NCERT Exemplar Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids
Chapter 9 NCERT Exemplar Class 11 Physics is devoted to the mechanical behaviour of solids and defines the reaction of materials to external forces. Some of the concepts presented in the chapter include stress, strain, elasticity and plasticity. Elasticity is the capacity of a material to resume its original shape after being stretched, and plasticity is the property that enables a material to experience irreversible material deformation. A real-life scenario is extending a bungee cord when one is jumping- the cord becomes longer because of the force that was used, and then goes back to its shape, which does not allow the person to land on the ground.
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
- NCERT Exemplar Class 11 Physics Solutions Chapter 9: Important Concepts and Formulas
- Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Class 11th Solutions
The NCERT Exemplar Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids present students with simple, but clear explanations on how the solids react to forces, giving students an easy time to comprehend the reactions of the solids to forces. These properly organised NCERT Exemplar Solutions assist students to understand such concepts as the Young modulus, shear modulus and stress-strain relationships. The ability to learn the mechanical properties of solids creates an excellent conceptual activity and forms the necessary problem-solving skills. These NCERT Exemplar Class 11 Solutions Physics Chapter 9 Mechanical Properties of Solids are very helpful in board tests and competitive tests like JEE Main and NEET, which assist students to become more accurate and confident in working with numerical and theoretical problems.
NCERT Exemplar Class 11 Physics Solutions Chapter 9: MCQ I
Section MCQ I - NCERT Exemplar Class 11 Physics Chapter 9 serves to reinforce conceptual ideas in the students about elasticity, stress, strain, and the modulus of elasticity (Young). These objective questions will be aimed at testing the basic ideas in a very clear and accurate way. Taking these MCQs using NCERT Exemplar Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids would help the students build confidence in board examinations and competitive examinations.
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 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 the 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 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 change.
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 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 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
$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 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 middle 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
Multiple Choice Questions (MCQ II) NCERT Exemplar Class 11 Physics Chapter 9 will be used to examine your comprehension level and formula usage of stress, strain, elasticity, Young's modulus, shear modulus, etc. Such NCERT Exemplar Class 11 Physics Solutions Chapter 9 can offer step-by-step solutions to MCQs, which can ensure students reinforce their problem-solving abilities and prepare well to take a board examination and competitive examination, such as the JEE and NEET.
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 answers are options (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 answers are options (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 answers are options (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 position. 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 answers are options (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 answers are options (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 the stresses are the 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
The VSA questions of Chapter 9, Mechanical properties of Solid,s assist students in testing the main concepts of stress, strain, modulus of elasticity and tensile strength. These short questions enhance conceptual knowledge and memorisation. The specific and clear solutions to quick revision are presented below.
Question:9.14
Answer:
Y = stress/strain and Y $\propto$ stressY (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
Answer:
Stress = force/areaThe 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
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
Short answer questions of the Mechanical Properties of Solids Class 11 NCERT Exemplar are used to help students know how solids behave under the forces imposed on them in a conceptual and very specific manner. These questions are questions of reasoning and application of formula, which develops sound problem solving in competition and board exams.
Question:9.19
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
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
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
Answer:
Both balls are dropped from the same height. This means the velocity with which they approach and strike the ground will be the 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 of energy transferred to the ivory ball is higher than that of the clay ball, and as a result, it rises higher.NCERT Exemplar Class 11 Physics Solutions Chapter 9: Long Answer
The questions given in the Mechanical Properties of Solids Class 11 NCERT Exemplar in the form of long answers are used to guarantee a better insight into the mechanical properties of solids. These questions will determine the skill of students in making real-life applications to the phenomena of stress, strain, Young's modulus and elasticity. With the solution of such long-answer questions, the students are not only able to sharpen their conceptual clarity, but they can dalso evelop their exam-oriented analytical ability.
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 the 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 the maximum 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
Answer:
An element dx is of mass dm taken from a wire of length L at 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 the 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
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 the 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
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
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 the 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
Answer:
The string is of length L, and the 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 the 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 balanced by the force of gravity, which makes mg = Kx
If we assume v to be equal to the 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 the point y=0,
Hence, $z_{0} = \left (L + \frac{mg}{K} \right )$
NCERT Exemplar Class 11 Physics Solutions Chapter 9: Important Concepts and Formulas
Solving real-life and numerical-based physics problems would require a clear understanding of the mechanical behaviour of solids. This chapter dwells on elasticity, stress, strain, Young's modulus, bulk modulus and shear modulus, among other concepts often examined in competitive examinations and school tests. The most essential formulas and concepts that are discussed in this chapter are as follows:
1. Stress (σ):
Stress is the restoring force per unit area applied to a body.
\begin{aligned}
&\sigma=\frac{F}{A}\\
&\text { Where } F=\text { applied force and } A=\text { area. }
\end{aligned}
2. Strain (ε):
Strain is the relative change in shape or size of the body.
$
\epsilon=\frac{\Delta L}{L}
$
It has no unit since it is a ratio.
3. Hooke’s Law:
For small deformations, stress is directly proportional to strain.
$
\sigma \propto \epsilon \Rightarrow \sigma=E \epsilon
$
4. Young’s Modulus (E):
It describes longitudinal elasticity when the force is applied along the length.
$
E=\frac{\sigma}{\epsilon}=\frac{F / A}{\Delta L / L}
$
5. Bulk Modulus (K):
It measures how the volume changes under uniform pressure.
$
K=-\frac{P}{\Delta V / V}
$
6. Shear Modulus (η or G):
It expresses resistance to shearing deformation.
$
\eta=\frac{\text { shear stress }}{\text { shear strain }}
$
7. Poisson’s Ratio:
Ratio of lateral strain to longitudinal strain.
$
\text{Poisson's ratio}=-\frac{\epsilon_{\text {lateral }}}{\epsilon_{\text {longitudinal }}}
$
8. Elastic Energy Stored in a Body (U):
$U=\frac{1}{2} \sigma \epsilon$
9. Stress-Strain Curve: Demonstrates elastic region, yield point, plastic region, and breaking point. Useful for identifying elastic behaviour.
10. Elastic Limit: The Maximum stress a solid can bear and still return to its original size once the force is removed.
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids
To master Class 11 Physics, it is necessary to know the mechanical behaviour of solids. The NCERT Exemplar Class 11 Physics Chapter 9 Solutions offer detailed explanations and proper problem-solving methods. Such NCERT Exemplar Class 11 Physics Chapter 9 Solutions lead students to reinforce the concepts of elasticity, stress, strain, and Young's modulus, among others, by engaging in systematic and exam-oriented practice.
- The solutions assist the students to develop a powerful conceptual base through explaining stress, strain, elasticity and the modulus of Elasticity of materials in simple and understandable language, giving clear reasoning.
- They are composed of a broad collection of solved problems, which help students to be confident about solving numerical and other theoretical problems in school tests and competitive examinations.
- These model solutions assist in enhancing analytical ability by offering problems of different levels of application, and in this way, the students can solve difficult problems.
- The step-by-step solutions provide better problem-solving methods and clarity to assist the students in avoiding typical errors in the process of examination.
- The solutions assist students in revising key formulas and concepts, which assist them in last-minute preparations.
- They address the whole chapter comprehensively without leaving out any vital topics in preparing the syllabus of CBSE Board tests and entrance tests.
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
The NCERT Exemplar Class 11 Physics solutions are organised into chapters to simplify and make learning easy and effective. Solutions to any chapter can be accessed easily by students, and they are able to revise and clarify their concepts. These are very useful in Board exams of CBSE and competitive examinations such as JEE and NEET.
NCERT Exemplar Class 11th Solutions
- NCERT Exemplar for Class 11 Physics
- NCERT Exemplar for Class 11 Maths
- NCERT Exemplar for Class 11 Biology
- NCERT Exemplar for Class 11 Chemistry
Check Class 11 Physics Chapter-wise Solutions
Also, Read NCERT Solution subject-wise -
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology
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
Also, Check NCERT Books and NCERT Syllabus here
Frequently Asked Questions (FAQs)
Q: Are NCERT Exemplar solutions effective in competitive exams?
A:
Absolutely. JEE and NEET test a lot of questions related to the fundamental principles of elasticity and mechanical properties. Exemplar problems practice provides a benefit.
Q: How many are the formula questions to be expected of this chapter?
A:
Many of the questions are formula-based since students are often examined on the relationship of the elastic modulus, stress-strain curve and Hooke's law.
Q: What will be the benefits of studying from the Class 11 Physics NCERT Exemplar solutions Chapter 9?
A:
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 it necessary to solve the numerical problems, in this chapter?
A:
Yes, mathematical questions of Young's modulus, bulk modulus and stress-strain relationship often appear in exams, answering them helps gain better accuracy and understanding.
Q: What is the significance of NCERT Exemplar solutions to this chapter?
A:
The solutions provide stepwise discussion, which will enable the students to reinforce concepts, bust myths, and acquire problem-solving techniques utilised in competitive examinations such as JEE Main and NEET.
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