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Edited By Safeer PP | Updated on Aug 08, 2022 05:09 PM IST

The chapter revolves around all the mechanical properties of the solid particles. NCERT Exemplar Class 11 Physics solutions chapter 9, explains in brief about all the mechanical properties any solid would contain. Elasticity is the qualitative measure to stretch an object, and plasticity is termed as the property of a solid to regain its original shape. A team of experts makes NCERT Exemplar Class 11 Physics solutions chapter 9 that helps the students gain knowledge by easy explanation and concept clearance. If any student wants to prepare for competitive exams such as JEE Main or NEET, the NCERT Exemplar Class 11 Physics solutions chapter 9, will help the student perform very well in the examinations.

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This Story also Contains

- NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ I
- NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ II
- NCERT Exemplar Class 11 Physics Solutions Chapter 9: Very Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 9: Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 9: Long Answer
- Main Subtopics in NCERT Exemplar Class 11 physics Solutions Chapter 9:
- What Will Students Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 9?
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Class 11 Solutions
- Important Topics To Cover For Exams From NCERT Exemplar Class 11 Physics Solutions Chapter 9

Question:9.1

Modulus of rigidity of ideal liquid is

a) infinity

b) zero

c) unity

d) some finite small non-zero constant value

Answer:

The answer is the option (b) zero.Question:9.2

The maximum load a wire can withstand without breaking when its length is reduced to half of its original length, will

a) be doubled

b) be half

c) be four times

d) remain same

Answer:

The answer is the option (d) remains same.Stress = force/area

The breaking stress is not dependent on the length, and hence if the cross-sectional area is changed, it will not affect the breaking force.

Question:9.3

The temperature of a wire is doubled. The Young’s modulus of elasticity

a) will also double

b) will become four times

c) will remain same

d) will decrease

Answer:

The answer is the option (d) will decrease.So as the young’s modulus increases, elasticity decreases.

Question:9.4

A spring is stretched by applying a load to its free end. The strain produced in the spring is

a) volumetric

b) shear

c) longitudinal and shear

d) longitudinal

Answer:

The answer is the option (c) longitudinal and shearThe strain produced is shearing and longitudinal in this case as the spring when stretched by a load, the shape and length changes.

Question:9.7

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports. It can be done in one of the following three ways:

The tension in the strings will be

a) the same in all cases

b) least in a)

c) least in b)

d) least in c)

Answer:

The answer is the option (c) least in b)Let us have a look at the free body diagram for the scenario.

The net forces will amount to zero.

Vertical components

Hence tension in all the cases is different.

And, option c is the correct choice.

Question:9.8

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods.

a) both the rods will elongate but there shall be no perceptible change in shape

b) the steel rod will elongate and change shape but the rubber rod will only elongate

c) the steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse

d) the steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre

Answer:

The answer is the option (d) the steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

M is the mass which is being positioned at the mid of the rods made of rubber and steel

Y steel > Y rubber

Hence, (rubber) will be larger for the same . in the case of steel is insignificant. But, in rubber is significant as the shape changes and hence d is the correct option.

Question:9.9

The stress-strain graphs for two materials are shown in Figure (assume same scale).

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle.

(b) Material (i) and (ii) have the same elasticity and the same brittleness.

(c) Material (ii) is elastic over a larger region of strain as compared to (i).

(d) Material (ii) is more brittle than material (i).

Answer:

The answer is the option (c) and (d)If we compare the tensile strength, (ii) > (i)

So (ii) is more elastic as compared to (i) and hence option c is correct.

The fracture point for the material (ii) is closer than that of material (i). hence (ii) is more brittle than (i) and hence, option d is correct.

Question:9.10

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.

a) tensile stress at any cross-section A of the wire is

b) tensile stress at any cross-section is zero

c) tensile stress at any cross-section A of the wire is

d) tension at any cross-section A of the wire is F

Answer:

The answer is the option (a) and (d)We know that stress = and hence option a is correct

Force balances the tension, which means T=F and hence option d is correct.

Question:9.11

Rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths. The cross-sectional areas of wires A and B are 1.0 mm^{2} and 2.0 mm^{2} respectively.

a) mass m should be suspended close to wire A to have equal stresses in both the wires

b) mass m should be suspended close to B to have equal stresses in both the wires

c) mass m should be suspended at the middle of the wires to have equal stresses in both the wires

d) mass m should be suspended close to wire A to have equal strain in both wires

Answer:

The answer is the option (b) and (d)------------------(1)

For wire A, stress

For wire B, stress

Since stress on steel = stress on Al

from point b

so, distance from A =

hence m is closer to B than A. so, option b is correct.

the rod remains in a horizontal and balanced. so, the strain will also be equal.

Strain (A) = Strain (B)

So, for an equal amount of strain, mass m will be closer to A

Hence option d is correct.

Question:9.12

For an ideal liquid

a) the bulk modulus is infinite

b) the bulk modulus is zero

c) the shear modulus is infinite

d) the shear modulus is zero

Answer:

The answer is the option (a) and (d).A property of an ideal liquid is that it is not compressible. (K)Bulk modulus =

and K = infinite for an ideal liquid.

No tangential forces act on this liquid, the shearing strain Δθ =0 and F =0

= indeterminate

hence a and d are true.

Question:9.13

A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1 cm. The two wires will have

a) the same stress

b) different stress

c) the same strain

d) different strain

Answer:

The answer is the option (a) and (d).As we know stress =

The areas of cross-sections of both wires are the same, and the force with which they get stretched is also the same. Hence the total stress for both is the same. Hence option a.

Strain = Stress/Y

As stresses are same on both,

Strain (steel) and Strain (Al)

Strain (steel)/ Strain (Al) = Y (Al) / Y(steel)

Now, Y(Al) < Y (Steel). Hence, strain(steel) < Strain (Al)

Hence, option d is correct.

Question:9.17

What is the Young’s modulus for a perfect rigid body?

Answer:

Young’s modulusSince the body is rigid, it cannot be deformed or reshaped.

Hence,

Y (rigid body) = = infinite

Question:9.18

What is the Bulk modulus for a perfect rigid body?

Answer:

Bulk modulus =

Since the body is rigid, it cannot be deformed or reshaped and cannot be stretched.

Hence,

B (rigid body) = = infinite.

Question:9.21

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%.

Answer:

B = Bulk modulus =p = density of water = 1000 kg/m

percentage change in volume =

let h be the depth to which the rubber ball is being taken = hpg

hence, h=100 m

Question:9.23

Answer:

Both balls are dropped from the same height. This means the velocity with which they approach and strike the ground will be same. But the ivory ball is more elastic in nature than the wet clay ball, so it tries to get back to its original form as quickly as possible as compared to the wet clay ball. Hence, the amount energy transferred to the ivory ball is higher than that of clay ball and as a result it rises higher.Question:9.24

Answer:

a) Tensile stress = normal forces to the surface of plane F/ area

A = area of cross-section which is perpendicular to the bar

A’ = the area of plane cut along aa’ of the cross section

The force perpendicular component along A or

Tensile stress =

For max value, sin θ should be equal to 1. Hence,

b) Shearing stress = forces along the plane F/ area =

for shearing stress to be maximum,

which gives us

Question:9.25

Answer:

An element dx is of mass dm taken from a wire of length L from a distance x. mass per unit length here is u.M = 25 kg, L =10m

a) The force that acts in the downward direction on dx = wt

dr is the increase in the length of wire

integrating both sides we get,

r (change in length of wire) = 1/AY [ug x2/g + Mgx] L0

extension in wire of length L =

extension=

extension = m mass m = volume x density =

b) at x = L, the wire experiences maximum tension

T(x) = ugx + Mg

T(L) = ugL + Mg

T = (m + M) g {as uL = m}

Yield force = Yield strength x A

Max tension = yield force

Question:9.26

Answer:

Dr element of the rod is at a distance of r from the centre. We assume Tr and T(r+dr) to be the tensions at points A and B.Dr has a centrifugal force acting on it equal to T(r+dr) - T(r)

Centrifugal force = -dT in an outward direction.

Centripetal force as a result of the rotation,

Hence,

If u is the mass per unit length, we have

On integrating both sides we get,

tension in rod,

--------------------------(1)

Y = stress/strain =

Since the extension is on both sides, total extension =

Question:9.27

Answer:

Differentiating on both sides of the equation,

--------------(1)

L1 = L2 = L3

also

substituting in 1, we get

Question:9.28

Answer:

Let us consider triangle ABC which is a right-angled triangle. We use Pythagoras theorem here.

------------(1)

we assume W to be the weight of trunk per unit volume.

Weight of trunk = total vol. x W =

Torque exerted by bending the trunk, T =

Given value of torque =

Equating both values we get,

Question:9.29

Answer:

The string is of length L, and stone is tied at P. The nail O is the point where the string is fixed. L is the height through which the stone is lifted. The stone tries to follow a path PP’ when it falls under the effect of gravity. But, due to elastic property of the string, it will follow a circular path from P to Q which gives rise to centrifugal force. The potential energy of the stone is converted to mechanical energy in the string with a spring constant K. PE of stone = Mechanical energy of the string

a)

b) At the lowest point, the acceleration is zero.

So, F = 0

Here the force of the string is balances by the force of gravity which makes, mg = Kx

If we assume v to be equal to maximum velocity then by the law of conservation of energy,

Kinetic energy of stone + Potential energy gained by the string = potential energy lost from P to Q’

c) From part a, at the lowest point

{through transformation of variables}

This is a second order differential equation of the kind of simple harmonic motion.

w is the angular frequency,

The solution of such differential equation is of type

Hence, the stone undergoes simple harmonic motion about point y=0,

Hence,

Also, check NCERT Solutions for Class 11 other subjects

Class 11 Physics NCERT Exemplar solutions Chapter 9 include the following topics:

- 9.1 Introduction
- 9.2 Elastic behaviour of solids
- 9.3 Stress and strain
- 9.4 Hooke’s law
- 9.5 Stress-strain curve
- 9.6 Elastic moduli
- 9.7 Applications of elastic behaviour of materials

Read NCERT Class 11 Physics Solutions

Here the topic might appear small and is relatively easy, but things are appearing easy to have difficult concepts at times. There are various mechanical properties of the solids, which are explained here in the chapter. NCERT Exemplar Class 11 Physics solutions chapter 9, will make the learning of the concepts an easier job. Factors such as elasticity, plasticity, types of stress- compressive, tensile, and longitudinal are mentioned in the chapter. Hooke’s law and Young’s modulus have been explained by practical experiments. The chapter says that deforming force in one direction can produce strains in another direction. Here, stress is said not to be a vector quantity because, unlike force, it cannot move in a specific direction. The difference between stress and strain is, restoring force per unit area and the fractional change in the dimensions, respectively.

It is important that students cover some key topics from this chapter for exams. Here are a few of them:

· NCERT Exemplar Class 11 Physics solutions chapter 9, covers all the important and needed topics in the chapter. It is explained with utter ease and various types of practical explanations.

· Hooke’s law tells us how for small deformations stress is directly proportional to strain for many materials.

· Young’s module is defined as the ratio of tensile (or compressive) stress (s) to the longitudinal strain (e).

· Basic terms of altitude, elasticity, strain, plasticity, stress, etc. are explained in detail.

· The actual meaning of elasticity is not of how long a material stretches but in fact something that extends lesser, in a given load.

**Also, Read NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, check NCERT Notes subject wise -**

- NCERT Notes Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

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1. How many questions are solved in the solution?

A total set of 21 questions are solved in the solution manual. The questions include long answer type, short answer type and MCQs

2. What will be the benefits of studying from the Class 11 Physics NCERT Exemplar solutions Chapter 9?

The benefit of studying from NCERT Exemplar Class 11 Physics Solutions Chapter 9, is that it would help students score better in their JEE Main and NEET examinations.

3. Is studying from the solutions reliable?

The solutions have been made by experts for easy understanding, thus, is a source of reliability.

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