NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations

# NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations

Edited By Safeer PP | Updated on Aug 09, 2022 11:41 AM IST

NCERT Exemplar Class 11 Physics solutions chapter 14 Oscillations is a chapter which is very commonly asked in the examinations. The students get to learn about the simple harmonic and uniform motion possessed by a body and various other terms related to the periodic motion of a body. NCERT Exemplar Class 11 Physics chapter 14 solutions revolve around different types of oscillations and waves and their respective characteristics. The students can also use NCERT Exemplar Class 11 Physics solutions chapter 14 pdf download for any future reference. This chapter of NCERT Class 11 Physics Syllabus consists of several relations & formulae and questions related to it that needs to be studied.

The NCERT Exemplar solutions Class 11 Physics chapter 14 comprises several exercises and their solutions that students should practice well to understand all its concepts. It is an important chapter from the competitive point of view so requires a deeper understanding.
Also, check NCERT Solutions for Class 11 other subjects

## NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQI

Question:14.1

The displacement of a particle is represented by the equation . The motion of the particle is
(a) simple harmonic with period
(b) simple harmonic with period
(c) periodic but not simple harmonic
(d) non-periodic

The answer is the option (b) Simple harmonic with period

Question:14.2

The displacement of a particle is represented by the equation . The motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period
(d) simple harmonic with period

The answer is the option (b) periodic but not simple harmonic

Question:14.3

The relation between acceleration and displacement of four particles are given below:
(a)
(b)
(c)
(d) $a_x = -2x$
Which one of the particle is exempting simple harmonic motion?

The answer is the option (d) $a_x = -2x$
Explanation: In simple harmonic motion,
Acceleration proportional (as well as opposite) to displacement.
Thus, opt (d)

Question:14.4

The motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid

The answer is the option (c) Simple harmonic and time period is independent of the density of the liquid.
Explanation: Let us take a test tube viz., filled with a liquid of density up to height ‘h’.
When the liquid is lifted in arm Q to a height ‘y’ from A to B, the liquid in arm P drops by the height ‘y’ from A’ to C’. The height difference between the two arms is 2y.
Here, the hydrostatic pressure provides the restoring force, thus,

, where,
A = Area of a cross-section of tube, &
Thus, it is simple harmonic motion.

Thus,
Thus, the motion is harmonic as the time period is independent of density.

Question:14.5

A particle is acted simultaneously by mutually perpendicular simple harmonic motion . The trajectory of motion of the particle will be
(a) an ellipse
(b) a parabola
(c) a circle
(d) a straight line

The answer is the option (c) a circle.
Explanation: We know that,
Resultant displacement = x + y

Thus,

……. (particle is acted simultaneously by mutually perpendicular direction)
……. Thus, the displacement can neither be a straight line nor a parabola
Now, let us square and add x & y,

Thus,
This is the equation of a circle, hence, opt (c)

Question:14.6

The displacement of a particle varies with time according to the relation

(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude
(d) The motion is SHM with amplitude $\sqrt{{a}^{2} + b^{2}}$

The answer is the option (d) The motion is SHM with amplitude
Explanation: Given:
Now, we know that,
The amplitude of motion,
Thus, opt (d).

Question:14.7

Four pendulums A, B, C and D are suspended from the same elastic support as shown in the figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
(a) D will vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude

The answer is the option (b) C will vibrate with maximum amplitude.
Explanation: If the pendulum vibrates with transverse vibration,
Time period, , where,
l = length of pendulum A & C.
Now, the disturbance produced is transmitted to all pendulum, i.e., B, C & D, where the time period (T) of C & A is the same. C will vibrate with a maximum in resonance, as the periodic force of period T produces resonance in C.
Hence, opt (b).

Question:14.8

Explanation: Circular motion is executed by P with radius B.
Here,
=
=
Thus,

$x = B \sin \frac{2 \pi }{30}t$
Thus, opt (a)

Question:14.9

The equation of motion of a particle is
The motion is
(a) periodic but not oscillatory.
(b) periodic and oscillatory.
(c) oscillatory but not periodic.
(d) neither periodic nor oscillatory.

The answer is the option (c) Oscillatory but not periodic.

Question:14.10

Explanation: Let us consider the equation of a SHM
Thus,

……. (i)

$= -a\omega^ 2 \sin \omega t$ ….. (ii)
…. (given)
From (i),

Thus, ….. (iii)
From (ii),
Now, …….. (iv)
Thus,
Therefore,
Now,
Thus, .
Hence opt (a).

Question:14.11

Explanation: If we connect a mass (m) to a spring on a frictionless horizontal surface, then, their frequencies will be –
&
…. (i) …. (ii)
Since the springs are parallel, their equivalent will be-

& frequency will be –

From (i),

&
Thus,

Hence, opt (b)

## NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQII

Question:14.12

The rotation of earth about its axis is
(a) periodic motion
(b) simple harmonic motion
(c) periodic but not simple harmonic motion
(d) non-periodic motion

The answer is the option (a) periodic motion and (c) periodic but not SHM
Explanation:
(i) The earth completes one revolution in a regular interval of time.
(ii) The motion of the earth is circular about its own axis.
(iii) But this motion is not SHM as we cannot measure its displacement since it is not about a fixed point. Also, it does not move both sides.

Question:14.13

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
(a) simple harmonic motion
(b) non-periodic motion
(c) periodic motion
(d) periodic but not SHM

The answer is the option (a) Simple harmonic motion and (b) non - periodic motion
Explanation: Lift the ball from point A to B and smoothly release it to reach C & then return to first A and then B. Thus, it is periodic motion. Here, mg sin ? balances the restoring force (R), also, a restoring force acts on the ball.
Thus,
or ….. (when the ball moves upward)

Thus,
Thus, it is a simple harmonic motion.
or
Thus, this motion is periodic as well as simple harmonic.

Question:14.14

Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements.
(a) Phase of the oscillator is same at t = 0s and t = 2 s
(b) Phase of the oscillator is same at t = 2s and t = 6 s
(c) Phase of the oscillator is same at t = 1s and t = 7 s
(d) Phase of the oscillator is same at t = 1s and t = 5 s

The answer is the option (b) Phase of the oscillator is same at t = 2s & t = 6s and (d) The phase of the oscillator is the same at t = 1s & t = 5s
Explanation: If the mode of vibration of two particles is the same, then they are said to be in the same phase. That means, their distance will be , where, n = 1, 2, 3,4, …..
(a) Here the particles are not in the same phase as the distance between them is .
(b) The particles are at a distance . Thus, they are in the same phase.
(c) The distance between particles is . Thus, they are not in the same phase.
(d) The distance between the particles is . Thus, they are in the same phase.

Question:14.15

Which of the following statements is/are true for a simple hannonic oscillator?
(a) Force acting is directly proportional to displacement from the mean position and opposite to it
(b) Motion is periodic
(c) Acceleration of the oscillator is constant
(d) The velocity is periodic

The answer is the option (a) Force acting is directly proportional to the displacement from the mean position and opposite to it, (b) Motion is periodic, and (d) The velocity is periodic.
Explanation: Let, , be a SHM …… (i)
…. (ii)

Thus, or A = -ω2x

Hence,
Now, x(t) = x(t + T)
Thus, the motion is periodic and simple harmonic.
Now, from (ii),
v(t) = v(t + T)
Hence, opt, (a), (b) & (d) are correct options.

Question:14.16

The displacement time graph of a particle executing S.H.M. is shown in Figure. Which of the following statement is/are true?

(a) The force is zero at
(b) The acceleration is maximum at
(c) The velocity is maximum at
(d) The P.E. is equal to K.E. of oscillation at

The answer is the option (a) The force is zero at , (b) The acceleration is maximum at , and (c) The velocity is maximum at
Explanation: (a) The particle is at its mean position at , so the force acting on it zero, but due to inertia of mass the motion continues
a = 0 thus, F = 0.
(b) Particles velocity changes increasing to decrease so maximum at the change in velocity, at . Thus, acceleration is maximum here.
(c) The velocity is maximum at its mean position at as there is no retarding force on it.
(d) K.E. = 0 at
Thus, P.E is not equal to kinetic energy.

Question:14.17

A body is performing SHM, then its
(a) average total energy per cycle is equal to its maximum kinetic energy
(b) average kinetic energy per cycle is equal to half of its maximum kinetic energy
(c) mean velocity over a complete cycle is equal to times of its maximum velocity
(d) root mean square velocity is times of its maximum velocity

The answer is the option (a) Average total energy per cycle is equal to its maximum kinetic energy, (b) Average kinetic energy per cycle is equal to half its maximum kinetic energy, and (d) Root square mean velocity is equal to times its maximum velocity.
Explanation: (a) let be a periodic SHM
Let m be the mass executing SHM

=

Now, = Total mechanical energy
Thus,
Thus,
Or we can say that the average total energy is K.E.max
(b) Let, amplitude = a
Angular frequency =
Thus, maximum velocity = αω, which varies according to the sine law.
Thus, the rms value of a complete cycle =
Thus, the average

=
=
=
=
(c)
$V_{mean} = \frac{(a\omega -a\omega )}{2}$
Vmean = 0 ….. since vmax is not equal to vmean
$V_{rms} = \sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2} }= \frac{a\omega }{\sqrt{2}}$
Thus,

Question:14.18

A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure). Take the direction from A to B as the positive direction and choose the correct statements. __ _
AO = OB = 5 cm
BC= 8 cm

(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive
(b) The sign of velocity of the particle at C going towards B is negative
(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative
(d) The sign of acceleration and force on the particle when it is at point B is negative

The answer is the option (a) The sign of velocity, acceleration & force on the particle when it is 3 cm away from A going towards B is positive, (c) The sign of velocity, acceleration & force on the particle when it is 4 cm away from B going towards A are negative, and (d) The sign of acceleration and force on the particle when it is at point B is negative.
Explanation: (a) The velocity of the particle increases up to 0 when it is 3 cm away from A and is going from A to B, i.e., in the positive direction. Thus, the velocity is positive. Also, acceleration in SHM is towards positive.
(b) Here the velocity is positive and not negative since the particle is going towards B.
(c) Here the particle is 4 cm away from B and is going towards A, i.e., the particle is going from B to A, i.e., in the negative direction. Hence, the velocity & acceleration towards mean position O is negative.
(d) Here, force and acceleration both are negative as the particle is at B, and they both are towards O.

## NCERT Exemplar Class 11 Physics Solutions Chapter 14: Very Short Answer

Question:14.19

Displacement versus time curve for a particle executing SHM is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.

(i) At A, C, E, and G the displacement is maximum. Hence, the velocity of the oscillator will also be maximum.
(ii) At B, D, F and H the displacement of the oscillator is zero. Thus, there is no restoring force. Hence, the speed of oscillator will be maximum.

Question:14.20

If we displace the mass ‘m’ frpm its equilibrium position towards the right by distance ‘x’, The spring B will be compressed by distance x, & let kx be the force applied on the mass ‘m’ towards left. If we apply the force kx on the mass, A will be extended by distance x towards left; and apply force kx towards left. Thus, restoring force will act on the block as net force towards left.
F = kx + kx
= 2kx
Thus, restoring force towards the left is 2kx.

Question:14.21

What are the two basic characteristics of a simple harmonic motion?

(i)The direction of acceleration is towards the mean position, and
(from mean position)
(ii) The direction of force and displacement are opposite, thus,

These are the two basic characteristics of SHM.

Question:14.22

When will the motion of a simple pendulum be simple harmonic?

Let us consider a pendulum whose,
Length = l
Mass of bob = m, viz., displaced by angle =
Now, Restoring force $= F=$
Also, if is small then
Sin =
$=\frac{arc}{radius}$
$=\frac{x}{l}$
Thus,

Or …………. (since m, g & l are constants)
Thus, simple harmonic for the small-angle

Question:14.21

What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Let us consider to be an SHM
Now,

Now, for ,
Thus,
Now,

Now, for ,

Thus,
Now,
.

Question:14.24

What is the ratio between the distance travelled by the oscillator in one time period and amplitude?

We already know that the distance travelled by an oscillator in a one-time period is equal to 4A,
Where A = amplitude of oscillation.
Thus, we know that the
Required ratio

=4:1

Question:14.25

At time = t, P’ is the foot perpendicular of the velocity vector of particle P.
This foot shifts from P to Q, i.e., towards the negative axis, when the particle moves from P to P1.
Thus, there is –ve sign of motion of P’.

Question:14.26

Let us consider to be a SHM …….. (i)
Now,

…….. [since ]
Thus,
From (i), we know that,
The phase of displacement =
& from (ii), we know that,
The phase of velocity =
Thus, the phase difference =

Question:14.27

Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.

Let us consider that a mass is lying on a horizontal frictionless surface, where spring constant = k.
If we displace the mass by distance A from its mean position, then it will execute SHM.
At this stretched position, P.E. of mass =
Now, at maximum stretch, i.e.,
At x = A, K.E. = 0
Now, at
P.E. = total energy
=
Now let’s consider that the mass is back at its mean position, now the restoring force acting on the particle will be

=
The restoring force constant of oscillator is
When

&

Now, when
 x K.E. P.E. T.E. 0 0 +A 0 -A 0
Now, at displacement ‘x’, T.E. will be,
E = K.E. + P.E.
=
Thus, with displacement ‘x’, E is constant.

Question:14.28

The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?

A second pendulum is a pendulum with time period (T) = 2 seconds

Thus, $T_{e}^{2} = 4\pi ^{2}{\frac{l_{e}}{g_{e}}}$ …… (i)
Now, $T_m = 2\pi \sqrt{ \frac{l_{m}}{g_{m}}}$
Thus, $T_{m}^{2} = 4\pi ^{2} 6\frac{ lm}{g_e}$ ……. (ii) (gm=ge/6)
Now, for the second pendulum,
$\frac{T_{m}^{2}}{T_{e}^{2}}= {6g_e} \frac{l_{e}}{g_{e}l_m}$
$1 = \frac{6lm}{l_{e}}$
……. (since )
Thus,

## NCERT Exemplar Class 11 Physics Solutions Chapter 14: Short Answer

Question:14.29

If we pull mass M & then release it, it oscillates with the pulley up & down. Let x0 be the extension of the string when loaded with M, Due to acceleration and the same amount of forces the extension and compression of the spring from initial position is larger and smaller, respectively. Hence, we can neglect the gravitational force here.
Now let us apply force ‘F’ to pull M downwards by displacement x. As the string cannot be extended, its extension will be 2x.
Thus, the total extension =
When we pull it downward by x

When we do not pull M,

F = 2T
Thus,
&

Now, restoring force,
Frest
=
=
Thus,

Thus,
Therefore, it is a simple harmonic motion.
Now,

Question:14.3

Show that the motion of a particle represented by$y = \sin wt - \cos wt$ is simple harmonic with a period of .

$= \sqrt{2}\ \left [sin \omega t \frac{1}{\sqrt{2}}- \cos \omega t \frac{1}{\sqrt{2}} \right ]$
$= \sqrt{2}\ sin\left [ \omega t- \frac{\pi }{4} \right ]$
Comparing this equation with standard SHM we get,

Or

Question:14.31

Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Let us consider an oscillator viz. at ,
displacement = x from its mean position,
Mass = m

Force constant of oscillator,
Thus,
At x = A, when K.E. = 0, P.E. will be maximum & it will be the total energy of the oscillator

At displacement ‘x’,
Thus,
=
&
Thus, when displacement = amplitude from mean position, P.E. will be half of the total energy.

Question:13.32

A body of mass m is situated in a potential field when and are constants. Find the time period of small oscillations.

Here, dW = F.dx
So, if W = U, dU = F.dx
Or ….. (since, restoring force here is opposite to displacement)

Now, is small for SHM, So sin will become …… (i)
Thus,
Now, since, & are constants,

Thus, the motion will be SHM.
From (ii)

Thus,

Thus, considering (i) time period is valid for the small-angle .

## NCERT Exemplar Class 11 Physics Solutions Chapter 14: Long Answer

Question:14.35

A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.
(a) Will there be any change in weight of the body, during the oscillation?
(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

Due to normal reaction ‘N’, there will be weight in weight machine.
Let us consider the top positions of the platform, both the forces, due to the weight of the person and oscillator acts downwards.
Thus, the motion will be downwards.
Let us consider, acceleration = a
ma = mg – N …….. (i)
Now, when the platform moves upwards from its lowest position,
ma = N – mg …….. (ii)
Now, acceleration of oscillator is

From (i),

Where,
Amplitude = A
Angular frequency = ω &
Mass of oscillator = m

A = 5cm

m = 50 kg

Thus, the minimum weight is 95.5 N.
From (ii),
N – mg = ma
Now, for upward motion from the lowest point of the oscillator,

Therefore, during oscillation, there is a change in the weight of the body.
Also, maximum weight = 885 N, when the platform moves to upward direction from lowest direction & minimum weight = 95.5 N, when the platform moves to downward direction from the highest point.

Question:14.36

(a) As no deforming force acts on the spring when mass ‘m’ is supported by hand extension in the spring. Let m reach its new position at displacement = x unit, from the previous one, then,
P.E. of the spring or mass = gravitational P.E. lost by man
P.E. = mgx
But due to spring,

Thus,
When extension is , the spring will be at the mean position by the block

F = mg
Thus, ……. (ii)
Now, from (i) & (ii)

=
Thus,

Thus, from the mean position, the amplitude of the oscillator is maximum.
(b) Now, we know that time period (T) does not depend on amplitude,

From (i)

Or,
Now,

Since the total extension in the spring is 4 cm when released & amplitude is 2 cm, the oscillator will not rise above 4cm, & hence, it will oscillate below the released position.

Question:14.37

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.

where m is mass of the body and ρ is density of the liquid.

If we press a log downward into the liquid, a buoyant force acts on it, and due to inertia it moves upwards from its mean position & comes down again due to gravity.
Thus, the restoring force on the block = Buoyant force (B.F.) – mg
Volume of liquid displaces by block = V
When it floats,
mg = BF
Or,
…….. (i)
Area of crossection = A
Height of liquid block =
When pressed in water, the total height of the block in water =
Thus, net restoring force =
Frest …….. ( since BF is upward & x is downward)
Frest proportional to-x
Hence, the motion here is SHM.
Now,

Thus,
Frest

Thus,

Thus,

Question:14.38

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Let initial height = h0 of liquid in both columns; due to pressure difference, if the liquid in arm A is pressed by x, then liquid in arm B will rise by x.
Let us consider dx as an element of height, then,
Mass where, Area of a cross-section of tube = A
PE of left dm element columns = (dm) gh
& PE of dm in left columns =
Thus, total PE in left column =

From the figure

Thus, PE in left column = PE in right column =
Thus, total potential energy =
Let the element move towards right by y units due to the pressure difference, then,
Liquid column in left arm = (l – y)
Liquid column in right arm = (l + y)
PE in left arm =
PE in right arm = $A\rho g (l + y)^{2} \sin^{2} 45^{\circ}$
Total
Final

Change PE = Final PE – Initial PE

If there is change in velocity(v) of liquid column,

Thus,
Now, change in total energy =
Total change in energy,

Thus,

is not equal to 0, thus,

Now, let us differentiate w.r.t. t,
$g \left [0 + 2 \times 2\times2 \frac{dy}{dt} \right ] + 2l.2v.\frac{dv}{dt} = 0$
$\frac{d^{2}y}{dt^{2}} + \frac{g}{l} .y= 0$
Since 4v is not equal to 0,

Thus,

Thus,
$T=2\pi\sqrt{ \frac{l}{g}}$

Question:14.39

A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

We know that,
Acceleration due to gravity inside the earth = g’

=
Now, R-d = y
Thus,
On both of them, force at depth ‘d’ will be –

Thus, in the tunnel, the motion of the body is SHM.
We can say,

$a =\frac{ -g}{R }y$
$\omega ^{2}y =\frac{ -g}{R}y$
Thus,
Or, .

Question:14.4

Let us consider the diagram at which,
t = 0

Now, at t = t1,

Thus,
…….. (Given: T = 1sec)
Thus,

Or,

Now, at

Now,
i.e.,
Thus,
By the –ve sign it is clear that the bob’s motion is towards left.
Now,

=

Let H’ be the vertical distance covered by vy,

Thus, ……. (Given: )

We weill neglect since is very small.
Thus,
Thus,
H’’<<H’ as is very small.
Thus, H = H’
Thus, vxt = distance covered in horizontal

…….. (Given: )
Thus,
The bob was at a distance of from A at the time of snapping.
Thus, the distance of bob from A where it meet the ground =

## Class 11 Physics NCERT Exemplar Solutions Chapter 14 Includes the Following Topics:

• 14.1 Introduction
• 14.2 Periodic and Oscillatory Motions
• 14.3 Simple Harmonic Motion
• 14.4 Simple Harmonic Motion and Uniform Circular Motion
• 14.5 Velocity and Acceleration in Simple Harmonic Motion
• 14.6 Force Law for Simple Harmonic Motion
• 14.7 Energy in Simple Harmonic Motion
• 14.8 Systems executing SHM
• 14.9 Damped Simple Harmonic Motion
• 14.10 Forced Oscillations and Resonance

Also, Read NCERT Solution subject wise -

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## What will students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 14?

• Students will be able to gain knowledge about oscillations, it’s types and properties such as amplitude, frequency and time with the help of suitable diagrams and graphs that give a deeper understanding of a concept.
• NCERT Exemplar Class 11 Physics chapter 14 solutions will also enlighten the students about other important topics related to waves travelling in different directions as well as their equations.
• One can also learn about vibrations in open and closed pipes and their practical applications.
• The students can refer to Class 11 Physics NCERT Exemplar solutions chapter 14 for any tricky or difficult questions they may come across while solving the exercises.

## NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

 Chapter 2 Units and Measurement Chapter 3 Motion in a Straight Line Chapter 4 Motion in a Plane Chapter 5 Laws of Motion Chapter 6 Work, Energy, and Power Chapter 7 Systems of Particles and Rotational Motion Chapter 8 Gravitation Chapter 9 Mechanical Properties of Solids Chapter 10 Mechanical Properties of Fluids Chapter 11 Thermal Properties of Matter Chapter 12 Thermodynamics Chapter 13 Kinetic Theory Chapter 15 Waves

## Important Topics To Cover for Exams From NCERT Exemplar Class 11 Physics Solutions Chapter 14

• NCERT Exemplar Solutions for Class 11 Physics chapter 14 introduces new principles like the Doppler Effect and its special cases and numerical related to it for different positions of source and listener.
• The students get to learn about the different characteristics and also about simple pendulum and its related formulae, that should be well known while appearing for any examination.
• NCERT Exemplar Class 11 Physics solutions chapter 14 also gives detailed information about the oscillations of vertical and horizontal spring with the help of diagrams to explain its concepts and derived formulae.

## NCERT Exemplar Class 11 Solutions

 NCERT Exemplar Class 11 Mathematics Solutions NCERT Exemplar Class 11 Chemistry Solutions NCERT Exemplar Class 11 Physics Solutions NCERT Exemplar Class 11 Biology Solutions

### Check Class 11 Physics Chapter-wise Solutions

 Chapter 1 Physical world Chapter 2 Units and Measurement Chapter 3 Motion in a straight line Chapter 4 Motion in a Plane Chapter 5 Laws of Motion Chapter 6 Work, Energy and Power Chapter 7 System of Particles and Rotational motion Chapter 8 Gravitation Chapter 9 Mechanical Properties of Solids Chapter 10 Mechanical Properties of Fluids Chapter 11 Thermal Properties of Matter Chapter 12 Thermodynamics Chapter 13 Kinetic Theory Chapter 14 Oscillations Chapter 15 Waves

### Also Check NCERT Books and NCERT Syllabus here

1. Where do we find the answers to exemplar exercise questions of Class 11 Physics chapter oscillations?

In this, NCERT Exemplar Class 11 Physics solutions Chapter 14 Oscillations, students can find the answers of all the exemplar questions of chapter 14.

2. What kind of questions are asked in examinations regarding Class 11 Physics topic?

The question can be of MCQ, short answer or long answer type. All these types of questions are given after each chapter, the students need to go through them once before appearing for the examination.

3. What all topics are covered in NCERT Exemplar Class 11 Physics solutions Chapter 14?

Topics such as oscillations, it’s types, SHM and its characteristic features are covered in it.

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Exam Date:26 September,2024 - 26 September,2024

Application Date:30 September,2024 - 30 September,2024

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9