ncert solutions class 11 physics

There are many different topics that are discussed in physics. This makes this subject very different and unique from other subjects. It covers a variety of topics that a student can learn. Thus, it gives students an opportunity to choose on their own which topics they should prepare well. With NCERT solutions for class 11 physics, you will get the choice of choosing your topics. You will have individual chapters that you can pick from and start learning. NCERT solutions for class 11 physics are perfect for you to develop more understanding of this subject.

NCERT Solutions for Class 11 Physics Chapter 15 Waves

NCERT Solutions for Class 11 Physics Chapter 15 Waves are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 15 Waves.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 15 Waves:

Section NameTopic Name
15Waves
15.1Introduction
15.2Transverse and longitudinal waves
15.3Displacement relation in a progressive wave
15.4The speed of a travelling wave
15.5The principle of superposition of waves
15.6Reflection of waves
15.7Beats
15.8Doppler effect

NCERT Solutions Class 11 PhysicsPhysics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 15. 1.  A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q1

Question 15. 2. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 ms-1 ? (g = 9.8 ms-2 )
Answer:  Here, h = 300 m, g = 9.8 ms-2  and velocity of sound, v = 340 ms-1 Let t1 be the time taken by the stone to reach at the surface of pond.
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q2

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Question 15. 3.  A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 2 0°C = 340 ms-1.
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q3

Question 15. 4.
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q4
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q4.1

Question 15. 5. You have learnt, that a travelling wave in one dimension is represented by a function y = f(x, t), where x and t must appear in the combination x-vtorx + vt i.e., y =f (x ± vt). Is the converse true? That is, does every function of (x – vt) or (x + vt) represent a travelling wave? Examine, if the following functions for y can possibly represent a travelling wave?
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q5
Answer: No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y none satisfies this condition. Therefore, none can represent a travelling wave.

Question 15. 6.  A bat emits ultrasonic sound of frequency 1000 kHz in air. If this sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air = 340 ms-1 and in water = 1486 ms-1.
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q6

Question 15. 7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km s-1? The operating frequency of the scanner is 4.2 MHz.
Answer: Here speed of sound => υ = 1.7 km s-1 = 1700 ms-1 and
frequency υ= 4.2 MHz = 4.2 x 106 Hz
.’. Wavelength, A = υ/V = 1700/(4.2 x 106) =4.1 x 10-4 m.

Question 15. 8. A transverse harmonic wave on a string is described by
y(x, t) = 3.0 sin (36 t + 0.018 x + π/4)
where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation ?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q8

NCERT Solutions for Class 11 Physics Chapter 15 Waves Q8.1

Question 15. 9. For the wave described in Exercise 8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shape of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another : amplitude, frequency or phase?
Answer: The transverse harmonic wave is
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q9
Using the values of t and y (as in the table), a graph is plotted as under The graph obtained is sinusoidal.
Similar graphs are obtained for y x = 2 cm and x = 4 cm. The (incm) oscillatory motion in the travelling wave only differs in respect of phase. Amplitude and frequency of oscillatory motion remains the same in all the cases.
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q9.1

Question 15. 10. For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m (b) 0.5 m
(c) λ/2 (d) 3λ/4.
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q10

Question 15. 11. The transverse displacement of a string (clamped at its two ends) is given by y(x, t) = 0.06 sin 2π /3 x cos (120π t)
where x, y are in m and t in s. The length of the string is 1.5 m and its mass is 3 x 10-2 kg. Answer the following:
(i) Does the function represent a travelling or a stationary wave?
(ii) Interpret the wave as a superimposition of two waves travelling in opposite directions. What are the wavelength, frequency and speed of propagation of each wave?
(iii) Determine the tension in the string.
Answer:  The given equation is
y(x, t) = 0.06 sin 2π/3 x cos 120 πt …(1)
(i)As the equation involves harmonic functions of x and t separately, it represents a stationary wave.
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q11

Question 15. 12. (i) For the wave on a string described in Question 11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers, (ii) What is the amplitude of a point 0.375 m away from one end?
Answer: (i) For the wave on the string described in questions we have seen that l = 1.5 m and λ = 3 m. So, it is clear that λ = λ /2 and for a string clamped at both ends, it is possible only when both ends behave as nodes and there is only one antinode in between i.e., whole string is vibrating in one segment only.
(a) Yes, all the sring particles, except nodes, vibrate with the same frequency v = 60 Hz.
(b) As all string particles lie in one segment, all of them are in same phase.
(c) Amplitude varies from particle to particle. At antinode, amplitude = 2A = 0.06 m. It gradually falls on going towards nodes and at nodes, and at nodes, it is zero.
(ii) Amplitude at a point x = 0.375 m will be obtained by putting cos (120 πt) as + 1 in the wave equation.
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q12

Question 15. 13. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q13
Answer: (a) It represents a stationary wave.
(b) It does not represent either a travelling wave or a stationary wave.
(c) It is a representation for the travelling wave.
(d) It is a superposition of two stationary wave.

Question 15. 14. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10-2 kg and its linear mass density is 4.0 x 10-2 kg m-3. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q14

Question 15. 15. A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a turning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q15
[Approximation has been used because edge effect is being ignored. Moreover, we know that in the case of a closed organ pipe, the second resonance length is three times the first resonance length.]
On simplification, n = 1
Now, (2n-1)υ/4l1 = 340. Substituting values
(2 x 1 -1) υ x 100/4 x 25.5 = 340 or υ =346.8 ms-1

Question 15. 16. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 k Hz. What is the speed of sound in steel?
Answer:  Here, L = 100 cm = 1m, v = 2.53 k Hz = 2.53 x 103 Hz
When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, a node is formed at the middle and ant mode is formed at each end.
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q16

Question 15. 17. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 ms-1).
Answer:  Here length of pipe, 1 = 20 cm = 0.20 m, frequency v = 430 Hz and speed of sound in air υ = 340 ms-1
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q17

Question 15. 18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Answer: Let υ1 and υ2 be the frequencies of strings A and B respectively.
Then, υ1 = 324 Hz, υ2 = ?
Number of beats, b = 6
υ2 = υ1 ± b = 324 ± 6 !.e., υ2 = 330 Hz or 318 Hz
Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency υ1 will be reduced i.e., number of beats will increase if υ2 = 330 Hz. This is not so because number of beats become 3.
Therefore, it is concluded that the frequency υ2 = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ2 = 318 Hz.

Question 15. 19. Explain why (or how):
(a) in a sound wave, a displacement node is a pressure antinode and vice versa.
(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”.
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes.
(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:  (a) In a sound wave, a decrease in displacement i.e., displacement node causes an increase in the pressure there i.e., a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure.
(b) Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles.
(c) The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear and used to differentiate between the two notes.
(d) This is due to the fact that gases have only the bulk modulus of elasticity whereas solids have both, the shear modulus as well as the bulk modulus of elasticity.
(e) A pulse of sound consists of a combination of waves of different wavelength. In a dispersive medium, these waves travel with different velocities giving rise to the distortion in the wave.

Question 15. 20. A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 ms~1. (b) recedes from the platform with a speed of 10 ms-1(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms-1   .
Answer: Frequency of whistle, v = 400 Hz; speed of sound, υ= 340 ms-1 speed of train, υs= 10 ms1
(i) (a) When the train approaches the platform (i.e., the observer at rest),
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q20

NCERT Solutions for Class 11 Physics Chapter 15 Waves Q20.1
(ii) The speed of sound in each case does not change.It is 340 ms-1 in each case.

Question 15. 21. A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 ms-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms-1? The speed of sound in still air can be taken as 340 ms-1?
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q21

Question 15. 22. A travelling harmonic wave on a string is described by y (x, t) = 7.5 sin (0.0050x + 12t + π /4)
(a) what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacement and velocity as the x = 1 cm point at t = 2s, 5s and 11s.
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q22

NCERT Solutions for Class 11 Physics Chapter 15 Waves Q22.1

Question 15. 23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium, (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:  (a) In a non dispersive medium, the wave propagates with definite speed but its wavelength of frequency is not definite.
(b) No, the frequency of the note is not 1/20 or 0.50 Hz. 0.005 Hz is only the frequency ‘ of repetition of the pip of the whistle.

Question 15. 24. One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer: Here, mass per unit length, g = linear mass density = 8 x 10-3 kg m-1;
Tension in the string, T = 90 kg = 90 x 9.8 N= 882 N;
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q24

Question 15. 25. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 ms-1.
Answer:
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q25

Question 15. 26. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s-1 . A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer: Here speed of S wave, υs = 4.0 km s-1 and speed of P wave, υp = 8.0 km s-1. Time gap between P and S waves reaching the resimograph, t = 40 min = 240 s.
Let distance of earthquake centre = sKm
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q26

Question 15. 27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall ?
Answer:  Here, the frequency of sound emitted by the bat, υ = 40 kHz. Velocity of bat, υs = 0.03 υ, where υ is velocity of sound. Apparent frequency of sound striking the wall
NCERT Solutions for Class 11 Physics Chapter 15 Waves Q27

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NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 14 Oscillations.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 14 Oscillations:

Section NameTopic Name
14Oscillations
14.1Introduction
14.2Periodic and oscilatory motions
14.3Simple harmonic motion
14.4Simple harmonic motion and uniform circular motion
14.5Velocity and acceleration in simple harmonic motion
14.6Force law for simple harmonic motion
14.7Energy in simple harmonic motion
14.8Some systems executing Simple Harmonic Motion
14.9Damped simple harmonic motion
14.10Forced oscillations and resonance

NCERT Solutions Class 11 PhysicsPhysics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 14. 1. Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer:  (a) It is not a periodic motion. Though the motion of a swimmer is to and fro but will not have a definite period.
(b) Since a freely suspended magnet if once displaced from N-S direction and released, it oscillates about this position, it is a periodic motion.
(c) The rotating motion of a hydrogen molecule about its centre of mass is periodic.
(d) Motion of an arrow released from a bow is non-periodic.

Question 14. 2. Which of the following examples represent (nearly) simple harmonic motion and which represent
periodic but not simple harmonic motion?
(a) the rotations of earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Answer: (a) Since the rotation of earth is not to and fro motion about a fixed point, thus it is periodic but not S.H.M.
(b) It is S.H.M.
(c) It is S.H.M.
(d) General vibrations of a polyatomic molecule about its equilibrium position is periodic but non SHM. In fact, it is a result of superposition of SHMs executed by individual vibrations of atoms of the molecule.

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Question 14. 3. Fig. depicts four x-t plots for linear motion of a particle. Which of ike plots represent periodic motion? What is the period of motion (in case of periodic motion)?
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q3

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q3.1
Answer: Figure (b) and (d) represent periodic motions and the time period of each of these is 2 seconds, (a) and (c) are non-periodic motions.

Question 14. 4. Which of the following function of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (to is any positive constant).
(a) sin wt – cos wt (b) sin2 wt (c) 3 cos -2 cos (π/4-2 wt) (d) cos wt + cos 3 wt + cos 5 wt
(e) exp (- w2t2) (f) 1 + wt + w2t2.
Answer:  The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent S.H.M if it can be written uniquely in the form of a cos
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q4
(e) e-w2t2 . It is an exponential function which never repeats itself. Therefore it represents non-periodic motion.
(f) 1 + wt + w2t2 also represents non periodic motion.

Question 14. 5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and (f)  at 4 cm away from B going towards A.
Answer:  In the fig. (given below), the points A and B, 10 cm apart, are the extreme positions of the particle in SHM, and the point O is the mean position. The direction from A to B is positive, as indicated.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q5
(a) At the end A, i.e., extreme position, velocity is zero, acceleration and force are directed towards O and are positive.
(b) At the end B, i.e., second extreme position, velocity is zero whereas the acceleration and force are directed towards the point O and are negative.
(c) At the mid point O, while going towards A, velocity is negative and maximum. The acceleration and force both are zero.
(d) At 2 cm away from B, that is, at C and going towards A: v is negative; acceleration and F, being directed towards O, are also negative.
(e) At 3 cm away from A, that is, at D and going towards B: v is positive; acceleration and F, being directed towards O, are also positive.
(f)  At a distance of 4 cm away from A and going towards A, velocity is directed along BA, therefore, it is positive. Since acceleration and force are directed towards OB, both of them are positive.

Question 14. 6. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x (b) a = – 200 x2
(c) a = – 10x (d) a = 100 x3
Answer:  Only (c) i.e., a = – 10x represents SHM. This is because acceleration is proportional and opposite to displacement (x).

Question 14. 7. The motion of a particle executing simple harmonic motion is described by the displacement function. x(t) = A cos (wt +Ф ).If the initial (t = 0) position of the particle is 1 cm and its initial velocity is w cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (wt + α), what are the amplitude and initial phase of the particle with the above initial conditions?
Answer: 
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q7

Question 14. 8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q8

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q8.1

Question 14. 9. A spring having with a spring constant 1200 Nm-1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q9
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q9.1

Question 14. 10. In Exercise 9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x – axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a)at the mean position,(b)at the maximum stretched position, and (c)at the maximum compressed position.In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer: 
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q10

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q10.1
Note: The functions neither differ in amplitude nor in frequency . They differ only in initial phase.

Question 14. 11. The following figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e., clockwise or anticlockwise) are indicated on each figure.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q11
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case.
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q11.1

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q11.2

Question 14. 12. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise in every case:
(x is in cm and t is in s)
(a) x = – 2 sin (3t + π /3)
(b) x = cos (π /6 – t)
(c) x = 3 sin (2πt + π /4)
(d) x = 2 cos π t.
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q12

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q12.1
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q12.2

Question 14. 13. Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Figure – (b) is stretched by the same force F.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q13
(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released free, what is the period of oscillation in each case?
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q13.1

Question 14. 14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rev/min, what its maximum speed?
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q14

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q14.1

Question 14. 15. The acceleration due to gravity on the surface of moon is 1.7 ms-2. What is the time period of a simple pendulum on the surface of moon if its time-period on the surface of Earth is 3.5 s? (g on the surface of Earth is 9.8 ms-2 .)
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q15

Question 14. 16.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q16
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q16.1
(c) The wristwatch uses an electronic system or spring system to give the time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.
(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.

Question 14. 17. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
Answer: In this case, the bob of the pendulum is under the action of two accelerations.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q17

Question 14. 18.  A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1 . The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q18
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q18.1

Question 14. 19. One end of a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q19
A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Answer: The suction pump creates the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be expressed as:
Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.
Let P = density of the mercury.
L = Total length of the mercury column in both the limbs.
A = internal cross-sectional area of U-tube. m = mass of mercury in U-tube = LAP.
Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q’.
.’. Difference in levels in the two limbs = P’ Q’ = 2y.
:. Volume of mercury contained in the column of length 2y = A X 2y
.•. m – A x 2y x ρ.
If W = weight of liquid contained in the column of length 2y.
Then W = mg = A x 2y x ρ x g
This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q19.1

Question 14. 20. An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction (Fig.). Shaw that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q20
Answer:  Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure.
Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.
There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q20.1

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q20.2
Note. If the ball oscillates in the neck of chamber under isothermal conditions, thru E = P = picture of air inside the chamber, when ball is at equilibrium position. If the ball oscillate in the neck of chamber under adiabatic conditions, then E = gP. where g = Cp/Cv.

Question 14. 21. You are riding an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg. g = 10 m/s2.
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q21

Question 14. 22.  Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Answer:  Let the particle executing SHM starts oscillating from its mean position. Then displacement equation is
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q22

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q22.1

Question 14. 23. A circular disc, of mass 10 kg, is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations of found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant) a is defined by the relation J = – αθ, where J is the restoring couple and θ the angle of twist).
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q23

Question 14. 24. A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q24

Question 14. 25. A mass attached to a spring is free to oscillate, with angular velocity w, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity vat time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters w, x0 and v0.
Answer:
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations Q25

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NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory:

Section NameTopic Name
13Kinetic Theory
13.1Introduction
13.2Molecular nature of matter
13.3Behaviour of gases
13.4Kinetic theory of an ideal gas
13.5Law of equipartition of energy
13.6Specific heat capacity
13.7Mean free path

NCERT Solutions Class 11 PhysicsPhysics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 13. 1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP.’ Take the diameter of an oxygen molecule to be 3 A.
Answer: Diameter of an oxygen molecule, d = 3 A = 3 x 10-10 m. Consider one mole of oxygen gas at STP, which contain total NA = 6.023 x 1023 molecules.
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q1

Question 13. 2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Answer:
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q2

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Question 13. 3. Following figure shows plot of PV/T versus P for 1.00 x 10-3 kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true : T1> T2 or T1< T2?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 x 10-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for the low-pressure high-temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mol-1 K-1.)
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q3
Answer: (a) The dotted plot corresponds to ‘ideal’ gas behaviour as it is parallel to P-axis and it tells that value of PV/T remains same even when P is changed.
(b) The upper position of PV/T shows that its value is lesser for T1 thus T1 > T2. This is because the curve at T1 is more close to dotted plot than the curve at T2 Since the behaviour of a real gas approaches the perfect gas behaviour, as the temperature is increased.
(c) Where the two curves meet, the value of PV/T on y-axis is equal to μR. Since ideal gas equation for μ moles is PV = μRT
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q3.1

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q3.2

Question 13. 4. An oxygen cylinder of volume 30 Hire has an initial gauge pressure of 15 atmosphere and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atmosphere and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder. (R = 8.31 J mol-1 K-1, molecular mass of O2 = 32 u.)
Answer:
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q4

Question 13. 5.  An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C.
Answer:
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q5

Question 13. 6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Answer:
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q6

Question 13. 7. Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer:
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q7

Question 13. 8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
Answer:  Equal volumes of all the gases under similar conditions of pressure and temperature contains equal number of molecules (according to Avogadro’s hypothesis). Therefore, the number of molecules in each case is same.
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q8

Question 13. 9. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:  Let C and C’ be the rms velocity of argon and a helium gas atoms at temperature T K and T K respectively.
Here, M = 39.9; M’ = 4.0; T =?; T = -20 + 273 = 253 K
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q9

Question 13. 10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer:
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q10

Question 13. 11. A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer:  When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.
.’. P1 = 76 cm and V1 = 15 cm3
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q11
When the tube is held vertically, 15 cm air gets another 9 cm of air (filled in the right handside in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure. Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q11.1

Question 13. 12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s-1. Identify the gas.
Answer:  According to Graham’s law of diffusion of gases, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.
If R1 and R2 be the rates of diffusion of two gases having molecular masses Mand M2 respectively, then
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q12

Question 13. 13. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n2 = n1 exp [ – mg (h2 – h1)/kBT]
where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column :
n2 = n1 exp [ – mg NA(ρ – P) (h2 – h1 )/(ρ RT)] where ρ is the density of the suspended particle, and ρ that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.]
[Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Answer:  Considering the particles and molecules to be spherical, the weight of the particle is
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q13

Question 13. 14.  Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q14
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few A].
Answer:  In one mole of a substance, there are 6.023 x 1023 atoms
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory Q14.1

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NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics:

Section NameTopic Name
12Thermodynamics
12.1Introduction
12.2Thermal equilibrium
12.3Zeroth law of thermodynamics
12.4Heat, internal energy and work
12.5First law of thermodynamics
12.6Specific heat capacity
12.7Thermodynamic state  variables and equation of state
12.8Thermodynamic processes
12.9Heat engines
12.10Refrigerators and heat pumps
12.11Second law of thermodynamics
12.12Reversible and irreversible processes
12.13Carnot engine

QUESTIONS FROM TEXTBOOK

Question 12. 1 A geyser heats water flowing at the rate of 3.0 litres per minute from 2 7°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 104 J/g?
Answer: Volume of water heated = 3.0 litre per minute Mass of water heated, m = 3000 g per minute Increase in temperature,
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q1

Question 12. 2 What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q2

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Question 12. 3 Explain why
(a) Two bodies at different temperatures T1 and T2, if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2)/2 ?
(b) The coolant in a chemical or nuclear plant (i.e., the liquid used to prevent different parts of a plant from getting too hot) should have high specific heat. Comment.
(c) Air pressure in a car tyre increases during driving. Why?
(d) The climate of a harbour town is more temperate (i.e., without extremes of heat and cold) than that of a town in a desert at the same latitude. Why?
Answer: (a) In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures become equal. The final temperature can be the mean temperature (T1+ T2)/2 only when thermal capacities of the two bodies are equal.
(b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.
(c) When car is driven, some work is being done on types in order to overcome dissipative forces of friction and air resistance etc. This work done is transformed into heat, due to which temperature of the car types increases.
(d) The climate of a harbour town is more temperate (neither too hot nor too cool) due to formation of sea breeze at day time and land breeze at night time as already explained in Chapter 11.

Question 12. 4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q4

Question 12. 5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q5

Question 12. 6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completey evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B ?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie of its P-V-T Surface?
Answer:  (a) Since the final temperature and initial temperature remain the same,
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q6
(b) Since the temperature of the system remains unchanged, change in internal energy is zero.
(c) The system being thermally insulated, there is no change in temperature (because of free expansion)
(d) The expansion is a free expansion. Therefore, the intermediate states are non equilibrium states and the gas equation is not satisfied in these states. As a result, the gas can not return to an equilibrium state which lie on the P-V-T surface.

Question 12. 7 A steam engine delivers 5.4 x 108 J of work per minute and services 3.6 x 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q7

Question 12. 8 An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q8

Question 12. 9 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q9

Question 12. 10 A refrigerator is to maintain eatables kept inside at 9 °C, if room temperature is 36 °C. Calculate the coefficient of performance.
Answer:
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Q10

NCERT Solutions for Class 11 Physics All Chapters

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter:

Section NameTopic Name
11Thermal Properties of matter
11.1Introduction
11.2Temperature and heat
11.3Measurement of temperature
11.4Ideal-gas equation and absolute temperature
11.5Thermal expansion
11.6Specific heat capacity
11.7Calorimetry
11.8Change of state
11.9Heat transfer
11.10Newton’s law of cooling

QUESTIONS FROM TEXTBOOK

Question 11. 1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer: The relation between kelvin scale and Celsius scale is TK – 273.15 =T=> TC=TK– 273.15
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q1
Question 11. 2.  Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ?
Answer:  As we know, triple point of water on absolute scale = 273.16 K, Size of one degree of kelvin scale on absolute scale A
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q2

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Question 11. 3.  The-electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R0 [1 + α (T – T0)].
The resistances is 101.6 Ωat the triple-point of water 273.16 K, and 165.5 Ωat the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Answer: Here, R0 = 101.6 Ω; T0 = 273.16 K Case (i) R1= 165.5 Ω; T1 = 600.5 K, Case (ii) R2 = 123.4 , T2 = ?
Using the relation R = R0[1 + α (T – T0)]
Case (i) 165.5 = 101.6 [1 + α (600.5 – 273.16)]
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q3

Question 11. 4. Answer the following:
(a) The triple-point of water is a standard fixed point in modem thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) Scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale tc = T – 273.15
Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?
Answer: (a) Triple point of water has a unique value i.e., 273.16 K. The melting point and boiling points of ice and water respectively do not have unique values and change with the change in pressure.
(b) On Kelvin’s absolute scale, there is only one fixed point, namely, the triple-point of water and there is no other fixed point.
(c) On Celsius scale 0 °C corresponds to the melting point of ice at normal pressure and the value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.
(d)The Fahrenheit scale and Absolute scale are related as
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q4

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q4.1

Question 11. 5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q5
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?
Answer: 
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q5.1
(b) The value of the melting point of sulphur found from the two thermometers differ slightly due to the reason that in practice, the gases do not behave strictly as perfect gases i.e., gases are not perfectly ideal.
To reduce the discrepency, readings should be taken for lower and lower pressures and the plot between temperature measured versus absolute pressure of the gas at triple point should be extrapolated to obtain the temperature in the limit pressure tends to zero (if P —> 0), when the gases approach ideal gas behaviour.

Question 11. 6.  A steel tape 1 m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = 1.20 x 10-1K-1.
Answer:  On a day when the temperature is 27 °C, the length of 1 cm division on the steel tape is exactly 1 cm, because the tape has been calibrated for 27 °C.When the temperature rises to 45 °C (that is, ΔT = 45 – 27 = 18 °C), the increase in the length of 1 cm division is Δl = αlΔT = (1.2 x 10-5C-1) x 1 cm x 18 °C = 0.000216 cm Therefore, the length of 1 cm division on the tape becomes 1.000216 cm at 45 °C. As the length of the steel rod is read to be 63.0 cm on the steel tape at 45 °C, the actual length of the rod at 45 °C is 63.0 x 1.000216 cm = 63.0136 cm The length of the same rod at 27 °C is 63.0 cm, because 1 cm mark on the steel tape is exactly 1 cm at 27 °C.

Question 11. 7.  A large steel wheel is to befitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is, 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range αsteel= 1-20 x 10-5K-1.
Answer: 
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q7

Question 11. 8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C ? Coefficient of linear expansion of copper = 1.70 x 10-5K-1.
Answer:
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q8

Question 11. 9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of – 39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 x 10-5K-1; Young’s modulus of brass = 0.91 x 1011 Pa
Ans.
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q9

Question 11. 10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 x 10-5 °C-1, steel = 1.2 x 10-5 °C-3.
Ans.
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q10

Question 11. 11. The coefficient of volume expansion of glycerine is 49 x 10-5K-1. What is the fractional change in its density for a 30 °C rise in temperature ?
Ans.
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q11

Question 11. 12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 J g-1 K-1 .
Answer: Power = 10 kW = 104 W
Mass, m=8.0 kg = 8 x 103 g
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q12

Question 11. 13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? Specific heat of copper is 0.39 Jg-1°C-1. Heat of fusion of water = 335 Jg-1.
Answer: 
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q13

Question 11. 14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40° C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Answer: Mass of metal block, m = 0.20 kg = 200 g
Fall in the temperature of metal block,
ΔT = (150 – 40) °C = 110 °C
If C be the specific heat of metal, then heat lost by the metal block = 200 x C x 110 cal Volume of water = 150 cm3
mass of water = 150 g
Increase in temperature of water = (40 – 27) °C = 13°C
Heat gained by water = 150 x 13 cal Water equivalent of calorimeter, w = 0.025 kg = 25g
Heat gained by calorimeter,
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q14

Question 11. 15. Given below are observations on molar specific heats at room temperature of some common gases.
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q15
The measured molar specific heats of these gases are markedly different from those for mono atomic gases. Typically, molar specific heat of a mono atomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
Answer: The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat =5/2 R = 5/2 x 1.98 = 4.95, which agrees fairly well with all observations listed in the table except for chlorine. A mono atomic gas molecule has only the translational motion. A diatomic gas molecule, apart from translational motion, the vibrational as well as rotational motion is also possible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, heat is to be supplied to increase not only translational energy but also rotational and vibrational energies. Hence, molar specific heat of a diatomic gas is greater than that for mono atomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. shows that for chlorine molecule, at room temperature vibrational motion also occurs along with translational and rotational motions, whereas other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This is the reason that chlorine has somewhat larger value of molar specific heat.

Question 11. 16. (a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium ?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2 ?
(c) What are the critical temperature and pressure for CO2 ? What is their significance 1
(d) Is CO2 solid, liquid or gas at (a) – 70 °C under 1 atm (b) – 60 °C under 10 atm (c) 15°C under 56 atm?
Answer:  (a) At the triple point, temperature = – 56.6 °C and pressure = 5.11 atm.
(b) Both the boiling point and freezing point of CO2 decrease if pressure decreases.
(c) The critical temperature and pressure of CO2 are 31.1°C and 73.0 atm respectively. Above this temperature, COwill not liquefy/even if compressed to high pressures.
(d) (i) The point (- 70 °C, 1.0 atm) lies in the vapour region. Hence, CO2 is vapour at this point.
(ii) The point (- 60 °C, 10 atm) lies in the solid region. Hence, CO2 is solid at this point.
(iii) The point (15 °C, 56 atm) lies in the liquid region. Hence, CO2 is liquid at this point.
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q16

Question 11. 17. Answer the following questions based on the P – T phase diagram of CO2 (Fig. of question 17 given above)
(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?
(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure ?
(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature – 65 °C as it is heated up at room temperature at constant pressure.
(d) CO2 is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe ?
Answer:  (a) No, the CO2 does not go through the liquid phase. The point (1.00 atm, – 60°C) is to the lift of the triple-point O and below the sublimation curve OA. Therefore, when CO2 is compressed at this point at constant temperature, the point moves perpendicular to the temperature-axis and enters the solid phase region. Hence, the COvapour condenses to solid directly without going through the liquid phase.
(b) CO2 at 4.0 atm pressure and room temperature (say, 27 °C) is in vapour phase. This point (4.0 atm, 27°C) lies below the vaporation curve OC and to the right of the triple point O. Therefore, when CO2 is cooled at this point at constant pressure, the point moves perpendicular to the pressure-axis and enters the solid phase region. Hence, the CO2 vapour condenses directly to solid phase without going through the liquid phase.
(c) When the solid CO2 at – 65 °C is heated at 10 atm pressure, it is first converted into liquid. A further increase in its temperature brings it into the vapour phase. If a horizontal line at P = 10 atm is drawn parallel to the T-axis, then the points of intersection of line with the fusion and vaporization curve give the fusion and boiling points at 10 atm.
(d) Above 31.1°C, the gas cannot be liquefied. Therefore, on being compressed isothermally at 70°C, there will be no transition to the liquid region. However, the gas will depart, more and more from its perfect gas behaviour with the increase in pressure.

Question 11. 18. A child running a temperature of 101°F is given an antipyrin (i.e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98° F in 20 minutes, what is the average rate of extra evaporation caused by the drug ? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g-1.
Answer:
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q18

Question 11. 19. A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and coefficient of thermal conductivity of thermacole is 0.01 Js-1 m-1 °C-1 [Heat of fusion of water = 335 x 103 J kg-1 ].
Answer:  Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.
.•. Total heat entering the icebox in 6 h is given by
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q19

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q19.1

Question 11. 20. A brass boiler has a base area 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/ min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js-1 m-1 K-1.(Heat of vaporization of water = 2256 x 103 J kg-1 )
Answer:
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q20

Question 11. 21. Explain why:
(a) a body with large reflectivity is a poor emitter.
(b) a brass tumbler feels much colder than a wooden tray on a chilly day.
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
(d) the earth without its atmosphere would be inhospitably cold.
(e) heat systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
Answer: (a) According to Kirchh off’s law of black body radiations, good emitters are good absorbers and bad emitters are bad absorbers. A body with large reflectivity is a poor absorber of heat and consequently, it is also a poor emitter.
(b) Brass is a good conductor of heat, while wood is a bad conductor. When we touch the brass tumbler on a chilly day, heat starts flowing from our body to the tumbler and we feel it cold. However, when the wooden tray is touched, heat does not flow from our hands to the tray and we do not feel cold.
(c) An optical pyrometer is based on the principle that the brightness of a glowing surface of a body depends upon its temperature. Therefore, if the temperature of the body is less than 600°C, the image formed by the optical pyrometer is not brilliant and we do not get the reliable result. It is for this reason that the pyrometer gives a very low value for the temperature of red hot iron in the open.
(d) The lower layers of earth’s atmosphere reflect infrared radiations from earth back to the surface of earth. Thus the heat radiation received by the earth from the sun during the day are kept trapped by the atmosphere. If atmosphere of earth were not there, its surface would become too cold to live.
(e) Steam at 100°C possesses more heat than the same mass of water at 100°C. One gram of steam at 100°C possesses 540 calories of heat more than that possessed by 1 gm of water at 100°C. That is why heating systems based on circulation of steam are more efficient than those based on circulation of hot water.

Question 11. 22. A body cools from 80 °C to 50°C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30°C. The temperature of the surroundings is 20 °C.
Answer:
NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q22

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter Q22.1

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NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties Of Fluids:

Section NameTopic Name
10Mechanical Properties Of Fluids
10.1Introduction
10.2Pressure
10.3Streamline flow
10.4Bernoulli’s principle
10.5Viscosity
10.6Reynolds number
10.7Surface tension

QUESTIONS FROM TEXTBOOK

Question 10. 1. Explain why
(a) The blood pressure in humans is greater at the feet than at the brain.
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer: (a) The height of the blood column is more for the feet as compared to that for the brain.
Consequently, the blood pressure in humans is greater at the feet than at the brain.
(b) The variation of air-density with height is not linear. So, pressure also does not reduce linearly with height. The air pressure at a height h is given by P = P0eαh where P0 represents the pressure of air at sea-level and α is a constant.
(c) Due to applied force on liquid, the pressure is transmitted equally in all directions inside the liquid. That is why there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar quantity.

Question 10. 2. Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface.
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape.
Answer: (a) Let a drop of a liquid L be poured on a solid surface S placed in air A. If TSL,and TSA be the surface tensions corresponding to solid-liquid layer, liquid-air layer and solid-air layer respectively and θ be the angle of contact between the liquid and solid, then
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q2
TLA Cos θ + TSL =TSA
=>Cos θ=TSA-TSL/TLA
For the mercury-glass interface, TSA< TSL. Therefore, cos 0 is negative. Thus θ is an obtuse angle. For the water-glass interface, TSA > TSL. Therefore cos 0 is positive. Thus, θ is an acute angle.
(b) Water on a clean glass surface tends to spread out i.e., water wets glass because force of cohesion of water is much less than the force of adhesion due to glass. In case of mercury force of cohesion due to mercury molecules is quite strong as compared to adhesion force due to glass. Consequently, mercury does not wet glass and tends to form drops.
(c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since h as force is independent of the area of liquid surface therefore, surface tension is also independent of the area of the liquid surface.
(d) We know that the clothes have narrow pores or spaces which act as capillaries. Also, we know that the rise of liquid in a capillary tube is directly proportional to cosθ (Here θ is the angle of contact). As θ is small for detergent, therefore cos θ will be large. Due to this, the detergent will penetrate more in the narrow pores of the clothes.
(e) We know that any system tends to remain in a state of minimum energy. In the absence of any external force for a given volume of liquid its surface area and consequently. Surface energy is least for a spherical shape. It is due to this reason that a liquid drop, in the absence of an external force is spherical in shape.

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Question 10. 3. Fill in the blanks using the words from the list appended with each statement:
(a) Surface tension of liquids generally…………….with temperature. (increases/decreases)
(b) Viscosity of gases………………..with temperature, whereas viscosity of liquids…………..with temperature. (increases/decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to…………………..while for fluids it is proportional to…………. (shear strain/rate of shear strain)
(d) For a fluid in steady flow, the increases inflow speed at a constriction follows from…………………………. while the decrease of pressure there follows from………………….(conservation of mass/Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a…………….speed than the critical speed for turbulence for an actual plane. (greater/smaller)
Answer: (a), decreases
(b) increases; decreases
(c) shear strain; rate of shear strain
(d) conservation of mass; Bernoulli’s principle
(e) greater.

Question 10. 4. Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it.
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
(c) The size of a needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
(d) A fluid flowing out of a small hole in a vessel results in a backward thurst on the vessel.
(e) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer:  (a) When we blow over the piece of paper, the velocity of air increases. As a result, the pressure on it decreases in accordance with the Bernoulli’s theorem whereas the pressure below remains the same (atmospheric pressure). Thus, the paper remains horizontal.
(b) By doing so the area of outlet of water jet is reduced, so velocity of water increases according to equation of continuity av = constant.
(c) For a constant height, Bernoulli’s theorem is expressed as P +1/2 ρ v= Constant
In this equation, the pressure P occurs with a single power whereas the velocity occurs with a square power. Therefore, the velocity has more effect compared to the pressure. It is for this reason that needle of the syringe controls flow rate better than the thumb pressure exerted by the doctor.
(d) This is because of principle of conservation of momentum. While the flowing fluid carries forward momentum, the vessel gets a backward momentum.
(e) A spinning cricket ball would have followed a parabolic trajectory has there been no air. But because of air the Magnus effect takes place. Due to the Magnus effect the spinning cricket ball deviates from its parabolic trajectory.

Question 10. 5. A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q5

Question 10. 6. Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q6

Question 10. 7. A vertical off-shore structure is built to withstand a maximum stress of 10Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
Answer: Here, Maximum stress = 109 Pa, h = 3 km = 3 x 103 m;
p (water) = 103 kg/m3 and g = 9.8 m/s2.
The structure will be suitable for putting upon top of an oil well provided the pressure exerted by sea water is less than the maximum stress it can bear.
Pressure due to sea water, P = hρg = 3 x 103 x 103x 9.8 Pa = 2.94 x 107 Pa
Since the pressure of sea water is less than the maximum

Question 10. 8. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?
Answer: 
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q8
This is also the maximum pressure that the smaller piston would have to bear.

Question 10. 9. A U tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the relative density of spirit?
Answer: For water column in one arm of U tube, h1 = 10.0 cm; ρ1 (density) = 1 g cm-3
For spirit column in other arm of U tube, h2 = 12.5 cm; ρ2 =?
As the mercury columns in the two arms of U tube are in level, therefore pressure exerted by each is equal.
Hence h1ρ1g = h2ρ2g or ρ2 = h1ρ1/h2 =10 x 1/12.5 = 0.8 g cm-3
Therefore, relative density of spirit = ρ21 = 0.8/1 = 0.8

Question 10. 10. In Q.9, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Relative density of mercury = 13.6)
Answer: 
Height of the water column, h1 = 10 + 15 = 25 cm
Height of the spirit column, h2 = 12.5 + 15 = 27.5 cm
Density of water, ρ1 = 1 g cm–3
Density of spirit, ρ2 = 0.8 g cm–3
Density of mercury = 13.6 g cm–3
Let h be the difference between the levels of mercury in the two arms.
Pressure exerted by height h, of the mercury column:
g
h × 13.6g … (i)
Difference between the pressures exerted by water and spirit:
ρ1h1g – ρ2h2g
g(25 × 1 – 27.5 × 0.8)
= 3g … (ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3g
h = 0.220588 ≈ 0.221 cm
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.

Question 10. 11. Can Bernoulli’s equation be used to describe the flow of water through a rapid motion in a river? Explain.
Answer: Bernoulli’s theorem is applicable only for there it ideal fluids in streamlined motion. Since the flow of water in a river is rapid, way cannot be treated as streamlined motion, the theorem cannot be used.

Question 10. 12. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer: No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 10. 13.  Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 x 10-3 kg s-1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 103 kg m-3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct],
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q13

Question 10. 14.
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q14
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q14.1

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q14.2

Question 10. 15. Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures in incorrect? Why?


Answer: Figure (a) is incorrect. It is because of the fact that at the kink, the velocity of flow of liquid is large and hence using the Bernoulli’s theorem the pressure is less. As a result, the water should not rise higher in the tube where there is a kink (i.e., where the area of cross-section is small).

Question 10. 16. The cylindrical tube of a spare pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min-1, what is the speed of ejection of the liquid through the holes?
Answer: 
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q16

Question 10. 17. A U-shaped wire is dipped in a soap solution, and removed. A thin soap film formed between the wire and a light slider supports a weight of 1.5 x 10-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Answer: In present case force of surface tension is balancing the weight of 1.5 x 10-2 N, hence force of surface tension, F = 1.5 x 10-2 N.
Total length of liquid film, l = 2 x 30 cm = 60 cm = 0.6 m because the liquid film has two surfaces.
Surface tension, T = F/l =1.5 x 10-2 N/0.6m =2.5 x 10-2 Nm-1

Question 10. 18. Figure (a) below shows a thin film supporting a small weight = 4.5 x 10-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) Explain your answer physically.


Ans. (a) Here, length of the film supporting the weight = 40 cm = 0.4 m. Total weight supported (or force) = 4.5 x 10-2 N.
Film has two free surfaces, Surface tension, S =4.5 x 10-2/2 x 0.4 =5.625 x 10-2 Nm-1
Since the liquid is same for all the cases (a), (b) and (c), and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 x 10-2. In Fig. 7(b), 38(b) and (c), the length of the film supporting the weight is also the saihe as that of (a), hence the total weight supported in each case is 4.5 x 10-2 N.

Question  10. 19. What is the pressure inside a drop of mercury of radius 3.0 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 x 10-1 Nm-1. The atmospheric pressure is 1.01 x 105 Pa. Also give the excess pressure inside the drop.
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q19
Since data is correct up to three significant figures, we should write total pressure inside the drop as 1.01 x 105  Pa.

Question 10. 20. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 x 10-2  Nm-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 x 105  Pa).
Answer: Here surface tension of soap solution at room temperature
T = 2.50 x 10-2 Nm-1, radius of soap bubble, r = 5.00 mm = 5.00 x 10-3 m.
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q20

Question 10. 21. A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. Compute the force necessary to keep the door close.
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q21

Question 10. 22. A manometer reads the pressure of a gas in an enclosure as shown in Fig. (a) When a pump removes some of the gas, the manometer reads as in Fig. (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) is poured into the right limb of 1 the manometer? Ignore the small change in the volume of the gas.


Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q22.1

Question 10. 23. Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Answer:  Pressure (and therefore force) on the two equal base areas are identical. But force is exerted by water on the sides of the vessels also, which has a non-zero vertical component when sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on the sides of the vessel is greater for the first vessel than the second. Hence, the vessels weigh different even when the force on the base is the same in the two cases.

Question 10. 24. During blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? Given: density of whole blood = 1.06 x 103 kg m-3.
Answer: h=P/ρg =200/(1.06 x 103 x 9.8) =0.1925 m
The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than 0.1925 m i.e„ 0.2 m.

Question 10. 25. In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy, (a) What is the largest average velocity of blood flow in an artery of diameter 2 x 10-3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Answer:  (a) If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces, due to which the pressure drop becomes large.
(b) The dissipative forces become more important with increasing flow velocity, because of turbulence.

Question 10. 26. (a) What is the largest average velocity of blood flow in an artery of radius 2 x 103 m if the flow must remain laminar?
(b) What is the corresponding flow rate? Take viscosity of blood to be 2.084 x 10-3 Pa-s. Density of blood is 1.06 x 103 kg/m3 .
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q26

Question 10. 27.  A plane is in level flight at constant speed and each of its wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg/m3), g = 9.8 m/s2.
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q27

Question 10. 28. In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10-5 m and density 1.2 x 103 kg m-3. Take the viscosity of air at the temperature of the experiment to be 1.8 x 10-5 Pa-s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Answer: Here radius of drop, r = 2.0 x 10-5 m, density of drop, p = 1.2 x 103 kg/m3, viscosity of air TI = 1.8 x 10-5 Pa-s.
Neglecting upward thrust due to air, we find that terminal speed is
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q28

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q28.1

Question 10. 29. Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.0 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 Nm-2. Density of mercury = 13.6 x 10 kg m-3
Answer:
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q29

Question 10. 30. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 x 10-2 Nm-2. Take the angle of contact to be zero and density of water to be 1.0 x 103 kg m-3(g = 9.8 ms-2).
Answer: Let rx be the radius of one bore and r2 be the radius of second bore of the U-tube. The, if h1 and h2 are the heights of water on two sides, then
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q30

Question 10. 31. (a) It is known that density p of air decreases with height y as

where ρ0 = 1.25 kg m-3 is the density at sea level, and y0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also, assume that the value of g remains constant.

(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains a constant radius as it rises. How high does it rise?[Take y0 = 8000 m and ρHe = 0.18 kg m-3].
Answer:  (a) We know that rate of decrease of density p of air is directly proportional to the height y. It is given as   dρ/dy = – ρ/y0
where y is a constant of proportionality and -ve sign signifies that density is decreasing with increase in height. On integration, we get
NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids Q31.1

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NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Solids:

Section NameTopic Name
9Mechanical Properties Of Solids
9.1Introduction
9.2Elastic behaviour of solids
9.3Stress and strain
9.4Hooke’s law
9.5Stress-strain curve
9.6Elastic moduli
9.7Applications of elastic behaviour of materials

NCERT Solutions Class 11 PhysicsPhysics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 9. 1. A steel wire of length 4.7 m and cross-sectional area 3.0 x 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10-5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q1

Question 9. 2. Figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q2
(a)  Young’s modulus of the material (Y) is given by
Y =Stress/Strain
=150 x 106/0.002
150 x 106/2 x 10-3
=75 x 109 Nm-2
=75 x 1010 Nm-2
(a)Yield strength of a material is defined as the maximum stress it can sustain. From graph, the approximate yield strength of the given material
= 300 x 106 Nm-2
= 3 x 108 Nm-2 .

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Question 9. 3. The stress-strain graphs for materials A and B are shown in figure.
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q3
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Answer: (a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus =(Stress /Strain) is greater for A than that of B.
(b) Strength of a material is determined by the amount of stress required to cause fracture. This stress corresponds to the point of fracture. The stress corresponding to the point of fracture in A is more than for B. So, material A is stronger than material B.

Question 9. 4. Read the ‘allowing two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Answer: (a) False. The-Young’s modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.
(b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance, as well as shape of helicals of the coil change and it, involves shear modulus.

Question 9. 5. Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m.Young’s modulus of steel is 2.0 x 1011 Pa. Compute the elongations of steel and brass wires. (1 Pa = 1 N m2).
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q5
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q5.1

Question 9. 6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Answer: Here, side of cube, L = 10 cm =10/100= 0.1 m
.•. Area of each face, A = (0.1)2 = 0.01 m2
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q6

Question 9. 7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. Young’s modulus, Y = 2.0 x 1011 Pa.
Answer: 
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q7

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q7.1

Question 9. 8. A piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Shear modulus of elasticity of copper is 42 x 109 N/m2.
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q8

Question 9. 9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 Nm-2 what is the maximum load the cable can support ?
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q9

Question 9. 10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Answer: Since each wire is to have same tension therefore, each wire has same extension. Moreover, each wire has the same initial length.So, strain is same for each wire.
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q10

Question 9. 11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1 m, is whirled in a vertical circle with an angular velocity of 2 rero./s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the zvire when the mass is at the lowest point of its path. Ysteel = 2 x 1011 Nm-2.
Ans.Here, m = 14.5 kg; l = r = 1 m; v = 2 rps; A = 0.065 x 10-4 m2 Total pulling force on mass, when it is at the lowest position of the vertical circle is F = mg + mr w2 = mg + mr 4,π2 v2
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q11

9.12.Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 x 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q12
The ratio is too large. This is due to the fact that the strain for air is much larger than for water at the same temperature. In other words, the intermolecular distances in case of liquids are very small as compared to the corresponding distances in the case of gases. Hence there are larger interatomic forces in liquids than in gases.

Question 9. 13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 x 103 kg m-3?
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q13

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q13.1

Question 9. 14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q14

Question 9. 15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 x 106 Pa.
Answer: Here a side of copper cube a = 10 cm, hence volume V = a3 = 10-3 m3, hydraulic pressure applied p = 7.0 x 106 Pa and from table we find that bulk modulus of copper B = 140 G Pa = 140 x 109 Pa.
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q15

Question 9. 16. How much should be pressure the a litre of water be changed to compress it by 0.10 %? Bulk modulus of elasticity of water = 2.2 x 109 Nm-2.
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q16

Question 9. 17.  Anvils made of single crystals of diamond, with the shape as shown in figure are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of50,000 N. What is the pressure at the tip of the anvil?
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q17
Answer:
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q17.1

Question 9. 18. A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A ) and aluminium (wire B) of equal lengths as shown in figure.
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q18
The cross-sectional areas of wires A and B are 1.0 mmand 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.
Answer: For steel wire A, l1=l; Az = 1 mm2; Y1= 2 x 1011 Nm-2
For aluminium wire B, l2 = l; A2 = 2mm2; Y2 = 7 x 1010 Nm-2
(a) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let Fand F2 be the tensions in two wires and there is equal stress in two wires, then
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q18.1
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F1 and F2 be the tension in the wires and there is equal strain in the two wires i.e.,
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q18.2

Question 9. 19. A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Answer: Let AB be a mild steel wire of length 2L = lm and its cross-section area A = 0.50 x 10-2 cm2. A mass m = 100 g = 0.1 kg is suspended at mid-point C of wire as shown in figure. Let x be the depression at mid-point i.e., CD = x
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q19

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q19.1

Question 9. 20. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 x 107 Pa? Assume that each rivet is to carry one-quarter of the load.
Answer:  Diameter = 6mm; Radius, r = 3 x 10-3 m;
Maximum stress = 6.9 x 107 Pa
Maximum load on a rivet
= Maximum stress x cross-sectional area
= 6.9 x 107 x 22/7 (3 x 10-3)2 N = 1952 N
Maximum tension = 4 x 1951.7 N = 7.8 x 103 N.

Question 9. 21. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 x 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
Answer: 
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Q21

NCERT Solutions for Class 11 Physics All Chapters

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 8 Gravitation.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 8 Gravitation:

Section NameTopic Name
8Gravitation
8.1Introduction
8.2Kepler’s laws
8.3Universal law of gravitation
8.4The gravitational constant
8.5Acceleration due to gravity of the earth
8.6Acceleration due to gravity below and above the surface of earth
8.7Gravitational potential energy
8.8Escape speed
8.9Earth satellite
8.10Energy of an orbiting satellite
8.11Geostationary and polar satellites
8.12Weightlessness

NCERT Solutions Class 11 PhysicsPhysics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 8. 1. Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small spaceship orbiting around the Earth cannot detect gravity. If the space station orbiting around the Earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the Earth due to the Sun to that due to the Moon, you would find that the Sun’s pull is greater than the Moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the Moon’s pull is greater than the tidal effect of Sun. Why?
Answer: (a) No. Gravitational forces are independent of medium. A body cannot be shielded from the gravitational influence of nearby matter.
(b) Yes. If the size of the spaceship is extremely large, then the gravitational effect of the spaceship may become measurable. The variation in g can also be detected.
(c) Tidal effect depends inversely on the cube of the distance, unlike force which depends inversely on the square of the distance. Since the distance of moon from the ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of Moon’s pull is greater than the tidal effect of the sun.

Question 8. 2. Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of the mass of the Earth/mass of the body.
(d) The formula – GMm (1/r2-1/r1) is more/less accurate than the formula mg (r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the Earth.
Answer: (a) decreases
(b) decreases
(c) mass of the body
(d) more

More Resources for CBSE Class 11

Question 8.3. Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q3

Question 8.4. Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 108 m. Show that the mass of Jupiter is about one-thousandth that of the Sun.
Answer. For a satellite of Jupiter, orbital period, T1 = 1.769 days = 1.769 x 24 x 60 x 60 s Radius of the orbit of satellite, r1 = 4.22 x 108 m
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q4

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q4.1

Question 8.5. Let us assume that our galaxy consists of 2.5 x 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky way to be 105 ly.
Answer: Here, r = 50000 ly = 50000 x 9.46 x 1015 m = 4.73 x 1020 m
M = 2.5 x 1011 solar mass = 2.5 x 1011 x (2 x 1030) kg = 5.0 x 1041kg
We know that
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q5

Question 8. 6. Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Answer:  (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.

Question 8.7. Does the escape speed of a body from the Earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q7
(a) The escape speed of a body from the Earth does not depend on the mass of the body.
(b) The escape speed does not depend on the location from where a body is projected.
(c) The escape speed does not depend on the direction of projection of a body.
(d) The escape speed of a body depends upon the height of the location from where the body is projected, because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also.

Question 8. 8. A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Answer: (a) The linear speed of the comet is variable in accordance with Kepler7s second law. When comet is near the sun, its speed is higher. When the comet is far away from the sun, its speed is very less.
(b) Angular speed also varies slightly.
(c) Comet has constant angular momentum.
(d) Kinetic energy does not remain constant.
(e) Potential energy varies along the path.
(f) Total energy throughout the orbit remains constant.

Question 8. 9.Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.
Answer: (a) The blood flow in feet would be lesser in zero gravity. So, the astronaut will not get swollen feet.
(b) In the conditions of weightlessness, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face.
(c) Due to more blood supply to face, the astronaut may get headache.
(d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 8. 10. In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig.)
(i) a, (ii) b, (iii) c, (iv) 0.
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q10
Answer: At all points inside a hollow spherical shell, potential is same. So, gravitational intensity, which is negative of gravitational potential gradient, is zero. Due to zero gravitational intensity, the gravitational forces acting on any particle at any point inside a spherical shell will be symmetrically placed. It follows from here that if we remove the upper hemispherical shell, the net gravitational force acting on a particle at P will be downwards. Since gravitational intensity is gravitational force per unit mass therefore, the direction of gravitational intensity will be along c. So, option (iii) is correct.

Question 8. 11. For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii), e, (iii) f (iv) g.
Answer: Using the explanation given in the solution of the previous problem, the direction of the gravitational field intensity at P will be along e. So, option (ii) is correct.

Question 8. 12. A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 x 1030 kg, mass of the earth = 6 x 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 1011 m).
Answer: Mass of Sun, M = 2 x 1030 kg; Mass of Earth, m = 6 x 1024 kg Distance between Sim and Earth, r = 1.5 x 1011 m
Let at the point P, the gravitational force on the rocket due to Earth
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q12

Question 8. 13. How will you ‘weigh the sun’, that is, estimate its mass? The mean orbital radius of the earth around the sun is 1.5 x 108 km.
Answer: The mean orbital radius of the Earth around the Sun
R = 1.5 x 108 km = 1.5 x 1011 m
Time period, T = 365.25 x 24 x 60 x 60 s
Let the mass of the Sun be M and that of Earth be m.
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q13

Question 8. 14. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 x 108 km away from the Sun?
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q14

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q14.1

Question 8. 15. A body weighs 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Answer:  Let gh be the acceleration due to gravity at a height equal to half the radius of the Earth (h = R/2) and g its value on Earth’s surface. Let the body have mass m.
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q15

Question 8. 16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q16

Question 8. 17. A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 x 1024 kg; mean radius of the earth = 6.4 x 106 m; G = 6.67 x 10-11 N m2 kg-2.
Answer:  Initial kinetic energy of rocket = 1/2 mv2 = 1/2 x m x (5000)2 = 1.25 x 107 mJ
At distance r from centre of earth, kinetic energy becomes zero
.•. Change in kinetic energy = 1.25 x 107 – 0 = 1.25 x 107 m J
This energy changes into potential energy.
Initial potential energy at the surface of earth = GMem/’r
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q17

Question 8. 18. The escape speed of a projectile on the Earth’s surface is 11.2 km s-1. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q18

Question 8. 19. A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 x 1024 kg; radius of the earth = 6.4 x 106 m; G = 6.67 x 10-11 N m2 kg-2.
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q19

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q19.1

Question 8. 20. Two stars each of one solar mass (=2 x 1030 kg) are approaching each other for a head on collision.When they are at a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Answer: Here, mass of each star, M = 2 x 1030 kg
Initial potential between two stars, r = 109 km = 1012 m.
Initial potential energy of the system = -GMm/r
Total K.E. of the stars = 1/2Mv2 + 1/2Mv2
where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r’ = 2 R.
:. Final potential energy of two stars = -GMm/2R
Since gain in K.E. is at the cost of loss in P.E
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q20

Question 8. 21. Two heavy spheres each of mass 100 kg and radius 0.10 mare placed 1.0 m apart on ahorizontal table. What is the gravitational field and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable or unstable?
Answer: Here G = 6.67 x 10-11 Nm2 kg-2; M = 100 kg; R = 0.1 m, distance between the two spheres, d = 1.0 m
Suppose that the distance of either sphere from the mid-point of the line joining their centre is r. Then r=d/2=0.5 m. The gravitational field at the mid-point due to two spheres will be equal and opposite.
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q21

Question 8. 22. As you have learnt in the text, a geostationary satellite orbits the Earth at a height of nearly 36,000 km from the surface of the Earth. What is the potential due to Earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the Earth = 6.0 x 1024 kg, radius = 6400 km.
Answer: Distance of satellite from the centre of earth = R = r + x
= 6400 + 36000 = 42400 km = 4.24 x 107 m
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q22

Question 8. 23. A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 x 1030 kg).
Answer: Acceleration due to gravity of the star,g= GM/R2 …………(i)
Here M is the mass and R is the radius of the star.
The outward centrifugal force acting on a body of mass m at the equator of the star =mv2/R =mR w2——-(ii)
From equation (i), the acceleration due to the gravity of the star
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q23
Since the inward force due to gravity on a body at the equator of the star is about 2.2 million times more than the outward centrifugal force, the body will remain stuck to the surface of the star.

Question 8. 24. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000 kg, Mass of the Sun = 2 x 1030 kg. Mass of the Mars = 6.4 x 1023 kg, Radius of Mars = 3395 km. Radius of the orbit of Mars = 2.28 x 1011 m, G = 6.67 x 10-11 N m2 kg-2.
Answer:  Let R be the radius of orbit of Mars and R’ be the radius of the Mars. M be the mass of the Sun and M’ be the mass of Mars. If m is the mass of the space-ship, then Potential energy of space-ship due to gravitational attraction of the Sun = – GM m/R Potential energy of space-ship due to gravitational attraction of Mars = – G M’ m/R’ Since the K.E. of space ship is zero, therefore,
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q24

Question 8. 25. A rocket is fired ‘vertically’ from the surface of Mars with a speed of 2 km s-1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of Mars = 6.4 x 1023 kg; radius of Mars = 3395 km; G = 6.67 x 10-11 N m2 kg-2
Answer:
NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q25

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Q25.1

NCERT Solutions for Class 11 Physics All Chapters

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

NCERTap Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion:

NCERT Solutions Class 11 PhysicsPhysics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 7. 1. Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Answer: In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

Question 7. 2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 A (1 A = 10-10m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Answer: Let us choose the nucleus of the hydrogen atom as the origin for measuring distance. Mass of hydrogen atom,m1= 1 unit (say) Since cholorine atom is 35.5 times as massive as hydrogen atom,
.•. mass of cholorine atom, m2 = 35.5 units
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q2

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q2.1

Question 7. 3. A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Answer: When the child gets up and runs about on the trolley, the speed of the centre of mass of the trolley and child remains unchanged irrespective of the manner of motion of child. It is because here child and trolley constitute one single system and forces involved are purely internal forces. As there is no external force, there is no change in momentum of the system and velocity remains unchanged.

Question 7. 4.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q4
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q4.1

Question 7. 5.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q5
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q5.1

Question 7. 6. Find the components along the x, y, z-axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z- component.
Answer: 
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q6

Question 7. 7. Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system the same whatever be the point about which the angular momentum is taken.
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q7

Question 7. 8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are 36.9° and ‘ 53.2° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q8
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q8.1

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q8.2

Question 7. 9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Answer:  Let F1 and F2 be the forces exerted by the level ground on front wheels and back wheels respectively.Considering rotational equilibrium about the front wheels,F2 x 1.8 = mg x 1.05 or F2 = 1.05/1.8 x 1800 x 9.8 N =10290 N Force on each back wheel is =10290/2 N or 5145 N.
Considering rotational equilibrium about the back wheels.
F1 x 1.8 = mg (1.8 – 1.05) = 0.75 x 1800 x 9.8
or F1=0.75 x 1800 x 9.8/1.8 = 7350 N
Force on each front wheel is 7350/2 N or 3675 N.

Question 7. 10. (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.
Answer:  (a) Moment of inertia of sphere about any diameter = 2/5 MR2
Applying theorem of parallel axes,Moment of inertia of sphere about a tangent to the sphere = 2/5 MR2 +M(R)2 =7/5 MR2
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.

Question 7. 11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Answer:  Let M be the mass and R the radius of the hollow cylinder, and also of the solid sphere. Their moments of inertia about the respective axes are I1 = MR2 and I2 = 2/5 MR2
Let τ be the magnitude of the torque applied to the cylinder and the sphere, producing angular accelerations α1and α2 respectively. Then τ=I1 α1 = I2 α2
The angular acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular speed after a given time.

Question 7. 12. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad 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?
Answer:  M = 20 kg
Angular speed, w = 100 rad s-1; R = 0.25 m
Moment of inertia of the cylinder about its axis
=1/2 MR2 = 1/2 x 20 (0.25)2 kg m2 = 0.625 kg m2
Rotational kinetic energy,
Er = 1/2 Iw2 = 1/2 x 0.625 x (100)2 J = 3125 J
Angular momentum,
L = Iw = 0.625 x 100 Js= 62.5 Js.

Question 7. 13. (a) A child stands at the centre of a turntable with his arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction, (b)  Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account of this increase in kinetic energy?
Answer: (a) Suppose, initial moment of inertia of the child is I1 Then final moment of inertia,
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q13

Question 7. 14. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Answer:  Here, M = 3 kg, R = 40 cm = 0.4 m
Moment of inertia of the hollow cylinder about its axis.
I = MR2 = 3(0.4)2 = 0.48 kg m2
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q14

Question 7. 15. To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?
Note: Uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter fricitional torque). Assume that the engine is 100 efficient.
Answer: Here, a = 200 rad s-1; Torque, τ= 180 N-m
Since,Power, P = Torque (τ) x angular speed (w)
= 180 x 200 = 36000 watt = 36 KW.

Question 7. 16. From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc.Locate the centre of gravity of the resulting flat body.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q16
Answer: Let from a bigger uniform disc of radius R with centre O a smaller circular hole of radius R/2 with its centre at O1 (where R OO1 = R/2) is cut out. Let centre of gravity or the centre of mass of remaining flat body be at O2, where OO2 = x. If σ be mass per unit area, then mass of whole disc M1 = πR2σ and mass of cut out part
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q16.1
i.e., O2 is at a distance R/6 from centre of disc on diametrically opposite side to centre of hole.

Question 7. 17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Answer: Let m be the mass of the stick concentrated at C, the 50 cm mark, see fig.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q17
For equilibrium about C, the 45 cm mark,
10 g (45 – 12) = mg (50 – 45)
10 g x 33 = mg x 5
=> m = 10 x 33/5
or m = 66 grams.

Question 7. 18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination, (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
Answer:  (a) Using law of conservation of energy,
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q18
(c) Clearly, the solid sphere will take longer to roll down the plane with smaller inclination.

Question 7. 19. A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Answer:  Here, R = 2 m, M = 100 kg
v = 20 cm/s = 0.2 m/s
Total energy of the hoop =1/2Mv2 + 1/2Iw2
=1/2Mv2 + 1/2(MR2)w2
=1/2Mv2 +1/2Mv2 =Mv2
Work required to stop the hoop = total energy of the hoop W = Mv2 = 100 (0.2)2= 4 Joule.

Question 7. 20. The oxygen molecule has a mass of 5.30 x 10-26 kg and a moment of inertia of 1.94 x 10-45 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q20

Question 7. 21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Answer: Here, θ= 30°, v = 5 m/ s
Let the cylinder go up the plane up to a height h.
From 1/2 mv2 +1/2IW2 = mgh
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q21

Question 7. 22. As shown in Fig. the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied halfway up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be friction less and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m)(Hint: Consider the equilibrium of each side of the ladder separately.)
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q22
Answer: The forces acting on the ladder are shown in Fig. 7.14.Here, IV = 40 kg = 40 x 9.8 N = 392 N, AB = AC = 1.6 m, BD = 1/2 x 1.6 m = 0.8 m,
BF = 1.2 m and DE 0.5 m,
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q22.1
=(392 x (1-0.375))/1=245 N
Now, it can be easily shown that tension in the string T = NB – NC = 245 – 147 = 98 N.

Question 7. 23. A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minutes. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.(a) What is his new angular speed? (Neglect friction)(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Answer:  Here, I1= 7.6 + 2 x 5 (0.9)2 = 15.7 kg m2
w1 = 30 rpm
I2 = 7.6 + 2 x 5 (0.2)2 = 8.0 kg m2
w2 = ?
According to the principle of conservation of angular momentum,
I2w2=I1w1
w2= I1/I w1= 15.7 x 30 /8.0 = 58.88 rpm
No, kinetic energy is not conserved in the process. In fact, as moment of inertia decreases, K.E. of rotation increases. This change comes about as work is done by the man in bringing his arms closer to his body.

Question 7. 24. A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.(Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3.)
Answer: Angular momentum imparted by the bullet, L = mv x r
=(10 x 10-3) x 500 x 1/2 =2.5
Also,I=ML2/3=12 x (1.0)2/3=4 kg m2
Since L=Iw
w=L/I=2.5/4=0.625 rad / s

Question 7. 25. Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed w1 and w2 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 w1 not equal to w2.
Answer: (a) Let I1 and I2 be the moments of inertia of two discs having angular speeds w1, and w2 respectively. When they are brought in contact, the moment of inertia of the two-disc system will be I1 + I2. Let the system now have an angular speed w. From the law of conservation of angular momentum, we know that
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q25
Now, (w1 – w2)2 will be positive whether w1 is greater or smaller than w2.
Also,I1I2/2(I1+ I2) is also positive because I1 and I2 are positive.
Thus, k1– k2 is a positive quantity.
.-. k1 = k2 + a positive quantity or k1 > k2
.-. The kinetic energy of the combined system ( k2) is less than the sum of the kinetic energies of the two dies.The loss of energy on combining the two discs is due to the energy being used up because of the frictional forces between the surfaces of the two discs. These forces, in fact, bring about a common angular speed of the two discs on combining.

Question 7. 26. (a) Prove the theorem of perpendicular axes.
Hint: Square of the distance of a point (x, y) in the x-y plane from an axis through the origin perpendicular to the plane is x2 + y2]
(b) Prove the theorem of parallel axes.
Hint: If the centre of mass of chosen the origin [Σ miri= 0]
Answer: (a) The theorem of perpendicular axes: According to this theorem, the moment of inertia of a plane lamina (i.e., a two dimensional body of any shape/size) about any axis OZ perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually perpendicular axes OX and OY in the plane of lamina, meeting at a point where the given axis OZ passes through the lamina. Suppose at the point ‘R’ m{ particle is situated moment of inertia about Z axis of lamina
= moment of inertia of body about x-axis
= moment of inertia of body about the y-axis.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q26
(b) Theorem of parallel axes: According to this theorem, moment of inertia of a rigid body about
any axis AB is equal to moment of inertia of the body about another axis KL passing through
centre of mass C of the body in a direction parallel to AB, plus the product of total mass M of the
body and square of the perpendicular distance between the two parallel axes. If h is perpendicular distance between the axes AB and KL, then Suppose rigid body is made up of n particles m1, m2, …. mn, mn at perpendicular distances r1, r2, ri…. rn. respectively from the axis KL passing through centre of mass C of the body.
If h is the perpendicular distance of the particle of mass m{ from KL, then
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q26.1

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q26.2

Question 7. 27. Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by,v2=2gh/(1+k2/R2) using dynamical consideration (i.e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Answer: Let a rolling body (I = Mk2) rolls down an inclined plane with an initial velocity u = 0; When it reaches the bottom of inclined plane, let its linear velocity be v. Then from conservation of mechanical energy, we have Loss in P.E. = Gain in translational K.E. + Gain in rotational K.E.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q27

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q27.1
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q27.2

Question 7. 28. A disc rotating about its axis with angular speed wo is placed lightly (without any translational push) on a perfectly friction less 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 Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q28
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q28.1

Question 7. 29. Explain why friction is necessary to make the disc roll (refer to Q. 28) 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?
Answer: 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 rolling begins only when velocity of point B becomes zero. Also, force of friction would become zero at this stage.

Question 7. 30. A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10π rad/s. Which of two will start to roll earlier? The coefficient of kinetic friction is uk = 0.2.
Answer:  When a disc or ring starts rotatory motion on a horizontal surface, initial translational velocity of centre of mass is zero.
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q30
The frictional force causes the centre of mass to accelerate linearly but frictional torque causes angular retardation. As force of normal reaction N = mg, hence frictional force f = uk N = uk mg.
For linear motion f = uk . mg = ma ———-(i)
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q30.1

Question 7. 31. A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction us = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination O of the plane is increased, at what value of 8 does the cylinder begin to skid, and not roll perfectly?
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q31

Question 7. 32. Read each statement below carefully, and state, with reasons, if it is true or false:
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolloing is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Answer:  (a) True. When a body rolls without slipping, the force of friction acts in the same
direction as the direction of motion of the centre of mass of rolling body.
(b) True. This is because rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground. Hence its instantaneous speed is zero.
(c) False. This is because when the body is rotating, its instantaneous acceleration is not zero.
(d) True. For perfect rolling motion as there is no relative motion at the point of contact, hence work done against friction is zero.
(e) True. This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, it will simply slip under the effect of its own weight.

Question 7. 33. Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q33
Answer:
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q33.1

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q33.2
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q33.3
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q33.4
NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q33.5

NCERT Solutions for Class 11 Physics All Chapters

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power are part of Class 11 Physics NCERT Solutions. Here we have given NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power:

Section NameTopic Name
6Work Energy and power
6.1Introduction
6.2Notions of work and kinetic energy : The work-energy theorem
6.3Work
6.4Kinetic energy
6.5Work done by a variable force
6.6The work-energy theorem for a variable force
6.7The concept of potential energy
6.8The conservation of mechanical energy
6.9The potential energy of a spring
6.10Various forms of energy : the law of conservation of energy
6.11Power
6.12Collisions

QUESTIONS FROM TEXTBOOK

Question 6. 1. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket,
(b) Work done by gravitational force in the above case,
(c) Work done by friction on a body sliding down an inclined plane,
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer: Work done, W = T.S = Fs cos θ
(a) Work done ‘positive’, because force is acting in the direction of displacement i.e., θ = 0°.
(b) Work done is negative, because force is acting against the displacement i.e., θ = 180°.
(c) Work done is negative, because force of friction is acting against the displacement i.e., θ= 180°.
(d) Work done is positive, because body moves in the direction of applied force i.e., θ= 0°.
(e) Work done is negative, because the resistive force of air opposes the motion i.e., θ = 180°.

Question 6. 2. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) Work done by the applied force in 10 s
(b) Work done by friction in 10 s
(c) Work done by the net force on the body in 10 s
(d) Change in kinetic energy of the body in 10 s and interpret your results.
Answer: (a) We know that Uk = frictional force/normal reaction
frictional force = Uk x normal reaction
= 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 N
net effective force = (7 – 1.96) N = 5.04 N
acceleration = 5.04/2 ms-2 = 2.52 ms-2
distance, s=1/2x 2.52 x 10 x 10 = 126 m
work done by applied force = 7 x 126 J = 882 J
(b) Work done by friction = 1.96 x 126 = -246.96 J
(c) Work done by net force = 5.04 x 126 = 635.04 J
(d) Change in the kinetic energy of the body
= work done by the net force in 10 seconds = 635.04 J (This is in accordance with work-energy theorem).

More Resources for CBSE Class 11

Question 6.3. Given figures are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of some physical contexts for which these potential energy shapes are relevant.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q3
Answer: We know that total energy E = K.E. + P.E. or K.E. = E – P.E. and kinetic energy can never be negative.The object can not exist in the region, where its K.E. would become negative.
(a) In the region between x = 0 and x = a, potential energy is zero. So, kinetic energy is positive. In the region x > a, the potential energy has a value greater than E. So, kinetic energy will be negative in this region. Thus the particle cannot be present in the region x > a.
The minimum total energy that the particle can have in this case is zero.
(b) Here P.E. > E, the total energy of the object and as such the kinetic energy of the object would be negative. Thus object cannot be present in any region of the graph.
(c) Here x = 0 to x = a and x > b, the P.E. is more than E, so K.E. is negative. The particle can not exist in these portions.
(d) The object can not exist in the region between x = -b/2 to x =-a/2 and x = -a/2 to x = -b/2 .Because in this region P.E. > E.

Question 6.4.The potential energy function for a particle executing linear simple harmonic motion is given by V (x) – kx2/2, where k is the force constant of the oscillator. For k = 0.5 Nm-1 , the graph of V (x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q4
Answer: Here, force constant k = 0.5 Nm-1 and total energy of particle E = 1J. The particle can go up to a maximum distance xm, where its total energy is transformed into elastic potential energy.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q4.1

Question 6. 5. Answer the following:
(a) The casing of a rocket in flight bums up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atomosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In Fig.(i), the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q5
Answer: (a) Heat energy required for burning of casing of rocket comes from the rocket itself. As
a result of work done against friction the kinetic energy of rocket continuously decreases – and this work against friction reappears as heat energy.
(b) This is because gravitational force is a conservative force. Work done by the gravitational ‘ force of the sun over a closed path in every complete orbit of the comet is zero.
(c)  As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential decreases rapidly. As a result, kinetic energy of
satellite slightly increases i.e., its speed increases progressively.
(d) In Fig. (i), force is applied on the mass, by the man in vertically upward direction but distance is moved along the horizontal.
θ = 90°. W = Fs cos 90° = zero
In Fig. (ii), force is applied along the horizontal and the distance moved is also along the horizontal. Therefore, θ = 0°.
W = Fs cos θ = mg x s cos 0°
W = 15 x 9.8 x 2 x 1 = 294 joule.
Thus, work done in (ii) case is greater.

Question 6. 6. Point out the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b)Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces of the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Answer: (a) Potential energy of the body decreases because the body in this case goes closer to the centre of the force.
(b) Kinetic energy, because friction does its work against the motion.
(c) Internal forces can not change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/ total energy of the system of two bodies.

Question 6. 7. State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer: (a) False, the total momentum and total energy of the system are conserved.
(b) False, the external force on the system may increase or decrease the total energy of the system.
(c) False, the work done during the motion of a body over a closed loop is zero only when body is moving under the action of a conservative force (such as gravitational or electrostatic force). Friction is not a conservative force hence work done by force of friction (or work done on the body against friction) is not zero over a closed loop.
(d) True, usually in an inelastic collision the final kinetic energy is always less than the initial kinetic energy of the system.

Question 6. 8. Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer:  (a) In this case total kinetic energy is not conserved because when the bodies are in contact dining elastic collision even, the kinetic energy is converted into potential energy.
(b) Yes, because total momentum conserves as per law of conservation of momentum.
(c) The answers remain unchanged.
(d) It is a case of elastic collision because in this case the forces will be of conservative nature.

Question 6. 9.  A body is initially at rest. It undergoes a one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) t1/2 (ii) t (iii) t3/2 (iv) t2
Answer: (ii) From v = u + at
v = 0 + at = at
As power, p = F x  v
.’. p = (ma) x at = ma2t
Since m and a are constants, therefore, p α t.

Question 6. 10. A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2
Answer: (ii)p = force x velocity
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q10

Question 6. 11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q11
where i, j, k, are unit vectors along the x- y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Answer:
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q11.1

Question 6. 12. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10-31 kg, proton mass = 1.67 x 10-27 kg, 1 eV= 1.60 x 1019J).
Answer:
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q12

Question 6. 13. A raindrop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms-1?
Answer: Here, r = 2 mm = 2 x 10-3 m.
Distance moved in each half of the journey, S=500/2= 250 m.
Density of water, p = 103 kg/ m3
Mass of rain drop = volume of drop x density
m =4/3 π r2 x ρ =4/3 x 22/7 (2 x 10-3)3 x 103 = 3.35 x 10-5 kg
.-. W = mg x s = 3.35 x 10-5 x 9.8 x 250 = 0.082 J
Note: Whether the drop moves with decreasing acceleration or with uniform speed, work
done by the gravitational force on the drop remains the same.
If there was no resistive forces, energy of drop on reaching the ground.
E1= mgh = 3.35 x 10-5 x 9.8 x 500 = 0.164 J
Actual energy, E2 = 1/2mv2 = 1/2 x 3.35 x 10-5 (10)2 = 1.675 x 10-3J
Work done by the resistive forces, W =E1 – E2 = 0.164 – 1.675 x 10-3 W
= 0.1623 joule.

Question 6. 14. A molecule in a gas container hits a horizontal wall with speed 200 msA and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Answer:  Let us consider the mass of the molecule be m and that of wall be M. The wall remains at rest due to its large mass. Resolving momentum of the molecule along x-axis and y-axis, we get
The x-component of momentum of molecule
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q14

Question 6. 15. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Answer:  Here, volume of water = 30 m; t = 15 min = 15 x 60 = 900s
h = 40 m ; n= 30%
As the density of water = p = 103 kg m-3
Mass of water pumped, m = volume x density = 30 x 103 kg
Actual power consumed or output power p0 = W/t = mgh/t
=>p0=(30 x 103 x 9.8 x 40)/900=13070 watt
If pi is input power (required), then as
η=p0/pi=> pi=p0/η = 13070/(30/100)=43567 W =43.56 KW

Question 6. 16. Two identical ball bearings in contact with each other and resting on a friction less table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig.) is a possible result after collision?
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q16
Answer: Let m be the mass of each ball bearing. Before collision, total K.E. of the system
=1/2mv2 + 0 =1/2 mv2
After collision, K.E. of the system is
Case I, E1 = 1/2 (2m) (v/2)2 = 1/4 mv2
Case II, E2 = 1/2 mv2
Case III, E3 = 1/2(3m) (v/3)2 = 1/6mv2
Thus, case II is the only possibility since K.E. is conserved in this case.

Question 6.17. The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Answer: Since collision is elastic therefore A would come to rest and B would begin to move with the velocity of A.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q17
The bob transfers its entire momentum to the ball on the table. The bob does not rise at all.

Question 6. 18. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
Answer:  On releasing the bob of pendulum from horizontal position, it falls vertically downward by a distance equal to length of pendulum i.e., h = l = 1.5 m .
As 5% of loss in P.E. is dissipated against air resistance, the balance 95% energy is transformed into K.E. Hence,
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q18

Question 6. 19. A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a friction less track. After a while, sand starts leaking out of a hole on the trolley’s floor at the rate of 0.05 kg s-1. What is the speed of the trolley after the entire sand bag is empty?
Answer:  The system of trolley and sandbag is moving with a uniform speed. Clearly, the system is not being acted upon by an external force. If the sand leaks out, even then no external force acts. So there shall be no change in the speed of the trolley.

Question 6. 20. A particle of mass 0.5 kg travels in a straight line with velocity u = a x3/2, where a = 5 m-1/2 s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Answer: Here m = 0.5 kg
u=a x3/2, a = 5 m-1/2 s-1.
Initial velocity at x = 0, v1 = a x 0 = 0
Final velocity at x = 2, v2 = a (2)3/2 = 5 x (2)3/2
Work done = increase in K.E
= 1/2 m(v22-v12) = 1/2 x 0.5[(5 x (2)3/2)2 – 0] = 50 J.

Question 6. 21. The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m-3 . What is the electrical power produced?
Answer: (a) Volume of wind flowing per second = Av Mass of wind flowing per second = Avρ
Mass of air passing in second = Avρt
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q21

Question 6. 22. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 x 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Answer: Here, m = 10 kg, h = 0.5 m, n = 1000
(a) work done against gravitational force.
W = n(mgh) = 1000 x (10 x 9.8 x 0.5) = 49000J.
(b) Mechanical energy supplied by 1 kg of fat = 3.8 x 107 x20/100
= 0.76 x107 J/kg
.-. Fat used up by the dieter =1kg/(0.76 x 107) x 49000 = 6.45 x 10-3 kg

Question 6. 23. A family uses 8 kW of power, (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
Answer:  (a) Power used by family, p = 8 KW = 8000 W
As only 20% of solar energy can be converted to useful electrical energy, hence, power
8000 W to be supplied by solar energy = 8000 W/20 = 40000 W
As solar energy is incident at a rate of 200 Wm-2, hence the area needed
A=4000 W/200 Wm-2 =200 m2
(b) The area needed is camparable to roof area of a large sized house.

Question 6. 24. A bullet of mass 0.012 kg and horizontal speed 70 ms-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by thin wire. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer: Here, m1 = 0.012 kg, u1 = 70 m/s
m2 = 0.4 kg, u2 = 0
As the bullet comes to rest with respect to the block, the two behave as one body. Let v be the velocity acquired by the combination.
Applying principle of conservation of linear momentum, (m1 + m2) v = m1H1 + m2u2 = m1u1
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q24

Question 6. 25. Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig). Will the stones reach the bottom at the same time? Will they reach there at the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and h = 10 m, what are the speeds and times taken by the two stones?
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q25
Answer:
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q25.1

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q25.2

Question 6. 26. A 1 kg block situated on a rough incline is connected to a spring with spring constant 100 Nm-1 as shown in Figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has negligible mass and the pulley is frictionless.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q26
Answer:
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q26.1

Question 6. 27. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms-1. It hits the floor of the elevator (length of elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Answer:  P.E. of bolt = mgh = 0.3 x 9.8 x 3 = 8.82 J
The bolt does not rebound. So the whole of the energy is converted into heat. Since the value of acceleration due to gravity is the same in all inertial system, therefore the answer will not change even if the elevator is stationary.

Question 6. 28. A trolley of mass 200 kg moves with a uniform speed of 36 km h-1 on a friction less track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms-1 relative to the trolley in a direction opposite to the trolley’s motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Answer:  Let there be an observer travelling parallel to the trolley with the same speed. He will observe the initial momentum of the trolley of mass M and child of mass m as zero. When the child jumps in opposite direction, he will observe the increase in the velocity of the trolley by Δv.
Let u be the velocity of the child. He will observe child landing at velocity (u – Δu) Therefore, initial momentum = 0
Final momentum = MΔ v – m (u – Δv)
Hence, MΔ v – m (u – Δv) = 0
Whence Δv =mu/ M + m
Putting values Δv =4 x 20/ 20 + 220 = ms-1
.-. Final speed of trolley is 10.36 ms-1.
The child take 2.5 s to run on the trolley.
Therefore, the trolley moves a distance = 2.5 x 10.36 m = 25.9 m.

Question 6.29. Which of the following potential energy curves in Fig. cannot possibly describe the elastic collision of two billiard balls? Here r is distance between centres of the balls.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q29
Answer: The potential energy of a system of two masses varies inversely as the distance (r) between 1
them i.e., V (r) α 1/r. When the two billiard balls touch each other, P.E. becomes zero i.e., at r = R + R = 2 R; V (r) = 0. Out of the given graphs, curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.

Question 6. 30. Consider the decay of a free neutron at rest: n > p + e. Show that the two body decay of this type must necessarily give an electron of fixed energy, and therefore, cannot account for the observed continuous energy distribution in the β -decay of a neutron or a nucleus, Fig.
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q30
Answer: Let the masses of the electron and proton be m and M respectively. Let v and V be the velocities of electron and proton respectively. Using law of conservation of momentum. Momentum of electron + momentum of proton = momentum of neutron
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Q30.1
Thus, it is proved that the value of v2 is fixed since all the quantities in right hand side are constant. It establishes that the emitted electron must have a fixed energy and thus we cannot account for the continuous energy distribution in the β-decay of a neutron.