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

Q: 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

Q: 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.

Q: Is studying from the solutions reliable?

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

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

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.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | ALLEN

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

NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ I

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.
$Y \alpha \frac{1}{\Delta T}$
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 shear
The strain produced is shearing and longitudinal in this case as the spring when stretched by a load, the shape and length changes.

Question:9.5

A rigid bar of mass M is supported symmetrically by three wires each of length l . Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to
$(a)\frac{Y_{copper}}{Y_{iron}}$
$(b)\sqrt{\frac{Y_{iron}}{Y_{copper}}}$
$(c)\frac{Y^{2}_{iron}}{Y^{2}_{copper}}$
$(d)\frac{Y_{iron}}{Y_{copper}}$

Answer:

The answer is the option (b)
Y = stress/strain
$\frac{FL}{A\Delta L} = \frac{4FL}{\pi D^{2}\Delta L}$ $\frac{FL}{A}\Delta L = \frac{4FL}{\pi D^{2}\Delta L}$
$\Delta L$ copper = $\Delta L$ iron, F is same in both cases
Hence, $Y \propto \frac{1}{D^{2}}$
So, $\frac{D_{copper}}{D_{iron}}$ = $\sqrt{\frac{Y_{iron}}{Y_{copper}}}$ Hence, option d is correct.

Question:9.6

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars Figure mass m is suspended from the mid point of the wire. Strain in the wire is

$(a)\frac{x^{2}}{2L^{2}}$
$(b)\frac{x}{L}$
$(c)\frac{x^{2}}{L}$
$(d) \frac{x^{2}}{2L}$

Answer:

$\Delta L = (AO + BO - AB)$
$\Delta L = 2[AO - L]$
$\Delta L = 2 (L^{2} + x^{2})^{\frac{1}{2}}- 1)$

$\Delta L = 2L [1 +\frac{x^{2}}{2L^{2}} - 1] = \frac{x^{2}}{L}$
Strain = $\frac{\Delta L}{2L} = \frac{x^{2}}{2L^{2}}$ . Hence option a is correct.

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)
eer3e2

Let us have a look at the free body diagram for the scenario.
The net forces will amount to zero.
Vertical components
$2T sin \theta - mg =0$
$T =\frac{mg}{2\sin \theta }$
$T \propto \frac{1}{\sin \theta}$

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, $\frac{\Delta L}{L}$ (rubber) will be larger for the same $\frac{F}{A}$ . in the case of steel $\Delta L$ is insignificant. But, in rubber $\Delta L$ is significant as the shape changes and hence d is the correct option.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ II

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 $\frac{F}{A}$
b) tensile stress at any cross-section is zero
c) tensile stress at any cross-section A of the wire is $2\frac{F}{A}$
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 = $\frac{F}{A}$ 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 mm² and 2.0 mm² 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)

$T_b x = T_a (l-x)$
$\frac{T_{b}}{T_{a}} = (\frac{l}{x}-1)$ ------------------(1)
For wire A, stress
$=\frac{T_{a}}{T_{b}}$
For wire B, stress
$\frac{T_{b}}{A_{b}}=\frac{T_{b}}{2A_{a}}$
Since stress on steel = stress on Al
$\frac{T_{a}}{A_{a}}=\frac{T_{b}}{2A_{a}}\rightarrow T_{a}=\frac{T_{b}}{2}$
$\frac{T_{b}}{T_{a}}=2$
$\frac{l}{x}-1 = \frac{2}{1}$
$x = \frac{l}{3}$ from point b
so, distance from A = $l-x = l-\frac{l}{3} = \frac{2l}{3}$
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)
$\frac{S_{a}}{Y_{a}} =\frac{S_{b}}{Y_{b}}$
$Y steel / Y Al =$ $(\frac{T_a}{T_{b}}) (\frac{A_{b}}{A_{a}}) = (\frac{x}{l}-x) (\frac{2Aa}{Aa})$
$\frac{200\times10^{9}}{70\times10^{9}}=\frac{2x}{l-x}$
$14 x = 20l - 20x$
$34x = 20l$
$x = \frac{20l}{34} = \frac{10l}{17}$
$(l-x) = \frac{7}{17}$
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 = $\frac{-p(V)}{\Delta V}$
$\Delta V = 0$ and K = infinite for an ideal liquid.
No tangential forces act on this liquid, the shearing strain Δθ =0 and F =0
$n = \frac{\frac{F}{A}}{\Delta \theta} = \frac{0}{0}$ = 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 =
$\frac{F}{A}$
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 $\propto \frac{1}{Y}$ (steel) and Strain $\propto \frac{1}{Y}$ (Al)
Strain (steel)/ Strain (Al) = Y (Al) / Y(steel)
Now, Y(Al) < Y (Steel). Hence, strain(steel) < Strain (Al)
Hence, option d is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: Very Short Answer

Question:9.14

The Young’s modulus for steel is more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

Answer:

Y = stress/strain and Y $\propto$ stress
Y (steel)/ Y(rubber) = Stress(steel)/Stress(rubber)
We know that Y steel > Y rubber
Y steel/Y rubber > 1
Hence, stress (steel)/stress (rubber) > 1
So, stress (steel) > stress (rubber)

Question:9.15

Is stress a vector quantity?

Answer:

Stress = force/area
The directions of restoring and deforming forces are equal and opposite and hence there is no net direction of force. So, stress is not a vector quantity.

Question:9.16

Identical springs of steel and copper are equally stretched. On which, more work will have to be to done?

Answer:

$\frac{\Delta L_{1}}{\Delta L_{2}} =\frac{Ycu}{Ys}$
Work done $(WD) = F. \Delta L$
WD steel/ WD copper = Y cu/ Ys
Ycu/Ys < 1, so WD steel < WD copper

Question:9.17

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

Answer:

Young’s modulus
$Y=\frac{FL}{A\Delta L}$
Since the body is rigid, it cannot be deformed or reshaped.
Hence, $\Delta L = 0$
Y (rigid body) = $\frac{FL}{0}$ = infinite

Question:9.18

What is the Bulk modulus for a perfect rigid body?

Answer:

Bulk modulus = $\frac{-p(V)}{\Delta V}$
Since the body is rigid, it cannot be deformed or reshaped and cannot be stretched.
Hence, $\Delta V = 0$
B (rigid body) = $\frac{pV}{0}$ = infinite.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: Short Answer

Question:9.19

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of the same material of length 2L and radius 2r is pulled by a force 2f. Find the increase in length of this wire.

Answer:

$\frac{l}{\Delta L_2}=\left (\frac{\frac{F_{2}L_{2}}{A_{2}Y_{2}}}{\frac{F_{1}L_{1}}{A_{1}Y_{1}}} \right )=\frac{2f2L}{fL}\times \pi r^{2}\times \frac{Y}{4\pi r^{2}\times Y}$
$\frac{\Delta L_{2}}{l}=\frac{4}{4}= 1$
Hence, $\Delta L_2 = l$ , which means the change of length in the second wire is the same.

Question:9.20

A steel rod of length 1 m and area of cross-section 1 cm² is heated from $0^{\circ}C$ to $200^{\circ}C$ without being allowed to extend or bend. What is the tension produced in the rod?

Answer:

$L_t = L (1 + \alpha \Delta t)$
$\Delta L = 1 \times 10^{-5 }\times 200$
$Y = \frac{FL}{A\Delta L} (L = 1m, A = 0.0001 m^{2})$
$Y = 2 \times 10^{11 }N/m^{2}$
$F = 2 \times 10^{11} \times 10^{-5} \times 200 \times \frac{0.0001}{1} = 4 \times 10^{4} N$

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 = $9.8 \times 108 N/m^{2}$
p = density of water = 1000 kg/m³
percentage change in volume = $\frac{\Delta V}{V} \times 100 = 0.1$
let h be the depth to which the rubber ball is being taken = hpg
$p = B \times \frac{\Delta V}{V}$
$h \times 9.8 \times 10^3 = 9.8 \times 10^8 \times \frac{1}{1000}$
hence, h=100 m

Question:9.22

A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched?

Answer:

The cable length is 9.1 m, r = 5 mm,
Tension in cable = 800 N
$Y = 2 \times 10^{11} N/m^{2}$
$\Delta L =\frac{FL}{AY} = 800 \times \frac{9.10}{3.14\times 10 ^{-6}\times 25 \times 2 \times 10^{11}}$
$\Delta L = 4.64 \times 10^{-5} m$

Question:9.23

Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the flood and why?

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.

NCERT Exemplar Class 11 Physics Solutions Chapter 9: Long Answer

Question:9.24

Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar. Consider a plane making an angle $\theta$ with the length. What are the tensile and shearing stresses on this planet?

a) for what angle is the tensile stress a maximum?
b) for what angle is the shearing stress a maximum?

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
$\sin \theta = \frac{A}{A^{'}}$

The force perpendicular component along A or $AA^{'}=\sin \theta$
Tensile stress = $\left ( \frac{F\sin \theta}{A} \right ) \times \sin \theta$
For max value, sin θ should be equal to 1. Hence, $\theta =\frac{\pi}{2}$
b) Shearing stress = forces along the plane F/ area = $F\frac{\cos \theta}{A^{'}}=\left ( \frac{F}{2A} \right )\times 2 \sin \theta \cos \theta = \left ( \frac{F}{2A} \right )\sin 2\theta$
for shearing stress to be maximum, $\sin 2 \theta =1$
which gives us $\theta =\frac{\pi}{4}$

Question:9.25

a) A steel wire of mass μ per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg/m³.
b) If the yield strength of steel is 2.5 × 10⁸ N/m², what is the maximum weight that can be hung at the lower end of the wire?

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.
$dm = u.dx, r=0.1 cm = 0.001 m$
$A = \pi r^{2} = \pi (10)^{-6}$
M = 25 kg, L =10m

a) The force that acts in the downward direction on dx = wt
$T(x) = (x.u) g + Mg$
$Y = T\frac{\frac{x}{A}}{\frac{dr}{dx}}$
dr is the increase in the length of wire
$Y = \frac{T(x). dx}{A. dr}$
$dr = \frac{1}{AY} (xug + Mg) dx$
integrating both sides we get,
r (change in length of wire) = 1/AY [ug x2/g + Mgx] L0
extension in wire of length L =
$\frac{gL}{2 \pi r^{2}Y}[ul + 2M]$
extension=
$10\times 10 [0.25 + 2 \times 0.25] / (2\times 3.14 \times 10^{-6} \times 2 \times 10^{11})$
extension = $400 \times 10^{-5}$ m mass m = volume x density =
$(A.L) \times P = \pi r^{2} \times 10 \times 7860 = 0.25 kg$
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
$= 250 \times 3.14$
Max tension = yield force
$(m + M) g = 250 \times 3.14$
$M = \frac{785}{10} - 0.25 = 78.35 Kg$

Question:9.26

A steel rod of length 2l, cross-sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod.

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, $dr = dm rw^{2}$
Hence, $-dT = dm rw^{2}$
If u is the mass per unit length, we have
$-dT = u w^{2}r.dr$
On integrating both sides we get,
$-\int_{0}^{T}=dT=uw^{2}\int_{r}^{1}r.dr$
tension in rod,
$-T(r)=\frac{uw^{2}}{2}(l^{2}-r^{2})$ --------------------------(1)
Y = stress/strain =
$\frac{\frac{T(r)}{A}}{\frac{\delta r}{dr}}$
${\frac{\delta r}{dr}}=\frac{T(r)}{AY}=\frac{uw^{2}}{2AY}\left ( l^{2} -r^{2}\right )dr$
$\int_{0}^{\delta }\delta r=\int_{0}^{l}\frac{(uw^{2})}{2AY}(l^{2}-r^{2})dr$
$\delta =\left ( \frac{uw^{2}}{2AY} \right )\left ( \frac{2}{3} \right )l^{3}$
$\delta =\frac{uw^{2}}{3AY}(l^{3})$ Since the extension is on both sides, total extension =
$2\delta =2\frac{uw^{2}}{3AY}(l^{3})$

Question:9.27

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that the temperature of each rod increases by $\Delta$ T. Find the change in the angle ABC.

Answer:

$\cos \theta = \frac{l_1^{2}+l_3^{2}-l_2^{2}}{2l_1l_3}$

$2l_1l_3\cos \theta =l_1^{2}+l_3^{2}-l_2^{2}$

Differentiating on both sides of the equation,
$2[d (l_1 l_3) \cos \theta + l_1 l_3 d (\cos \theta)] = 2 l_1 dl_1 + 2 l_3 dl_3 - 2 l_2 dl_2$
$(l_1 dl_3 + l_3 dl_1) \cos \theta - l_1 l_3 \cos \theta d\theta = l_1 dl_1 + l_3 dl_3 - l_2 dl_2$ --------------(1)
$dl_1 = l_1 \alpha_ 1 \Delta t$
$dl_2 = l_2 \alpha_ 2 \Delta t$
$dl_3 = l_3 \alpha _3 \Delta t$
L1 = L2 = L3
$\\dl_1=dl_3=l\alpha_1\Delta t$ also $dl_2 = l\alpha _2 \Delta t$
substituting in 1, we get
$\cos \theta (l^{ 2} \alpha_ 1 \Delta t + l^{2} \alpha_ 1 \Delta t) - l^{2} \sin \theta d\theta = l^{2} \alpha_1 \Delta t + l^{2} \alpha_1 \Delta t - l^{2} \alpha_2 \Delta t$
$2 \alpha _1 \Delta t \times \frac{1}{2}- 2 \alpha_ 1 \Delta t + \alpha_ 2 \Delta t = \frac{\sqrt{3}}{2}d \theta$
$d\theta = 2 (\alpha 2-\alpha 1) \frac{\Delta t }{\sqrt{3}}$

Question:9.28

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{\pi r^{4}}{4R}$ . Y is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Answer:

Let us consider triangle ABC which is a right-angled triangle. We use Pythagoras theorem here.
$R^{2}=(R-d)^{2}+\left (\frac{h}{2} \right )^{2}$
$2Rd=\frac{h^{2}}{4}$
$d=\frac{h^{2}}{8R}$ ------------(1)
we assume W to be the weight of trunk per unit volume.
Weight of trunk = total vol. x W = $\pi r^{2}h\times W$
Torque exerted by bending the trunk, T = $\pi r^{2}h\times W \times d$
Given value of torque = $\frac{\pi r^{4}Y}{4R}$
Equating both values we get,
$\pi r^{2}h \times W \times \frac{h^{2}}{8R}=\frac{\pi r^{4}Y}{4R}$
$h^{3} = \pi r^{4}Y \times \frac{8R}{4R\pi r^{2}W}$
$h^{3} = [\frac{2Y}{W}]^{\frac{1}{3}}r^{2/3}$

Question:9.29

A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially, the stone is at the same level as the point P. The stone is dropped vertically from point P.
a) find the distance y from the top when the mass comes to rest for an instant, for the first time
b) what is the maximum velocity attained by the stone in this drop?
c) what shall be the nature of the motion after the stone has reached its lowest point?

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) $mgy = \frac{1}{2}K (y-L)^{2}$
$mgy = \frac{1}{2}K (y^{2}+L^{2}-2yL)$
$2mgy = Ky^{2} - 2KyL + KL^{2}$
$Ky^{2} - 2(KL+mg) y + KL^{2} = 0$

$Y=\frac{[KL+mg ]}{K}\pm \frac{\sqrt{mg(mg+2KL)}}{K}$

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’
$\frac{1}{2}mv2 + \frac{1}{2} Kx^{2} = mg (L+x)$
$mv^{2} + Kx^{2} = 2mg(L+x)$
$x = \frac{mg}{K}$
$mv^{2} = 2mgL + \frac{2m^{2}g^{2}}{K}-\frac{m^{2}g^{2}}{K}$
$v = [2gL + \frac{mg^{2}}{K}]^{\frac{1}{2}}$

c) From part a, at the lowest point
$m \frac{d^{2}z}{dt^{2}} = mg - K(y-L)$
$\frac{d^{2}z}{dt^{2}} - g +\frac{K}{m(y-L)}= 0$
$z=\left [ \left ( y-L \right ) -\frac{mg}{K}\right ]$ {through transformation of variables}
$\frac{d^{2}z}{dt^{2}} + \frac{k}{m} (z) = 0$
This is a second order differential equation of the kind of simple harmonic motion.
w is the angular frequency, $w=\sqrt{\frac{k}{m}}$
The solution of such differential equation is of type $z = A \cos (wt + \theta )$
$z= \left ( L+\frac{K}{m}g \right )+A^{'}\cos\left ( wt+\theta \right )$
Hence, the stone undergoes simple harmonic motion about point y=0,
Hence, $z_{0} = \left (L + \frac{mg}{K} \right )$

Also, check NCERT Solutions for Class 11 other subjects

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

Main Subtopics in NCERT Exemplar Class 11 physics Solutions Chapter 9:

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

What Will Students Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 9?

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.

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 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

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

Important Topics To Cover For Exams From NCERT Exemplar Class 11 Physics Solutions Chapter 9

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 -

Also, check NCERT Notes subject wise -

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

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

Frequently Asked Questions (FAQs)

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.

Also Read

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

Hydrogen Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 9 Notes - Download PDF

The S-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 10 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

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

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

Articles

How to Make APAAR ID via DigiLocker and School: A Complete Guide

Dec 17, 2024

Sainik School Entrance Exam 2025: Application, Exam Date, How to Apply, Syllabus

Dec 17, 2024

MVPP Scholarship Exam 2024: Mukhyamantri Vigyan Pratibha Pariksha Registration, Exam Date

Dec 16, 2024

Silverzone ISSO Answer Key 2024-25: Download PDF

Dec 16, 2024

Silverzone IOEL Answer Key 2024 – Download International Olympiad of English Language Solutions

Dec 16, 2024

Silverzone Olympiad IOM Answer Key 2024-25: Qualifying Marks, Score

Dec 16, 2024

Silverzone STEM Olympiad Answer Key 2024-25: Check Answer Key for all Classes

Dec 16, 2024

Silverzone ABHO Answer Key 2024 | Download Akhil Bhartiya Hindi Olympiad Solutions

Dec 16, 2024

Upcoming School Exams

National Institute of Open Schooling 10th examination

Application Date:21 November,2024 - 20 December,2024

Application Process

Exam Pattern

Mock Test

National Institute of Open Schooling 12th Examination

Application Date:21 November,2024 - 20 December,2024

Application Process

Exam Pattern

Admit Card

National Means Cum-Merit Scholarship

Admit Card Date:27 November,2024 - 22 December,2024

Eligibility Criteria

Application Process

Exam Pattern

West Bengal Board 12th Examination

Exam Date:02 December,2024 - 20 December,2024

Result

Exam Pattern

Admit Card

Gujarat Board Secondary School Certificate Examination

Late Fee Application Date:11 December,2024 - 20 December,2024

Exam Pattern

Admit Card

Result

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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

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

Check Class 11 Physics Chapter-wise Solutions

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

National Institute of Open Schooling 10th examination

National Institute of Open Schooling 12th Examination

National Means Cum-Merit Scholarship

West Bengal Board 12th Examination

Gujarat Board Secondary School Certificate Examination

Popular Questions

Colleges After 12th

Popular Course After 12th

Applications for Admissions are open.

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

TOEFL ® Registrations 2024

Pearson | PTE

JEE Main high scoring chapters and topics

JEE Main Important Mathematics Formulas

Explore on Careers360

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

Schools By Medium

By Ownership

By City

By State

NVS

JNVST

Download Careers360 App

All this at the convenience of your phone

Scan and download the app