jumps from the upper storey of a house with a load on his back. What is the force of the load on his back when the thief is in air?**

**Sol.**

**Q. Why does a swimmer push the water backwards?**

**Sol.**

**Q. Can a body in linear motion be in equilibrium?**

**Sol.**

**Q. A uniform rope of length and mass hung from**

**a support. What is the tension at a distance from the free end?**

**Sol.**

**Q. In the arrangement given, if the points and move down with a velocity , find the velocity of M?**

**Sol.**

**Q. A soda water bottle is falling freely. Will the bubbles of gas rise in the water of the bottle?**

**Sol.**

**Q. A bird is sitting on the floor of a wire cage and the cage is in the hands of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?**

**Sol.**

**Q. Air is thrown on a sail attached to a boat from an electric fan placed on the boat. Will the boat start moving?.**

**Sol.**

**Q. Is a ‘single isolated force’ possible in nature?**

**Sol.**

**Q. A disc of mass $m$ is placed on a table. A stiff spring is attached to it and is vertical. To the other end of the spring is attached a disc of negligible mass. What minimum force should be applied to the upper disc to press the spring such that the lower disc if lifted off the table when the external force is suddenly removed?**

**Sol.**

**Q. Explain how a man walks on the ground?**

**Sol.**

**Q. A man weights 70 kg. He stands on a weighting**

**scale in a lift which is moving.**

**(a) Upward with a uniform speed of $10 \mathrm{ms}^{-1}$.**

**(b) Downward with a uniform acceleration of ****$5 \mathrm{ms}^{-2}$ with a uniform acceleration of**

**(c) Upward with a uniform acceleration of $5 \mathrm{ms}^{-2}$.**

**What would be the reading on the scale in each ****case? What would be the reading if the lift ****mechanism failed, and it hurtled down freely ****under gravity? $\left[g=10 \mathrm{ms}^{-2}\right]$**

**Sol.**

**Q. It is easy to catch a table tennis ball than a cricket**

**ball, even when both are moving with the same**

**velocity. Why?**

**Sol.**

**[question] In which case will a rope have the greater tension **

**(a) Two men pull the ends of the rope with forces $F$ equal in magnitude but opposite in direction.**

**(b) One end of the rope is fastened to a fixed support and the other is pulled by a man with a force $2 F .$**

**Q. Ten one-rupee coins are put on the top of eachother on a table. Each coin has a mass $m$ kg. Give the magnitude and direction of**

**(a) The force on the $7^{\text {th }} \operatorname{coin}$ (counted from the bottom) due to all the coins on its top**

**(b) The force on the $7^{\text {th }}$ coin by the eighth coin**

**(c) The reaction of the $6^{\text {th }}$ coin on the $7^{\text {th }}$ coin**

[NCERT]

**Sol.**

**Q. A force acting on a material particle of mass $m$ first grows to a maximum value $F_{m}$ and then decreases to zero. The force varies with time according to a linear law, and total time of motion is $t_{m}$. What will be the velocity of the particle at the end of this interval if the initial velocity is zero**

**Sol.**

**Q. The assertion made by the Newton’s first law ofmotion that every body continues in its state of uniform motion in the absence of external force appears to be contradicted in everyday experience. Why?**

**Sol.**

**Q. A stone when thrown on a glass window smashes the window pane to piece, but a bullet from the gun passes through making a clean hole. Why?**

**Sol.**

**Q. A block is supported by a cord $C$ from a rigid support, and another cord $D$ is attached to the bottom of the block. If you give a sudden jerk to $D,$ it will break. But if you pull on $D$ steadily, $C$ will break. Why?**

**Sol.**

**Q.**

**Why do you fall forward when a moving train decelerates to a stop and fall backward when a train accelerates from rest? What would happen if the train rounded a curve at constant speed?**

[NCERT]

**Sol.**

**Q. “It is reasonable to consider the Earth as an inertial frame for many laboratory scale terrestrial experiments, but the Earth is a non-inertial frame of reference for astronomical observation.” Explain this statement. Do you see how the same frame of reference is approximately inertial for one purpose, and non-inertial for the other?**

[NCERT]

**Sol.**

**Q. A mass of $6 \mathrm{kg}$ is suspended by a rope of length $2 \mathrm{m}$ from a ceiling. A force of $50 \mathrm{N}$ in the horizontal direction is applied at the mid-point of the rope, as shown in the figure. What is the angle that rope makes with the vertical in equilibrium? Neglect the mass of the rope. Given : $\mathrm{g}=10 \mathrm{ms}^{-2}$.**

[NCERT]

**Sol.**

**Q. A piece of uniform string hangs vertically, so that its free end just touches the horizontal surface of a table. If the upper end is released, show that at any instant during fall of the string, the total force on the surface is three times the weight of that part of the string which is lying on the surface.**

[NCERT]

**Sol.**

**Q. With what acceleration ‘ $a^{\prime}$ should a box descend so that a block of mass $M$ placed in it exerts a force $\frac{\mathrm{Mg}}{4}$ on the floor of the box?**

**Sol.**

**Q. The pulley arrangements of figures $(a)$ and $(b)$ are identical. The mass of the rope is negligible. In figure $(a),$ the mass $m$ is lifted up by attactating a mass $2 m$ to the other end of the rope. In Figure (b), $m$ is lifted up by pulling the other end of the rope with a constant downward force $F=2 m g$. Calculate the accelerations in the two cases.**

**Sol.**

**Q. A monkey of mass 40 kg climbs on a rope which can withstand a maximum tension of $600 N .$ In which of the following cases will the rope break: the monkey**

**(a) Climbs up with an acceleration of $6 \mathrm{ms}^{-2}$**

**(b) Climbs down with an acceleration of $4 \mathrm{ms}^{-2}$**

**(c) Climbs up with a uniform speed of $5 \mathrm{ms}^{-1}$**

**(d) Falls down the rope nearly freely under gravity**

**Take $g=10 \mathrm{ms}^{-2}$. Ignore the mass of the rope.**

[NCERT] (new)

**Sol.**

**Q. Give the magnitude and direction of the net force acting on**

**(a) A drop of rain falling down with a constant speed**

**(b) A cork of mass 10 g floating on water**

**(c) A kite skilfully held stationary in the sky**

**(d) A car moving with a constant velocity of $30 \mathrm{km}$**

**(e) A high-speed electron in space far from all gravitating objects and free of electric and magnetic fields.**

**Sol.**

**Q. For ordinary terrestrial experiments, which of the observers below are inertial and which non-inertial? (a) A child revolving in a “giant wheel” (b) A driver in a sport car moving with a constant high speed of $200 \mathrm{km} \mathrm{h}^{-1}$ on a straight road,**

**(c) The pilot of an aeroplane which is taking off, (d) A cyclist negotiating a sharp turn,**

**(e) The guard of a train which is slowing down to stop at a station?**

[NCERT]

**Sol.**

**Q. Explain, why**

**(a) A horse cannot pull a cart and run in empty space.**

**(b)Is it easier to pull a lawn mower than to push it**

**(c) Does a cricketer move his hands backward when holding a catch.**

**Sol.**

**Q. Fig. Shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle $?$ What is the magnitude of each impulse?**

[NCERT]

**Sol.**

**Q. Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any loss of speed, as shown in the figure below. What is (i) the direction of the force on the wall due to each ball? and (ii) the ratio of the magnitudes of impulses imparted on the two balls by the wall?**

[NCERT]

**Sol.**

**Q. A light string carrying a small bob hangs from the roof of a vehicle as shown in the figure. If the vehicle moves in a horizontal straight line with a constant acceleration of $2.4 \mathrm{ms}^{-2}$ from left to right, determine the angle, which the string makes with the vertical.**

**Sol.**

**Q. Two masses 7 kg and 12 kg are connected at the two ends of a light inextensible strings that passes over frictionless pulley. Find the acceleration of the masses and the tension in the string, when the masses are released.**

**Sol.**

**Q. Two pulley of masses $12 \mathrm{kg}$ and $8 \mathrm{kg}$ are $\mathrm{d} \mathrm{e}$ connected by a fine string hanging over a fixed pulley as shown. Over the latter is hung a fined string with masses $4 \mathrm{kg}$ and $M .$ Over the $12 \mathrm{kg}$ pulley is hung another fine string with masses 3 $\mathrm{kg}$ and $6 \mathrm{kg}$. Calculate $M$ so that the string overthe fixed pulley remains stationary.**

**Sol.**

**Q. One end of a string is attached to a 6 kg mass on a smooth horizontal table. The string passes, the edge of the table, and to its other end is attached a light smooth pulley. Over this pulley passes another string to the ends of which are attached masses of $4 \mathrm{kg}$ and $2 \mathrm{kg}$ respectively as shown. Show that the $6 \mathrm{kg}$ mass moves with an acceleration of $\frac{8 \mathrm{g}}{17}$.**

**Sol.**

**Q. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass $25 \mathrm{kg}$ is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of $0.1 \mathrm{ms}^{-2}$. What is**

**the action of the block on the floor (a) before and**

**(b) after the floor yields? Take $\mathrm{g}=10 \mathrm{ms}^{-2}$.**

[NCERT]

**Sol.**

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Get linear momentum questions and answers for Board exams. View 11th Physics important questions for exam point of view. These important questions with answers will play significant role in clearing concepts of Conservation of Linear Momentum Principles. This question bank is designed by expert and experienced IITian faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions with answers for class 11 Physics chapters .

**Q. “The total momentum in the universe remains constant’.’. Is this statement true? If so, why?**

**Sol.**since no external force can be applied on the universe, therefore total momentum is conserved.

**Q. A box, initially sliding on the floor of a room, is eventually brought to rest by friction. Is the momentum of the box conserved? If not, does this process contradict the principle of conservation of momentum? What becomes of original momentum of the box?**

**Sol.**No; No; transferred to earth.

**Q. A meteorite burns in the atmosphere before it reaches the Earth’s surface. What happens to its momentum?**

**Sol.**The momentum will be transferred to the air particles and finally to the earth.

**Q. Does relative velocity of separation in one dimensional elastic collision change?**

**Sol.**In one dimensional elastic collision relative velocity of separation after collision is equal to relative velocity of approach before collision.

**Q. What happens when two bodies of equal masses undergo elastic collision in one dimension?**

**Sol.**Their velocities are just interchanged. This result has an important application in a nuclear reaction.

**Q. Is linear momentum of a system always conserved.**

**Sol.**No, only when the system is isolated.

**Q. The collision between two hydrogen atoms is perfectly elastic so the momentum is conserved do you agree with this statement.**

**Sol.**No, because momentum is conserved even when collision is inelastic.

**Q. Is it possible to have a collision in which the whole of K.E. is lost.**

**Sol.**Yes, for example in perfectly inelastic collision of two bodies moving toward each other with equal linear momentum

**Q. If two object collide and one is initially at rest**

**(a) Is it possible for both to be at rest after collision.**

**(b) Is it possible for any one to be at rest after collision.**

**Sol.**(a) No, because momentum will not be conserved in that case.

(b) Yes, when masses of two object are equal and collision is perfectly elastic.

**Q. A meterorite burn in the atmosphere before it reach the earth’s surface what happens to it momentum.**

**Sol.**The momentum is transferred to the air molecules and the earth.

**Q. Is centre of mass a reality?**

**Sol.**No, It is only a mathematical concept.

**Q. Give an example each for a body, where the centre of mass lies inside the body and outside the body.**

**Sol.**Following are the two examples – (i) In case of solid sphere the centre of mass lie inside the body. (ii) In case of hollow sphere or ring the centre of mass lie out side the body.

**Q. How does the centre of mass of an isolated system move?**

**Sol.**The centre of mass of an isolated system moves with a constant velocity.

**Q. A child sits stationary at one end of a long trolley moving uniformly with a speed $v$ on a smooth horizontal floor. If child gets up and runs about on the trolley in any manner, what is the speed of the C.M. of the (trolley + child) system?**

[NCERT]

**Sol.**The speed of the centre of mass of the system i.e. trolley and the child remains unchanged, if the child gets up and runs about on the trolley in any manner. It is because, the forces involved are from within the system, where as the state of the system can change only under the effect of an external force.

**Q. Briefly explain the motion of centre of mass of earth-moon system.**

**Sol.**The centre of mass of earth-moon system move in elliptical path round the sun. The system of earth- moon moves due to external force provided by the sun and the parts of the system under motion about COM is due to the internal force between them.

**Q. Two men stand facing each other on two boats floating on still water at a short distance apart. A rope is held at its ends by both. The two boats are found to meet always at the same point, whether each man pulls separately or both pull together. Why ? Will the time taken be different in the two cases?Neglect friction.**

**Sol.**The men on the two boats floating on water constitute a single system and the forces applied by them on each other are internal forces. When the man on the each boat pulls separately or both pull together, the center of mass of the system of boats remains fixed. It is because, centre of mass moves under the effect of external force only. Therefore, the two boats always meet at a fixed point, which is centre of mass of the system. It does not matter, whether each man pulls separately or both pull together. However, time taken in the two cases will not be same. The boats will take shorter time to meet, when both the men pull together.

**Q. From a uniform circular disc of diameter $d,$ a circular hole of diameter $\mathrm{d} / 6$ and having its centre at a distance of $\mathrm{d} / 4$ from the centre of the disc is scoped out. Find the centre of mass of the remaining portion.**

**Sol.**Let the mass per unit area of the disc be m. Then,

**Q. A projectile is fired with a velocity $u$ at an angle to the horizontal. At the highest point, it breaks into two equal parts. If one part, retraces its path, what is the direction and velocity of the other part? Ans**. Momentum at the highest point

**Sol.**

**Q. Locate the centre of mass of a system of three particles of masses $1.0 \mathrm{kg}, 2.0 \mathrm{kg}$ and $3.0 \mathrm{kg}$ placed at the corners of an equilateral triangle of $1 \mathrm{metre}$side.**

**Sol.**Let the triangle lie in the $X$ -Y plane, with its corner $m_{1}(1.0 \mathrm{kg})$ at the origin and the

side $m_{1} m_{2}$ along the $X$ -axis, Let $\left(x_{c m}, y_{c m}\right)$ be the co-ordinates of the centre of mass. By the

definition of the centre of mass. we have

**Q. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must be emitted in opposite directions.**

[NCERT]

**Sol.**Applying principle of conservation of momentum,

$m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}=0$ (Here, the letters have usual

meanings)

$\Rightarrow m_{2} \vec{v}_{2}=-m_{1} \vec{v}_{1}$

$\Rightarrow \vec{v}_{2}=-\frac{m_{1} \vec{v}_{1}}{m_{2}}$

The negative sign indicates that the products move in opposite directions.

**Q. (a) Two balls of masses $3 m$ and $m$ are separated by a distance $l .$ Find the position of the centre of mass.**

**(b) If these two balls are attached to a vertical axis by means of two threads whose combined length is $l$ and rotated in a horizontal plane with uniform angular velocity $\omega$ about an axis of rotation passing $\mathrm{g}$ in the $\mathrm{g}$ in the through centre of mass, prove that tensions in the two strings will become equal.**

**Sol.**(a) Let $x$ be the distance of centre of mass from the ball $A$ of mass $3 m[\text { Fig. }]$. Then, distance of the centre of mass from the ball $B$ will be $(l-x)$.

**Q. A bag of sand of mass $M$ is hanging from a rope. A bullet of mass $\frac{\mathrm{M}}{\mathrm{x}}$ is fired at it with a velocity $v .$ The bullet gets embedded into the bag. What is the velocity of the bag after the bullet gets embedded into it?**

**Sol.**

**Q. A machine gun has a mass of $20 \mathrm{kg}$. It fires $35 \mathrm{g}$ bullets at the rate of 4 bullets per second, with a speed of $400 \mathrm{ms}^{-1}$. What force must be applied to the gun to keep it in position?**

**Sol.**

Aliter. Number of bullets fired per second, $n=4$

Momentum given to the bullets per second $=n m v$

Same momentum per second is gained by the gun.

This gives the force acting on the gun and also the

force applied to keep the gun in position.

So, $F=n m v=4 \times 0.035 \times 400 N=56 N$

Aliter. Number of bullets fired per second, $n=4$

Momentum given to the bullets per second $=n m v$

Same momentum per second is gained by the gun.

This gives the force acting on the gun and also the

force applied to keep the gun in position.

So, $F=n m v=4 \times 0.035 \times 400 N=56 N$

**Q. A man of mass $m$ stands on a rope-ladder, which is tied to the free balloon of mass $M$ The balloon is initially at rest. In what direction and at what speed will the balloon move, if the man starts to climb the rope-ladder at a constant velocity $v$ relative to the ladder?**

**Sol.**Initially, when the man stands on the rope-ladder, the system (man, balloon and rope-ladder) is at rest i.e. initial momentum of the system is zero. When the man climbs up the ladder, the force acting on the balloon and rope-ladder is an internal force.Therefore, total momentum of the system should still remain zero. Since man is climbing up, the balloon and the rope-ladder would descend. Let $u$ be the velocity of the balloon and rope-ladder w.r.t. earth in downward direction, when the man starts climbing up. Since velocity of the man w.r.t the earth, $v^{\prime}=v-u$ According to the principle of conservation of momentum,

**Q. A rocket explodes in mid air how does this affect (a) Its total momentum (b) Its total kinetic energy.**

**Sol.**(a) Because no external force acts on the rocket its total momentum remains unchanged. (b) When the rocket explodes its fragments receive additional kinetic energy from explosion (Chemical, energy of fuel changes into kinetic energy). As a result of this the total kinetic energy gets increased.

**Q. The velocity of an aeroplane is doubled**

**(a) What will happen to the momentum, will the momentum remain conserved.**

**(b) What will happen to the kinetic energy will the energy remain conserved.**

**Sol.**(a) When the velocity of aeroplane is doubled. Its momentum gets doubled but the momentum of air also gets increased by the same amount in opposite direction so that total momentum of plane and air remains conserved.

(b) The kinetic energy will become four times additional energy will be obtained by burning of the fuel of the plane. In fact the total energy will remain conserved.

**Q. In an elastic collision what happened to mechanical energy?**

**Sol.**In an elastic collision the mechanical energy is not converted into any other form of energy and force of interaction are conservative in nature.

**Q. What is the work done by resistive force of air on vibrating pendulum in bringing it to rest?**

**Sol.**Work done is negative because the resistive force of air always act opposite of the direction of motion of the vibrating pendulum

**Q. Is whole of the kinetic energy lost in any perfectly inelastic collision.**

**Sol.**No, only that much amount of kinetic energy is lost as is necessary for the conservation of momentum.

**Q. What is the energy associated with a bird flying in**

**Sol.**A flying bird possesses both potential and kinetic energy as it is at a certain height above the groundand moving with certain velocity.

**Q. A cake of mud is thrown on a wall where it sticks what happens to its initial kinetic energy.**

**Sol.**A part of the kinetic energy is used deforming the cake and the remaining part converted into heat and sound energy.

**Q. A molecule in a gas container hits the wall with speed $100 \mathrm{m} / \mathrm{s}$ at an angle of $35^{\circ}$ with the normal and rebounds with the same speed. State whether linear momentum is conserved or not. What type of collision is it?**

**Sol.**Linear momentum is always conserved whether the collision is elastic or inelastic. Thus in the given situation, momentum is conserved. As the speed of the molecules remains same after the collision, so kinetic energy of the molecules is also conserved. Hence, the collision is elastic.

**Q. Is collision possible even without actual contact of the colliding particle?**

**Sol.**Yes, such collision is called collision at a distance. The collision between subatomic particles (like proton and nucleus) are example of such collisions.

**Q. The bob A of a simple pendulum released from $30^{\circ}$ to the vertical hits another bob $\mathrm{B}$ of the same mass at rest on a table as shown in figure. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be**

**elastic.**

**Sol.**The bob A shall not rise. This is because when two bodies of same mass undergo an elastic collision. Their velocities are interchanged after collision ball A will come to rest and the ball B would move with the velocity of A.

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**Q. At which point of the trajectory is the speed of motion minimum?**

**Sol.**

**Q. What is the angle of projection at which the and Range are equal.**

**Sol.**

**Q. What is the change in momentum between the initialand final points of the projectile path, if the range is maximum?**

**Sol.**

**Q. A hunter aims his gun at a monkey sitting on a tree and fires. Will the bullet hit the monkey?**

**Sol.**

**Q. Name a quantity which remains unchanged during the flight of an oblique projectile.**

**Sol.**

**Horizontal Component of velocity**

**Q. Name five physical quantities which change during the motion of an oblique projectile.**

**Sol.**

**Q. A projectile of mass is fired with velocity at an angle with the horizontal. What is the change in momentum as it rises to highest point of the trajectory?**

**Sol.**

**Q. A bomb is released from a horizontally flying bomber when it is vertically above the target. Shall it hit the target?**

**Sol.**

**Q. What is the path followed by a Javelin projected horizontally by an athlete?**

**Sol.**Projectile -parabolic path.

**Q. A projectile of mass is projected with velocity at an angle with the horizontal. What is the magnitude of the change in momentum of the projectile after time ?**

**Sol.**Change in momentum = Impulse = Force Time

**Q. Why does a tennis ball bounce higher on hills than in plains?**

**Sol.**

**Q. Give two applications and two examples of projectile motion.**

**Sol.**

**Q. A body is projected in horizontal direction with a uniform velocity from top of tower. Show that the path is parabola.**

**Sol.**

**Q. Galileo, in his book Two new sciences, stated that “for elevations which exceeds or fall short of ” by equal amounts, the ranges are equal.” Prove this statement.**

**Sol.**

**Q.**

**If the time of flight of a projectile projected with a velocity at an angle is find the condition for maximum range and its value?**

**Sol.**

**Q. Prove that the path of a projectile is a parabola.**

**Sol.**

**Q. The height and the distance along the horizontal plane of a projectile on a certain plane (with no atmosphere) are given by,**

**Sol.**

**Q. What are the possible angles of projection with same velocity to have same range?**

**Sol.**

**Q. A mass is projected horizontally with a velocity from a tower. Find the horizontal length it will cover from the foot of the tower?**

**Sol.**

**Q. Prove that a gun will shoot three times as high when its angle of elevation is as when it is but cover the same horizontal range.**

**Sol.**

**Q. When will the relative velocity of two moving objects be zero?**

**Sol.**

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## Motion in a Straight Line Class 11 Problems with Solutions

**Q. Can a body be said to be at rest as well as in motion at the same time.**

**Sol.**Yes, rest and motion are relative terms. A body at rest with respect to one body may be in motion with respect to another body.

**Q. Can a body have zero velocity but yet possesses acceleration?**

**Sol.**Yes, When a body is projected upward, at the highest point its velocity is zero but the acceleration is g.

**Q. Under what condition will the distance and displacement of a moving object will have the same magnitude?**

**Sol.**It indicates Instantaneous speed.

**Q. What does speedometer of a car indicate?**

**Sol.**It indicates Instantaneous speed.

**Q. As the body moves in a straight line, it moves smaller and smaller distances in equal intervals of time.**

**Sol.**The Body is Retarded

**Q. Derive an expression for stopping distance of a vehicle in terms of initial velocity and deceleration a.**

[NCERT]

**Sol.**Let $\mathrm{d}_{\mathrm{s}}$ be the distance travelled by a vehicle before

it stops.

Using $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{aS},$ we get, $0^{2}-\mathrm{v}_{0}^{2}=-2 \mathrm{ad}_{\mathrm{s}}$

$\mathrm{d}_{\mathrm{s}}=\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{a}}$

The stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increase the stopping distance by a factor $4,$ provided deceleration is kept the same.

**Q. Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the quality sign true?**

[NCERT]

**Sol.**(a) Consider a particle which moves form $A$ to $B$ and then back to $A$. In this case, the magnitude of displacement is zero. This is because the magnitude of displacement is equal to the shorteddistance between initial and final positions of the particle.

If $B$ is separated from $A$ by a distance $x$, then the total path length travelled by the particle is $2 x$.

(b) Average velocity $=\frac{\text { Displacement }}{\text { Time }}=\frac{0}{t}=0$

Average speed $=\frac{\text { Total path length }}{\text { Corresponding time interval }}=\frac{2 x}{t}$

**Q. Derive three equations of motion by calculus method. Express conditions under which they can be used.**

**Sol.**(1) $\mathrm{v}=\mathrm{u}+\mathrm{at}$ : Let any instant $t$, the velocity of a particle is $v$

$\therefore$ Acceleration, $a=\frac{d v}{d t} \Rightarrow d v=a d t$

Let at $t=0, v=u$

$\therefore$ Integrating equation (i), we get

$\int_{u}^{v} d v=\int_{0}^{t} a d t \Rightarrow \int_{u}^{v} d v=a \int_{0}^{t} d t \quad \therefore[v]_{u}^{v}=a[t]_{0}^{t}$

$\therefore(\mathrm{v}-\mathrm{u})=\mathrm{a}(\mathrm{t}-0) \Rightarrow \mathrm{v}-\mathrm{u}=\mathrm{at} \Rightarrow \mathrm{v}=\mathrm{u}+\mathrm{at}$

(2) $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}:$ Let a particle is moving with velocity $u,$ along $X$ -axis and at any instant $t,$ its displacement is $x$

$\therefore \mathrm{v}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} \Rightarrow \mathrm{dx}=\mathrm{v} \mathrm{dt}$

Also, we know that, $\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\therefore \mathrm{dx}=(\mathrm{u}+\mathrm{at}) \mathrm{dt}$ —–(i)

Now, at $\mathrm{t}=0, \mathrm{x}=\mathrm{x}_{0}$

$\therefore$ Integrating equation (i), we get

$\int_{x_{0}}^{x} d x=\int_{0}^{t}(u+a t) d t \Rightarrow[x]_{x_{0}}^{x}=\int_{0}^{t} u d t+\int_{0}^{t} a t d t$

$\Rightarrow \quad \mathrm{x}-\mathrm{x}_{0}=\mathrm{u}[\mathrm{t}]_{0}^{\mathrm{t}}+\mathrm{a}\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{t}}$

$\Rightarrow \quad \mathrm{x}-\mathrm{x}_{0}=\mathrm{u}(\mathrm{t}-0)+\mathrm{a}\left(\frac{\mathrm{t}^{2}}{2}-0\right)$

$\Rightarrow \mathrm{x}-\mathrm{x}_{0}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

But, $\mathrm{x}-\mathrm{x}_{0}=\mathrm{s} \quad \therefore \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

(3) $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \quad:$ Now we know that,

$\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}} \times \frac{\mathrm{dx}}{\mathrm{dt}}$

$\therefore \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \mathrm{v} \Rightarrow \mathrm{a} \mathrm{dx}=\mathrm{v} \mathrm{dv}\left[\because \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\right]$

Now limits, when $\mathrm{t}=0, \mathrm{x}=\mathrm{x}_{0}$ and $\mathrm{v}=\mathrm{u} \ldots \ldots$ –(i)

And when $\mathrm{t}=\mathrm{t}, \mathrm{x}=\mathrm{x}$ and $\mathrm{v}=\mathrm{v}$

Integrating the equation (i), we get

$\int_{x_{0}}^{x} a d x=\int_{u}^{v} v d v \Rightarrow a[x]_{x_{0}}^{x}=\left[\frac{v^{2}}{2}\right]_{u}^{v}$

$\Rightarrow \mathrm{a}\left(\mathrm{x}-\mathrm{x}_{0}\right)=\left[\frac{\mathrm{v}^{2}}{2}-\frac{\mathrm{u}^{2}}{2}\right]$ But, $\mathrm{x}-\mathrm{x}_{0}=\mathrm{s}$

$\therefore \quad \mathrm{as}=\frac{\mathrm{v}^{2}-\mathrm{u}^{2}}{2} \Rightarrow \mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}$

$\therefore \mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$

**Q. Can a body have constant velocity and still have varying speed?**

**Sol.**Not possible.

**Q. Are rest and motion absolute or relative terms?**

**Sol.**Both rest and motion are Relative Terms

**Q. Why the speed of the object can never be negative?**

**Sol.**Speed is distance covered per unit time. since distance cannot be negative there fore speed cannot be negative.

**Q. Define instantaneous velocity.**

Sol. Velocity of a body at a particular instant is called

Instantaneous velocity.

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\operatorname{Lt}_{\mathrm{st} \rightarrow 0}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)$

**Sol.**

**Q. What is average velocity?**

**Sol.**Avg. Velocity: The Ratio of Total Displacement and total time Taken is said as average Velocity.

i.e. Average velocity $=\frac{\text { Total displacement }}{\text { Total time }}$

If the position of a particle in time $t$ and

t’are $x(t)$ and $x\left(t^{\prime}\right),$ then, $V_{a v}=\frac{x\left(t^{\prime}\right)-x(t)}{t^{\prime}-t}=\frac{\Delta x}{\Delta t}$

**Q. The position of an object moving along -axis is given by where and is measured in second. What is it velocity at and . What is the average velocity between and**

[NCERT]

**Sol.**$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}+\mathrm{bt}^{2}\right)=2 \mathrm{bt}=5.0 \mathrm{tms}^{-1}$

At $\mathrm{t}=-\mathrm{s}, \mathrm{v}=0 \mathrm{ms}^{-1}$ and at $\mathrm{t}=2 \mathrm{s}, \mathrm{v}=10 \mathrm{ms}^{-1}$

Average Velocity =$\frac{x(4)-x(2)}{4-2}=\frac{a+16 b-a-4 b}{2}=6 b$

$=6 \times 2.5=15.0 \mathrm{ms}^{-1}$

**Q. A jet air plane travelling at the speed of ejects its products of combustion at the speed to relative to the jet plane. What is the speed of the combustion products w.r.t. an observer on the ground?**

[NCERT]

**Sol.**

**Speed of combustion product w.r.t. observer on the ground $=?$**

Velocity of jet air plane w.r.t. observer on ground $=500 \mathrm{kmh}^{-1}$

If $\overrightarrow{\mathrm{v}}_{\mathrm{j}}$ and $\overrightarrow{\mathrm{v}}_{\mathrm{o}}$ represent the velocities of jet and

observer respectively, then $\mathrm{v}_{\mathrm{j}}-\mathrm{v}_{\mathrm{o}}=500 \mathrm{kmh}^{-1}$

Velocity of jet air plane w.r.t. observer on ground $=500 \mathrm{kmh}^{-1}$

If $\overrightarrow{\mathrm{v}}_{\mathrm{j}}$ and $\overrightarrow{\mathrm{v}}_{\mathrm{o}}$ represent the velocities of jet and

observer respectively, then $\mathrm{v}_{\mathrm{j}}-\mathrm{v}_{\mathrm{o}}=500 \mathrm{kmh}^{-1}$

Similarly, If $\vec{v}_{\mathrm{c}}$ represents the velocity of the

combustion products $w \cdot r .$ jet plane, then,

$\mathrm{v}_{\mathrm{c}}-\mathrm{v}_{\mathrm{j}}=-1500 \mathrm{kmh}^{-1}$

The negative sign indicates that the combustion products move in a direction opposite to that of jet.

Speed of combustion products w.r.t. observer

$=v_{c}-v_{o}=\left(v_{c}-v_{j}\right)+\left(v_{j}-v_{o}\right)$

$=(-1500+500) \mathrm{kmh}^{-1}=-1000 \mathrm{kmh}^{-1}$

**Q. A police van moving on a highway with a speed of fires a bullet at a thief’s car speeding away in the same direction with a speed of . If the muzzle speed of the bullet is , with what speed does the bullet hit the thief’s car?**

[NCERT]

**Sol.**Speed of bullet, $v_{b}=$ speed of police vant speed with which bullet is actually fired

$\therefore \quad v_{b}=\left(\frac{25}{3}+150\right) \mathrm{ms}^{-1}=\frac{475}{3} \mathrm{ms}^{-1}$

Relative velocity of bullet w.r.t. thief’s car,

$\mathrm{v}_{\mathrm{bt}}=\mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{t}}=\left(\frac{475}{3}-\frac{160}{3}\right) \mathrm{ms}^{-1}=105 \mathrm{ms}^{-1}$”

**Q. Four persons K, L, M and start from the vertices of a square of side , simultaneously and move towards the neighbour in order always with the same speed of . When and where do they meet.**

[NCERT]

**Sol.**

**As $K, L, M$ and $N$ move towards the next person in order after a short time they will be at $K^{\prime}, L^{\prime}$ $M^{\prime}$ and $N^{\prime}$ respectively. The size to the square reduces. It indicates that they have come closer.**

After next short interval if they are at $K^{\prime \prime}, L^{\prime \prime}$

$M^{\prime \prime}$ and $N^{\prime \prime},$ the size of the square further reduces.

After next short interval if they are at $K^{\prime \prime}, L^{\prime \prime}$

$M^{\prime \prime}$ and $N^{\prime \prime},$ the size of the square further reduces.

Finally, they will follow a curvi-linear path and meet at $O,$ the centre of the square.

We know, the time taken $=\frac{\text { displacement }}{\text { Velocity in the direction of displacement }}$

$\therefore \quad \mathrm{t}=\frac{\mathrm{LO}}{\left(\frac{\mathrm{v}}{\sqrt{2}}\right)}=\frac{\frac{\mathrm{a}}{\sqrt{2}}}{\frac{\mathrm{v}}{\sqrt{2}}}=\frac{\mathrm{a}}{\mathrm{v}}$

**Q. On a two-lane road, car is travelling with a speed of . Two cars and approach car in opposite directions with a speed of each. At a certain instant, when distance. A B is equal to A C, both being decides to overtake before does. What minimum acceleration of is required to avoid**

**an accident?**

[NCERT]

**Sol.**$v_{A}=36 \mathrm{kmh}^{-1}=36 \times \frac{5}{18} \mathrm{ms}^{-1}=10 \mathrm{ms}^{-1}$

$v_{B}=v_{\mathrm{c}}=54 \mathrm{kmh}^{-1}=54 \times \frac{5}{18} \mathrm{ms}^{-1}=15 \mathrm{ms}^{-1}$

Relative velocity of $B$ w.t. $A, \quad v_{\mathrm{B}_{\mathrm{A}}}=5 \mathrm{ms}^{-1}$

Relative velocity of $C$ wr.t. $A, \mathrm{v}_{\mathrm{cA}}=25 \mathrm{ms}^{-1}$

Time taken by $C$ to cover distance

$\mathrm{AC}=\frac{1000 \mathrm{m}}{25 \mathrm{ms}^{-1}}=40 \mathrm{s}$ Now, for $B$

$1000=5 \times 40+\frac{1}{2} \mathrm{a} \times 40 \times 40$ On simplification

**Q. On a long horizontal moving belt, a child runs to and fro with a speed (with respect to the belt between his father and mother located apart on the moving belt. The belt moves with . speed of For an observer on a stationary . platform outside, what is the :**

**(a) speed of the belt child running in the direction of motion of the belt ? **

**(b) speed of the child running opposite to the direction of motion of the belt?**

**(c) time taken by**

**child in (a) and (b)? Which of the answer alter if motion is viewed by one of the parents?**

[NCERT]

**Sol.**Speed of child with respect to belt =9 $\mathrm{kmh}^{-1}$;

Speed of belt = 4 $\mathrm{kmh}^{-1}$

(a) When the child runs in the direction of motion Of the belt, then speed of child w.r.t. stationary

observer $=(9+4) \mathrm{kmh}^{-1}=13 \mathrm{kmh}^{-1}$

(b) When the child runs opposite to the direction of motion of the belt, then speed of child writ.

stationary observer $=(9-4) \mathrm{kmh}^{-1}=5 \mathrm{kmh}^{-1}$

(c) Speed of child w.rt. either parent $=9 \mathrm{kmh}^{-1}$

Distance to be covered

$=50 \mathrm{m}=0.05 \mathrm{km} ;$ Time $=\frac{0.05 \mathrm{km}}{9 \mathrm{kmh}^{-1}}=0.0056 \mathrm{h} \approx 20 \mathrm{s}$

If the motion is viewed by one of the parents, then the answers to (a) and (b) are altered but answer to (c) remains unaltered.

**Q. A man walks on a straight road from his home to**

**a market away with a speed of F**

**inding the market closed, he instantly turns and**

**walks back home with a speed of .**

**What is the ****(a) magnitude of average velocity, and (b) average ****speed of the man over the following intervals of ****time : (i) 0 to 30 min (ii) 0 to 50 min (iii) 0 to ****40 min?**

[NCERT]

**Sol.**

**Q. A boy standing on a stationary lift (open from**

**above) throws a ball upwards with the maximum**

**initial speed he can, equal to . (a) How**

**much time does the ball take to return to his hands**

** b) If the lift starts moving up with a uniform ****speed of , and the boy again throws the ****ball up with the maximum speed he can, how long ****does the ball take to return to his hands?**

**Sol.**(a) $\mathrm{v}(0)=49 \mathrm{ms}^{-1}, \mathrm{a}=-9.8 \mathrm{ms}^{-2}, \mathrm{v}(\mathrm{t})=0, \mathrm{t}=?$

$\mathrm{v}(\mathrm{t})=\mathrm{v}(0)+\mathrm{at}$

$\Rightarrow \quad 0=49-9.8 \mathrm{t} \Rightarrow 9.8 \mathrm{t}=49 \Rightarrow \mathrm{t}=\frac{49}{9.8} \mathrm{s}=5 \mathrm{s}$

This is the time taken by the ball to reach the maximum height. The time of descent is also 5 s. So, the total time after which the ball comes back is $5 s+5 s$ i.e., $10 s$

(b) The uniform velocity of the lift does not change the relative motion of ball and lift. So, the ball would take the same total time $i . e .,$ it would come back after 10 second.

**Q. A woman starts from her home at $9.00 A M,$ walks**

**with a speed pf on a straight road up to**

**her office away, stays at the office up to**

**and returns home by an auto with a speed**

**of . Choose suitable scales and plot the**

**graph of the motion.**

[NCERT]

**Sol.**

If $O$ is regarded as the origin for both time and

distance, then At $\mathrm{t}=9.00 \mathrm{AM}, \mathrm{x}=0$

and, at $\mathrm{t}=9.30 \mathrm{AM}, \mathrm{x}=2.5 \mathrm{km}$

OA is the $\mathrm{x}-\mathrm{t}$ graph of the motion when the

woman walks from her home to office. Her stay in the office from $9.30 \mathrm{AM}$ to $5.00 \mathrm{PM}$ is

represented by the straight line $A B$ in the graph.

Now, time taken to return home by an auto =

$\frac{2.5}{25} \mathrm{h}=\frac{1}{10} \mathrm{h}=6$ minuteSo, at $\mathrm{t}=5.06 \mathrm{PM}, \mathrm{x}=0$

These were Motion in a Straight Line Class 11 Problems with Solutions. These questions were important as per exams.

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**Q. What is the value of young’s modulus for perfectly rigid body.**

**Sol.**Infinite

**Q. What is the value of bulk modulus for an**

**incompressible liquid ?**

**Sol.**Infinite

**Q. What is the value of modulus of rigidity for any liquid ?**

**Sol.**Zero

**Q. How does young’s modulus change with rise in temperature.**

**Sol.**Young’s modulus of material decreases with rise in temperature.

**Q. Which type of strain is there, when a spiral spring is stretched by a force ?**

**Sol.**Shear strain.

**Q. The ratio stress/strain remains constant for small deformation. What will be effect on this ratio when the deformation made is very large ?**

**Sol.**When the deforming force is applied beyond elastic limit, the strain produced is more than that has been observed within elastic limit. Due to which the ratio stress/strain will decrease.

**Q. Which of the three Y,K, and h is possible in all the three states of matter (solid, liquid and gas)**

**Sol.**Bulk modulus of elasticity (K) only

**Q. Give the example of a body which is nearly perfectly elastic.**

**Sol.**Quartz fibre.

**Q. Give the example of a body which is nearly perfectly plastic**

**Sol.**Putty.

**Q. What do you mean by elastic hysteresis ?**

**Sol.**On reducing the stress to zero the material does not recover. Its unstrained state as some strain always remain. This phenomenon of lagging of strain behind the stress, is called elastic hysteresis.

**Q. What do you mean by an elastomer ?**

**Sol.**The materials which has large elastic limit but do not obey Hook’s law, are called elastomers.

Rubber is a very good example of such type of material.

**Q. Diamond is called a hard material. What does it mean in term of its modulus of elasticity?**

**Sol.**Modulus of elasticity becomes large. It means it becomes more elastic.

**Q. Paints and lubricating oils have low surface tension. Why?**

**Sol.**Being of low surface tension, they spread over a large surface area.

**Q. A piece of chalk immersed in water emits bubbles in all directions. Why?**

**Sol.**Chalk has pores all over its surface which acts as fine capillaries. When immersed in water, the water rises into these capillaries and forces the air out in the form of bubbles in water all around.

**Q. A mercury barometer reads slightly less than the actual pressure. Why?**

**Sol.**This is because, due to capillary action, mercury is depressed down in the barometer tube.

**Q. Some straw are spread on the surface of pure water filled in a vessel. On dropping a piece of sugar in water, the straw come nearer to the piece, but on dropping a piece of soap, they go away from it. Explain.**

**Sol.**The surface tension of sugar solution is greater than that of pure water. Therefore, the surface contracts near the piece of sugar and hence straw come nearer it. The surface tension of soap solution is less due to which the surface spreads near it and straw go away from it.

**Q. If the free surface of liquid is plane. Where does the force of surface tension act?**

**Sol.**The surface tension acts horizontally, and has no component normal to surface.

**Q. When pure water falls on a flat glass plate it spreads on the plate while the mercury when falls on a glass plate gets converted into small globules. Why?**

**Sol.**The adhesive force of mercury and glass is less than that of their cohesive force. Therefore, Hg breaks into droplets.

**Q. What are the properties of an ideal liquid ?**

**Sol.**An ideal liquid is one that is perfectly incompressible, non-viscous and unable to withstand shearing stress, however small.

**Q. Which one is most viscous among water, air, blood and honey ? Which one is the least viscous ?**

**Sol.**Honey is most viscous and air is least viscous.

**Q. When oil flows in a pipe, which layer moves fastest ?**

**Sol.**The axial layer moves fastest.

**Q. Hot liquid flows more rapidly than cold one, why?Ans**.The coefficient of viscosity of a liquid decreases with rise in temperature. Hence, hot liquid flows with larger speed than when it is cold.

**Sol.**

**Q. Why are machine parts usually jammed in winter?**

**Sol.**This is because the lubricant used in the machine becomes more viscous in winter.

**Q. According to Bernoulli’s theorem, the pressure of fluid should remain uniform in a pipe of uniform radius. But actually it goes on decreasing. Why is it so ?**

**Sol.**It is due to viscosity of fluid. When fluid flows, work is done against the viscous force, and this work is taken from the pressure energy. Hence pressure fluid falls.

**Q. If instead of fresh water, sea water is filled in the tank, will the velocity of efflux change ?**

**Sol.**No, velocity of efflux does not depend upon density of liquid.

**Q. Which is easier to lift in the air; 1 Kg steel or 1 Kg cotton ? What will be your answer if we lift these object in vacuum ?**

**Sol.**In case of air, it will be easier to lift 1 Kg of cotton because it will displace greater volume of air and lose more weight. In case of vacuum, as there will be no upward buoyant force on the objects, hence both will weight equally.

**Q. Which fundamental law forms the basis of equation of continuity ?**

**Sol.**Law of conservation of mass.

**Q. Why water is not used in barometers ?**

**Sol.**On account of following reasons, water is not used in barometers :

(i)Water sticks to the walls of the barometer tube.

(ii) Water has low density so height of water column becomes very large (11m) which is unmanegeable.

**Q. Why the boiling point of a liquid varies with pressure ?**

**Sol.**At boiling point, vapour pressure of liquid is equal to the atmospheric pressure. Hence, when the atmospheric pressure on the surface of liquid increases, the liquids boil at higher temperature to generate greater vapour pressure.

**Q. Why water does not flow out of a dropper unless the rubber bulb is pressed ?**

**Sol.**The upward air pressure at the dip of dropper is equal to the pressure of liquid column in it. When we press the rubber bulb, the inward pressure increases making the liquid flow out.

**Q. What is the fractional volume submerged of an ice cube in a pail of water placed in an enclosure which is falling freely under gravity ?**

**Sol.**Because the pail of water is falling freely under gravity, hence it will be in a state of weightlessness. Both the weight of the ice cube and the upthrust would be zero. So, the ice cube can float with any value of fractional volume submerged in water.

**Q. A wooden block is in bottom of a tank when water is poured in. The contact between the block and the tank is so good that no water gets between them. If there a buoyant force on the block?**

**Sol.**Since, there is no water under the block to exert an upward force on it, therefore there is no buoyant force.

**Q. What is hydrostatic paradox ?**

**Sol.**According to it the pressure exerted by a liquid depends only upon the height of liquid column and is independent of shape of the containing vessel.

**Q. Why the wings of an aeroplane are rounded outwards while flattened inwards ?**

**Sol.**Due to such shape of the wing, the velocity of air relative to the wind on the upper surface is larger than that of lower surface. This causes the pressure difference. The pressure below the wing is more than above the wing and hence causes upward lift on the aeroplane.

**Q. To keep a piece of paper horizontal, one should blow over it and not under it. Why ?**

**Sol.**When we blow air over a piece of paper, the velocity of air moving along the upper surface of paper is higher than that along the lower surface. Therefore, the air pressure on the upper surface of paper is lowered (according to Bernoulli’s theorem), while the pressure on the lower surface is still atmospheric. Due to higher pressure below the paper, the paper stays horizontal and does not fall under gravity.

**Q. A balloon filled with helium does not rise in air indefinitely but halts after a certain height (Neglect winds). Explain why ?**

**Sol.**A balloon filled with helium goes on rising in air so long as the weight of air displaced by it (i.e. upthrust) is greater than the weight of filled balloon. We know that the density of air decreases with height. Therefore, the balloon halts after attaining a height at which density of air is such that the weight of air displaced just equals the weight of filled balloon.

**Q. Two tooth picks floating on a water surface are parallel and close to each other. A hot needle is touched between them to the water surface. Explain why they fly apart ?**

**Sol.**When a hot needle is touched between them to the water surface, the surface tension of water in this region decreases due to increase in temperature. So, the outward pulling force due to surface tension of water on either side is more. Hence they fly apart.

**Q. The hot soup is tastier than the cold one. Why?**

**Sol.**Since, the surface tension of liquid decreases with rise in temperature. Therefore, the

surface tension of hot soup is less than that of cold soup. Hence, hot soup spreads over a larger area of the tongue and gives a tastier feeling.

**Q. Small insects swimming on water die when kerosene oil is added into the water. Why?**

**Sol.**Insects swim on water surface because of surface tension due to which surface behaves like a stretched membrane. On adding kerosene oil, the surface tension is appreciably reduced and the insects sink to death.

**Q. Discuss the various application of elasticity.**

**Sol.**Some of the important applications of the elasticity of the material are discussed as follow.

(1) Bridges are declared unsafe after long use.

(2) A metallic part of a machinery will get permanently deformed if subjected to a stress beyond the elastic limit. Therefore, the metallic part of the machinery is never subjected to the stress beyond the elastic limit.

(3) A crane is used for lifting and moving heavy load from one place to another. The crane makes use of a thick metallic rope. We have to specify the maximum load that will be lifted with the help of metallic rope. The maximum load should be such that the elastic limit of the material of the rope is not exceeded.

Let us suppose, the rope is required to lift a

maximum load of $10^{5} \mathrm{kg}$. Elastic limit of the steel

is $3 \times 10^{8} \mathrm{Nm}^{-2}$. Therefore maximum stress on

rope $=30 \times 10^{7} \mathrm{Nm}^{-2}$.

Let $r$ be radius of the required rope.

(4) To estimate the maximum height of a mountain.

(5) Bending of metallic bar. When a metallic bar of length l breadth b and thickness d placed on two wedges as shown in figure is loaded at the middle with a weight W = mg then depression in the bar is given by

$ \delta = \frac{m g l^{3}}{b d^{3} y}$

where y is the young’s modulus of the bar

**Q. Find the excess pressure inside a liquid drop.**

**Sol.**If the liquid surface is curved, then in equilibrium a net pressure acts on it towards the concave side or the pressure on the concave side is larger than that of convex side.

A liquid drop is spherical in shape and its outer surface is convex. Therefore, due to surface tension a net inward force acts on every molecule situated on the surface of the drop. As a result the pressure inside the drop is more than that of outside.

Let the radius of drop be R and the excess pressure inside it is P. Due to this excess pressure an outward force acts on the surface of the drop.

Let due to the excess of pressure the radius of

the drop increases from $R$ to $R+\Delta R$

Therefore, work done by excess of pressure in expansion of the drop.

$W=$ Force $\times$ displacement $=$ Pressure $\times$ area $\times$displacement

$W=P \times 4 \pi R^{2} \times \Delta R$

and increase in surface area of the drop $=$ Final

surface area – Initial surface area

$\therefore$ Increase in surface energy $=T \times 8 \pi R . \Delta R$

since, this increase in surface energy is due to

excess of pressure,

$\therefore P \times 4 \pi R^{2} \times \Delta R=T \times 8 \pi R . \Delta R \cdots P=\frac{2 T}{R}$

i.e. (i) $P \propto T \quad$ (ii) $P \propto \frac{1}{R}$

i.e. smaller the drop, greater is the excess pressure inside it.

Note : A bubble inside a liquid is similar to a liquid drop.

Let us consider an air bubble of radius R formed in a liquid of surface tension T. Like a liquid drop, the air bubble also has one surface in contact with the liquid. Hence proceeding in the same way as in case of liquid drop, we can derive an expression for excess pressure inside the air bubble.

$P=\frac{2 T}{R}$

**Q. Find excess pressure inside a soap bubble.**

**Sol.**Let us consider a thin soap bubble of radius R formed from a soap solution of surface tension T.

Due to air inside it the bubble has two free surfaces, one inside and other outside

Due to excess pressure p in it an outward force acts on the surface of the bubble. This is balanced by the force due to surface tension.

Now, the work done by excess pressure $P$ in

increasing radius from $R$ to $R+\Delta R$.

$W=$ Force $\times$ displacement $=P \times 4 \pi R^{2} \times \Delta R$

and increase in surface area of the bubble

$=2\left[4 \pi(R+\Delta R)^{2}-4 \pi R^{2}\right]$

$=2\left[4 \pi\left(R^{2}+2 . R . \Delta R+\Delta R^{2}\right)-4 \pi R^{2}\right]=16 \pi R \Delta R$

$\therefore$ Increase in surface energy $=T \times 16 \pi R \Delta R$

since, the increase in surface energy is due to

excess of pressure,

$\therefore P \times 4 \pi R^{2} \times \Delta R=T \times 16 \pi R . \Delta R \quad$ i.e. $\quad P=\frac{4 T}{R}$

**Q. Show that there is an excess of pressure on curved surface of liquid, which is more on concave face than that on convex face of curved surface of liquid.**

**Sol.**When the liquid surface is plane, the molecule on the liquid surface will be equally attracted on all the sides, and the forces (T, T) of surface tension will be acting tangentially to the liquid surface in opposite directions. Hence no resultant force is acting on the molecule. As a result no extra pressure exists to the inner side or outer side of liquid surface.

When the liquid surface is curved, the molecule O on the liquid surface is acted upon by the forces (T, T) due to surface tension along the tangents to surface. Resolving these forces into horizontal and vertical components, the horizontal components cancel each other whereas vertical components add up. Thus a resultant force acts on the curved surface of the liquid which acts towards its centre of curvature i.e. the resultant force is directed inwards in case of convex surface fig. (b) and is directed outwards in case of concave surface fig (c). In order that the curved surface of liquid may be in equilibrium, there must be an excess pressure on its concave side over that on the convex side, so that the forces of excess pressure may balance the resultant forces due to surface tension.

Hence, for the curved surface of liquid in equilibrium the pressure on concave side of liquid will be greater than pressure on its convex side.

**Q.**What is capillarity and deduce ascent formula.

**Sol.**The phenomenon of ascent or descent of liquid in a capillary tube is called capillarity or capillary action.

Generally those liquids which wet the glass surface rises up and the liquids which do not wet the glass surface descend in the capillary.

Ascent formulla : Let a uniform capillary tube of radius ‘r’ is dipped into a liquid in which liquid rises upto height ‘h’. Considering the meniscus (concave) radius nearly equal to the radius of capillary, the surface tension of liquid acts at the point of contact of meniscus along the direction of tangent making an angle q (angle of contact) with the wall of capillary tube. The horizontal and vertical components of T are Tsinq and Tcosq. The horizontal component Tsinq acting diametrically opposite around the meniscus cancel each other. But the vertical component Tcosq acts upwardly around the circumference of meniscus, due to which rises the liquid in the capillary tube.

\[\therefore \]Force which rises the liquid $=T \cos \theta .2 \pi r$

This force is balanced by the weight of the liquid.

Now, the volume of liquid raised = V = Volume of liquid upto height h + Volume of liquid in meniscus.

**Q. What do you understand by Reynold’s number? Give its physical significance.**

**Sol.**Reynold number is a pure number which determines the nature of flow of liquid through a pipe. According to Reynold, the critical velocity $v_{c}$ of a liquid flowing through a tube of diameter D is given by.

$v_{c}=\frac{k \eta}{\rho D} \Rightarrow K=\frac{\rho D V_{c}}{\eta}$

when, $\eta=$ coefficient of viscosity of liquid.

$K=A$ constant and is called Reynold’s number

$\rho=$ density of liquid

The value of K lies between 0 to 2000, for streamline or laminar flow and for value K above 3000, the flow of liquid is turbulent.

Physical significance : Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.

Let us consider a tube of small area of cross section A, through which a fluid of density r is flowing with velocity v.

The mass of fluid flowing through the tube per second,

$\Delta m=$ volume flowing per second $\times$ density.

$=A v \rho$

\[\therefore \]Inertial force per unit area

$=\frac{\text { rate of change of momentum }}{\text { Area }}$

$=\frac{(\Delta m) v}{A}=\frac{(A v \rho) v}{A}=v^{2} \rho$

and viscous force, $F=\eta A \frac{v}{r}$ (in magnitude)

where $r=$ radius of tube

$\frac{v}{r}=$ velocity gradient between the layers of this

liquid flow.

$\therefore$ viscous force per unit area, $=\frac{F}{A}=\frac{\eta v}{r}$

Hence, Reynold’s number,

$\mathrm{K}=\frac{\text { Inertial force per unit area }} {\text { viscous force per unit area }}=\frac{v^{2} \rho}{\eta v / r}=\frac{v \rho r}{\eta}$

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Get important questions from Thermodynamics and Heat for Class 11 exams. View all Physics important questions for all the chapters. These important questions from thermodynamics class 11 physics will play significant role in clearing concepts of Physics. This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions from heat and thermodynamics chapter.

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**Q. How is the mean kinetic energy of a gas related to its temperature?**

**Sol.**Mean kinetic energy is proportional to its absolute temperature.

**Q. Although the root mean square speed of gas molecules is of the order of the speed of sound in that gas, yet on opening a bottle of ammonia in one corner of a room its smell takes time in reaching the other corner. Explain why?**

**Sol.**The molecules of ammonia continuously collide with one another. Hence they are not able to advance in one particular direction speedly.

**Q. There are n molecules of gas in a box. If the number of molecules is increased to 2n. What will be the effect on the pressure of the gas? On the kinetic energy of the gas? On the root mean square speed of the molecules?**

**Sol.**

**Q. On**

**which factor does the average kinetic energy of gas molecules depend : nature of the gas, absolute temperature, volume.**

**Sol.**It depends only upon the absolute temperature

and is directly proportional to it $\left(=\frac{3}{2} K T\right)$

**Q. Equal masses of monoatomic gas and diatomic gas at the same temperature are given equal quantities of heat. Which gas will have greater temperature rise?**

**Sol.**Monoatomic gas will have greater temperature rise because a monoatomic gas has only translational energy but diatomic gas has both translational as well as rotational energy.

**Q. The graph shows the variation of the product PV with respect to the pressure (P) of given masses of three gases A, B and C. The temperature is kept constant, state with proper arguments which of these gases is ideal.**

**Sol.**Gas

*C*is ideal, because

*PV*is constant for this gas. It means the gas

*C*obeys Boyle’s law at all pressures.

**Q. There is a temperature known as Boyle’s temperature ( $\left.T_{\mathrm{B}}\right)$ for each real gas at which it behaves like an ideal gas. But if gas temperature T is less than $\mathrm{T}_{\mathrm{B}}$ then does it behaves like ideal gas.**

**Sol.**No, because gas does not behave like an ideal gas if $T>T_{B}$ and $T<T_{B}$.

**Q. What does universal gas constant signifies.**

**Sol.**Universal gas constant signifies the work done by (or on) a gas per mole per kelvin.

**Q. The molecules of a gas are 2n a state of continuous, rapid and random motion. They move in all directions with different speeds. What is the range of their speeds?**

**Sol.**Zero to infinite.

**Q. Two different gases have exactly the same temperature. Does this mean that their molecules have same r.m.s. speed?**

**Sol.**When the two gases have exactly the same temperature, the average kinetic energy per

molecule $\left(=\frac{1}{2} m c^{2}=\frac{3}{2} k T\right)$ for each gas is the

same. But as the different gases may have molecules of different masses, the r.m.s. speed (*c*) of molecules of different gases shall be different.

**Q. Distinguish between average speed and r.m.s. speed. If three molecules have speed u1, u2, u3 what will be their average speed and r.m.s. speed.**

**Sol.**Average speed is the arithmetic mean of the speeds of the molecules.

$\therefore$ Average speed $=\frac{u_{1}+u_{2}+u_{3}}{3}$

But r.m.s. speed is the root mean square speed and is defined as the square root of the mean of the squares of different speeds of the individual molecules.

**Q. A gas is filled in a cylinder fitted with a piston at a definite temperature and pressure. Explain on the basis of kinetic theory : on pulling the piston out, the pressure of gas decreases.**

**Sol.**On pulling the piston out, the volume of the cylinder for the given gas increases. Due to which the molecules of gas get more space to move about. As a result of which less molecules will collide with the wall of cylinder per second and hence less momentum is transferred to the wall per second. In addition to it, now these collisions take place on the larger area of the walls. Due to both these reasons, the pressure decreases.

**Q. State law of equipartition of energy.**

**Sol.**According to this law, for any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom and the energy associated with each molecule per degree of freedom is $\frac{1}{2} k T$, and hence a constant. (Here

*k*is Boltzmann constant and

*T*is temperature of the system).

**Q. Obtain the relation between degree of freedom of a gas and ratio of two principal specific heats of the gas.**

**Sol.**Suppose a polyatomic gas molecule has

*n*degrees of freedom.

Total energy associated with a gram molecule of the gas

This is the relation between $\gamma$ and *n*. Hence the value of for a polyatomic gas can be determined from degree of freedom *n*.

**Q. Why the molecular motion of the molecules ceases at zero kelvin?**

**Sol.**We know kinetic energy of a molecule is proportional to the absolute temperature i.e.

$\frac{1}{2} m c^{2} \propto T$

At $T=0$

$\frac{1}{2} m c^{2}=0$ since $\frac{1}{2} m \neq 0 \quad \therefore c=0$

Thus molecular motion ceases at zero kelvin.

**Q. Explain the concept of temperature on the basis of kinetic theory?**

**Sol.**According to kinetic theory of gases, the pressure exerted by the gas is given by $P=\frac{1}{3} \frac{m N}{V} c^{2} \ldots \ldots$ (i)

Where $m=$ Mass of one molecule, $N=$ Number of molecules present in volume $V$ and $\frac{1}{c^{2}}=$ Mean square velocity exerted by the gas is given by $P=\frac{1}{3} \frac{m N}{V} c^{2} \ldots \ldots$ (i)

Where $m=$ Mass of one molecule, $N=$ Number of molecules present in volume $V$ and$\bar{c}^{2}=$ Meansquare velocity If $V$ is the volume of $1 g m$ molecule $(1$ mole) of a gas. Then $N$ will be number of molecules present in volume $V$ i.e., $N$ will be Avogadro’s number. Therefore, $m N$will be gram molecular mass.

From eq. (i)$P V=\frac{1}{3} m N c^{2}$

But $m N=M=$ Gram molecular mass

$P V=\frac{1}{3} M c^{2}$ ……..(ii)

But from ideal gas equation, $P V=R T$ …..(iii)

From eq. (ii) and (iii)

$\frac{1}{3} M c^{2}=R T$ …….(iv)

$\overline{c^{2}}=\frac{3 R T}{M}$

$\overline{c^{2}} \propto T$ ………(v)

Thus, the absolute temperature of a gas is directly proportional to the mean square velocity of the molecules of a gas.

From eq. (iv)

$\frac{2}{3} \times \frac{1}{2} M c^{2}=R T$

$\Rightarrow \frac{1}{2} M c^{2}=\frac{3}{2} R T$

$\frac{1}{2} m N c^{2}=\frac{3}{2} R T$ or $\frac{1}{2} m \bar{c}^{2}=\frac{3}{2} \frac{R}{N} T$

or $\frac{1}{2} m c^{2}=\frac{3}{2} K T$ or $E=\frac{3}{2} K T$ ………(vi)

where $E=\frac{1}{2} m c^{2}=$ Mean kinetic energy of the

molecule and $=\frac{R}{N}=$ constant $k$ is called

Boltzmann’s constant.

From eq. (vi) $E \propto T$

Thus, the mean kinetic energy of a molecule is directly proportional to the absolute temperature of the gas.

If $T=0$ then from eq. (v), $\bar{c}^{2}=0$

and from eq. (vi) $E=0$

Thus, the absolute zero is that temperature at which the r.m.s. velocity of the gas becomes zero.

Since, can never be negative, therefore the value of absolute temperature *T* can never be negative. This is the reason due to which the lowest temperature on kelvin’s scale is assumed to be 0K.

**Q. At what temperature does all molecular motions cease?**

**Sol.**Zero kelvin (or Absolute zero $\left.=-273.13^{\circ} \mathrm{C}\right)$

**Q. What is the nature of graph, taking $0^{\circ} \mathrm{C}$ along the $\mathrm{y}$ -axis and ‘F along the x-axis. What is the slope of this graph?**

**Sol.**Straight line of the form :

$y=m x-c,\left[c=\frac{5}{9}(F-32)\right]$, having slope $=5 / 9$.

**Q. How is the molar gas constant related to the principal gas constant?**

**Sol.**

*R = Mr*.

**Q. Why temperature beyond $1200^{\circ} \mathrm{C}$ cannot be measured accurately by a platinum resistance thermometer?**

**Sol.**Beyond $1200^{\circ} \mathrm{C},$ platinum begins to evaporate.

**Q. Two bodies at different temperatures $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$if brought in thermal contact do not necessarily settle to the mean temperature $\left(\frac{T_{1}+T_{2}}{2}\right)$**

**Sol.**In thermal contact, heat flows from a body at higher temperature to the body at lower temperature till temperatures become equal. The final temperature can be the mean temperature of $\left(\frac{T_{1}+T_{2}}{2}\right)$ only

**Q. Why platinum wire can be sealed into glass but not the copper wire?**

**Sol.**Platinum has coefficient of expansion slightly less than that of crown glass but copper has greater thermal expansion coefficient. So, if a platinum wire is put into molten glass thus the platinum-wire fits into a hole in molten glass. Since glass contracts more than platinum, glass fixes on to the wire. But, since copper is more expansible than glass, the copper wire contracts more than the hole in the glass, hence the copper wire losens out.

**Q. What is the ice-point on kelvin scale? If the temperature of body increases by $1^{\circ} \mathrm{C},$ what will be change in temperature on kelvin scale?**

**Sol.**$273.15 K$ is the ice point on kelvin scale. The relation between kelvin scale and centigrade scale is given by $\left.T(K)=\theta^{(0} C\right)+273.15$. The increaments in the temperature on kelvin scale are the same as the those of the celsius scale. i.e. $1^{\circ} C=1 K$.

**Q.**Which single point is used as reference point in thermometer ?

**Sol.**This single reference point is triple point of the substance.

**Q. An electric refrigerator transfers heat from the cold cooling coils to the warm surroundings. Is it against the second law thermodynamics ? Justify your answer ?**

**Sol.**No, it is not against the second law. This is because external work is done by the compressor of this transfer of heat.

**Q. Can the carnot engine be realised in practice ?**

**Sol.**No, because carnot engine is an ideal heat engine, whose conditions are not realised in practice.

**Q. Does the mass of body change when it is heated or cooled ?**

**Sol.**Yes in heating or cooling, a body absorbs or loses energy. The mass will, therefore increase or decrease accordingly as per the relation $E=m c^{2}$

**Q. It is possible to convert internal energy into work ?**

**Sol.**Yes, for example, in explosion of a bomb, chemical energy (which is a form of internal energy) is converted into kinetic energy.

**Q. An ideal gas is compressed at constant temperature, will its internal energy increase or decrease ?**

**Sol.**No, because internal energy of an ideal gas depends only on temperature of the gas.

**Q. In summer, when the valve of a bicycle tube is opened, the escaping air becomes cold. Why?**

**Sol.**This happens due to adiabatic expansion of air of the tube of the bicycle.

**Q. Why does a gas get heated on compression ?**

**Sol.**Because work done in compressing the gas increases the internal energy of gas.

**Q. Why are the brake-drums of a car heated when the car moves down a hill at constant speed ?**

**Sol.**Since the speed of car is not increasing, the gravitational potential energy is converted into internal energy of the break drums which are heated.

**Q. If an electric fan be switched in a closed room, will the air of the room be cooled ? If not, why have feel cold ?**

**Sol.**The air will not be cooled, but will be heated because due to motion of the fan, the speed of air molecules will increase. In fact, we feel cold due to evaporation of sweat.

**Q. Is superheating of steam an isobaric process or isothermal process, and why ?**

**Sol.**Isobaric, because during heating the temperature of steam does not remain constant.

**Q. What are the limitations of first law of thermodynamics ?**

**Sol.**The limitations are :

(i)There is no information about the direction in which change takes place.

(ii) It does not give any idea about the extent to which the change takes place.

**Q. A gas does work during isothermal expansion ? What is the source of mechanical energy so produced ?**

**
**

**
**

**The energy required for doing work during isothermal expansion is acquired as heat by the gas from surroundings. Here $\quad \Delta Q=\Delta W i . e$.**

**Sol.**$\Delta U=0$

**Q. What is the valve of work-done in a cyclic process ?**

**Sol.**The work-done in a cyclic process is equal to the area of the closed curve on a

*P-V*-diagram.

**Q. Name the sink in case of steam engine ?**

**Sol.**Atmosphere is the sink in most of the cases.

**Q. Why can a ship not use the internal energy of sea water to operate its engine ?**

**Sol.**For operation of a heat engine we require a sink at temperature lower than the source and of sufficiently high thermal capacity, which is not possible in sea.

**Q. What do you understand by triple point of water ? Why it is unique ?**

**Sol.**It is the temperature at which three phases of water namely ice, water and water vapour are equally stable and exist simultaneously. It is unique because it occurs at one single temperature

$(273.16 K)$ at one single pressure of about $0.46 \mathrm{cm}$

of $\mathrm{Hg}$. column.

**Q. Two cylinders A and B of equal capacity are connected to each other via a stop-cock.**

**A contains a gas at standard temperature and pressure, B is completely evacuated. The entire system is thermally insulated. The stop-cock is suddenly opened. Answer the following **

**(i) What is the final pressure of the gas in A and B?**

**(ii) What is the change in internal energy of the gas ?**

**(iii) What is the change in temperature of the gas?**

**(iv) Do the intermediate state the system (before setting to final equilibrium state) be on its P-V-T surface ?**

**Sol.**(i) Because an opening the stop-cock, the volume of gas is doubled at constant temperature as the system is thermally insulated, as the gas expands against zero pressure, so, in accordance to first law of thermodynamics,

$\Delta U=0$. Further internal energy of an ideal gas depends on temperature only. Hence, temperature of gas will remain constant. Therefore pressure is halved *i.e.* pressure will reduce to 0.5 atm.

(ii) An explained above, there is no change in internal energy.

(iii)As explained above, there is also no change in temperature.

(iv)No, since the process (called free expansion) is rapid and cannot be controlled. The intermediate states are non-equilibrium states and do not satisfy the gas equation. After a short time interval, the gas returns to equilibrium states which lies on its P-V-T surface.

**Q. A Thermos bottle containing coffee is vigorously shaken. Considering coffee as a system**

**(a) has heat been added to it? (b) has work been done on it? (c) has its internal energy changed? (d) does its temperature rise ?**

**Sol.**(a) No, heat has not been transferred to the coffee as the coffee is in thermos flask which is insulated from the surroundings

*i.e.*$\Delta Q=0$

(b) Yes, in shaking work is done on the coffee against viscous force $i . e . \Delta W$ is negative.

(c) According to first law, $\Delta Q=\Delta U+\Delta W,$ however as here and negative, so be positive i.e. internal energy of coffee will increase.

(d) Now, as internal energy of system depends upon its temperature *i.e.* , so temperature of the system (coffee) will also increase.

**Q. Why is the conversion of heat into work not possible without a sink at lower temperature ?**

**Sol.**If we want to convert heat energy continuously into work, a part of heat energy absorbed from the source has to be rejected. The rejection of heat energy is possible only if there is a body having temperature less than that of source. This body at lower temperature is called sink.

**Q. What is meant by reversible engine ? Explain why the efficiency of reversible engine is maximum ?**

**Sol.**The engine in which the process can be retraced at any stage of its operation by reversing the boundary conditions is called reversible. Its efficiency is maximum because in such a device dissipation of energy takes place against friction etc. only.

**Q.**

*Show that an adiabatic curve is always steeper than an isothermal curve.*

**Sol.**Isothermal process : That thermal process in which the pressure and volume of the system (gas) change but temperature remains constant is called isothermal process.

This process is achieved

(a) if the gas is enclosed in a vessel whose walls are made of a highly conducting material and

(b) the gas is compressed or allowed to expand very slowly.As temperature remains constant in an isothermal process, the gas equation is, *PV* = constant (*K*)

Adiabatic process

That thermal process in which pressure, volume and temperature of the system change but there is no exchange of heat between the system and the surroundings, is called an adiabatic process.

This process is achieved

(a) if the gas in enclosed in a vessel whose walls are made of a highly insulating material and

(b) the gas is compressed or allowed to expand very quickly.

The gas equation is, $P V^{\gamma}=$ constant,

Slopes of isothermal and adiabatic curves

The slope of an isothermal or adiabatic curve is

given by $\frac{d P}{d V}$.

(a) For an isothermal process, $P V=k$

Differentiating on both sides, we get,

$P d V+V d P=0$

$\Rightarrow \frac{d P}{d V}=-\frac{P}{V}$ is slope of isothermal curve,

$\left(\frac{d P}{d V}\right)=-\frac{P}{V}$ ………….. (1)

(b) For an adiabatic process, $P V^{\gamma}=K$ (constant)

Differentiating on both sides, we get,

$P_{\gamma} v^{\gamma-1} . d V+V^{\gamma} . d P=0$

$\Rightarrow V^{\gamma} d P=-\gamma P V^{\gamma-1} d V |$

$\therefore \frac{d P}{d V}=-\frac{\gamma P V^{\gamma-1}}{V^{\gamma}}=-\frac{\gamma P}{V}$

i.e. slope of adiabatic curve,

$\left(\frac{d P}{d V}\right)_{A}=-\frac{\gamma P}{V}$

From equation ( 1) and $(2),$ we have,

$\left(\frac{d P}{d V}\right)_{A}=\gamma\left(-\frac{P}{V}\right)=\gamma\left(\frac{d P}{d V}\right)_{I}$

As $\gamma$ is always greater than one, hence slope of adiabatic curve is greater than the slope of isothermal curve. In other words, an adiabatic curve is sleeper than an isothermal curve.

**Q. State and explain first law of thermodynamics. Establish the relation between two principal specific heats of a gas on the basis of this law.**

**Sol.**First law of thermodynamics : It is a statement of conservation of energy in thermodynamical process.

According to it “when some quantity of heat (*dQ*) is supplied to system capable of doing external work, then the quantity of heat absorbed by the system (*dQ*) is equal to the sum of the increase in internal energy of the system (*dU*) and external work-done by the system (*dW*) ie.

$d Q=d U+d W$

Relation between two principal specific heat of a gas : A gas has two specific heats

Relation between two principal specific heat of a

gas: A gas has two specific heats

(i) Specific heat at constant volume $\left(C_{V}\right):$ It is the amount of heat required to raise the temperature of unit mass of a gas by $1^{\circ} C$ at constant volume. It is denoted by $C_{V}$

(ii) Specific heat at constant pressure $\left(\mathrm{C}_{P}\right):$ It is the amount of heat required to raise the temperature of unit mass of gas by $1^{\circ} \mathrm{C}$ at constant pressure. It is denoted by $C_{P}$. Let 1 mole of a gas is supplied $\Delta Q$ amount of heat at constant volume so that the temperature of the gas rises by $\Delta T$.

$\therefore \Delta Q=1 \times C_{V} \times \Delta T=C_{V} \Delta T$ ……………(1)

$\begin{array}{ll}{\text { As volume remains constant, }} \\ {d V=0} & {\therefore \Delta W=P \Delta V=0} \\ {\text { According to first law of thermodynamics, }} \\ {\Delta Q=\Delta U+\Delta W} \\ {\Rightarrow \Delta U=C_{V} \Delta T}\end{array}$

Now, let 1 mole of same gas is given an amount of heat $\Delta Q$ at constant pressure so as to increase the temperature by $\Delta T,$ then,

$\Delta Q=1 \times C_{P} \times \Delta T \Rightarrow \Delta Q=C_{P} \Delta T$ ………….(3)

If $\Delta V$ is the change in volume at constant pressure,

hen,

$\Delta W=P \Delta V$

From first law, $\Delta Q=\Delta W+\Delta U$

$\Rightarrow \Delta Q=P \Delta V+\Delta U \quad \Rightarrow C_{P} \Delta T=P \Delta V+\Delta U$

$\Rightarrow P \Delta V+C_{V} \Delta T=C_{P} \Delta T[\text { From equation }(2)]$

$\Rightarrow\left(C_{P}-C_{V}\right) \Delta T=P \Delta V$ ………..(4)

$\begin{array}{l}{\text { But from ideal gas equation, }} \\ {P V=R T \quad \text { ideal }} \\ {\text { From equation }(4), \text { we have, }}\end{array} \Rightarrow P \Delta V=R \Delta T$

$\left(C_{P}-C_{V}\right) \Delta T=R \Delta T$

This is known as Mayor’s equation.

Note

(1) In above equation units of all quantities $\left(C_{P}, C_{V} \text { and } R\right)$ are Joule $m o l^{-1} K^{-1} .$ If $C_{P}$ and $d$ $C_{V}$ are expressed in units calorie $m o l^{-1} K^{-1},$ then $C_{P}-C_{V}=\frac{R}{J}$

(2) In above relation is expressed for one gram of gas, then, $C_{P}-C_{V}=\frac{r}{J}$

(3) As value of molar gas constant $R$ is $(+v e) 8.31$ Jmol $^{-1} K^{-1}$, it implies that $C_{P}>C_{V}$ and their difference in equal to $8.31 J \mathrm{mol}^{-1} K^{-1}$.

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