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CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Physics
Mechanical Properties Of Solids
Quiz 12
A wire of density 9$$gm/cm^2$$ is stretched between two clamps 1.00 m apart subjected extension of 0.05 cm. The lowest frequency of transverse vibrations in the wire is
(Assume Young's modulus Y = $$9* 10^10 N/m^2$$)
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35 Hz
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45 Hz
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75 Hz
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90 Hz
Explanation
$$\begin{array}{l} The\, \, mass\, \, per\, \, unit\, \, length, \\ m=\dfrac { M }{ l } =\dfrac { { Al\rho } }{ l } =A\rho \\ ules\, \, of\, \, elasticity \\ y=\dfrac { { \dfrac { T }{ A } } }{ { \dfrac { { \Delta l } }{ l } } } \\ T=\dfrac { { Y\Delta lA } }{ l } \\ Hence\, \, lowest\, \, frequency\, \, of\, \, vibration \\ n=\dfrac { 1 }{ { 2l } } \sqrt { \dfrac { T }{ m } } \\ =\dfrac { 1 }{ { 2l } } \sqrt { \dfrac { { y\left( { \dfrac { { \Delta l } }{ l } } \right) A } }{ { A\rho } } } =\dfrac { 1 }{ { 2l } } \sqrt { \dfrac { { y\Delta l } }{ { l\rho } } } \\ n=\dfrac { 1 }{ { 2\times 1 } } \sqrt { \dfrac { { 9\times { { 10 }^{ 10 } }\times 4.9\times { { 10 }^{ -4 } } } }{ { 1\times 9\times { { 10 }^{ 3 } } } } } \\ =35Hz \end{array}$$
The isothermal bulk modulus of an ideal gas at pressure 'P" is
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P
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yP
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P/2
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P/r
Explanation
$$ \begin{array}{c} \text { for iso thermal brocess of an ideal gas } \\ p v=k \quad \text { (constant) } \end{array} $$
$$ \begin{array}{l} P V=K \quad \text { (constant) } \\ P+V \cdot \frac{d P}{d V}=0 \\ V \frac{d P}{d V}=-P \end{array} $$
$$ \text { Bulk } \operatorname{modulus}(\beta)=-\frac{\theta d P}{\frac{d V}{V}}=-V \frac{d P}{d V}=P $$
The relation between Young's modulus $$(Y)$$, bulk modulus $$(K)$$ and modulus of elasticity $$(n)$$ is
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$$1/y = 1/k = 3/n$$
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$$3/y = 1/n + 1/3k$$
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$$1/y = 3/n + 1/3k$$
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$$1/n = 3/y + 1/3k$$
Explanation
$$ \begin{aligned} \frac{3}{y} &=\frac{1}{n}+\frac{1}{3 \beta} \\ \text { where } & \text { Y- Ycoung modulus } \\ n \text { - rigidity modulus } \\ B &=\text { Bulk modulus } \end{aligned} $$
The ratio of lateral contraction strain to the longitudinal elongation strain of a stretched wire when the volume of the wire remains constant is :
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1
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1/2
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2
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3/2
Explanation
A sphere contracts in volume by 0.01% when taken to the bottom of sea 1 km deep. Find Bulk modulus of the material of the sphere.
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$$1.2\times { 10 }^{ 10 }\quad N/{ m }^{ 2 }$$
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$$9.8\times { 10 }^{ 11 }\quad N/{ m }^{ 2 }$$
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$$9.8\times { 10 }^{ 10 }\quad N/{ m }^{ 2 }$$
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$$9.8\times { 10 }^{ 6 }\quad N/{ m }^{ 2 }$$
Explanation
$$ \begin{array}{l} \text { pressure difference }(d p)=p g h \\ \text { Bulk modulus }(\beta)=-v \frac{d p}{d v}=-\frac{d p}{\left(\frac{d v}{v}\right)} \end{array} $$
$$\begin{aligned} \beta &=\frac{10^{3} \times 9.8 \times 10^{3}}{10^{-1}} \\ \beta &=9.8 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \end{aligned}$$
In an experiment, bras and steel wires of length lm each with areas of cross section $$1 mm^2$$ are used. the wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of $$0.2 mm$$ is :
(Given, the Young's Modulus for steel and brass are respectively, $$120 \times 10^9 N/m^2$$ and $$60 \times 10^9 N/m^2$$)
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$$0.2 \times 10^6 N/m^2$$
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$$4.0 \times 10^6 N/m^2$$
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$$1.8 \times 10^6 N/m^2$$
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$$1.2 \times 10^6 N/m^2$$
Explanation
$$k_1 = \dfrac{y_1 A_1}{\ell_1} = \dfrac{120 \times 10^9 \times A}{1}$$
$$k_2 = \dfrac{y_2 A_2}{\ell_2} = \dfrac{60 \times 10^9 \times A}{1}$$
$$k_{eq} = \dfrac{k_1 k_2}{k_1 \times k_2} = \dfrac{120 \times 60}{180} \times 10^9 \times A$$
$$k_{eq} = 40 \times 10^9 \times A$$
$$F = k_{eq} (x)$$
$$F = (40 \times 10^9) A. (0.2 \times 10^{-3})$$
$$\dfrac{F}{A} = 8 \times 10^6 N/m^2$$
No option is matching.Hence question must be bonus.
$$10^{22}$$ particle each of mass $$10^{-26}Kg$$ are striking perpendicularly on a wall of area $$1m^2$$ with speed $$10^4m/s$$ in $$1sec$$. The pressure on the wall if collisions are perfectly elastic is:
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$$2N/m^2$$
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$$4N/m^2$$
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$$6N/m^2$$
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$$8N/m^2$$
Explanation
$$v=10^4m/s$$
$$m=10^{-26}$$
$$n=10^{22}$$
$$A=1m^2$$
$$\Delta p=2mnv$$
$$\Delta p=2\times 10^{22}\times 10^{-26}\times 10^{4}=2$$
$$P=\dfrac{F}{A}=2N/m^2$$
Two elastic wire $$A \& B$$ having length $$\ell_A = 2 \,m$$ and $$\ell_B = 1.5 \,m$$ and the ratio of young's modules $$Y_A : Y_B$$ is $$7:4$$. If radius of wire $$B (r_B)$$ is $$2 mm$$ then choose the correct value of radius of wire $$A$$. Given that due to application of the same force charge in length in both $$A \& B$$ is same
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$$1.7 \,mm$$
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$$1.9 \,mm$$
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$$2.7 \,mm$$
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$$2 \,mm$$
Explanation
$$Y = \dfrac{F\ell}{\pi r^2 \Delta \ell}$$
$$\dfrac{Y_A}{Y_B} = \dfrac{F\ell_A}{\pi r^2_A \Delta \ell} \times \dfrac{\pi r^2_B \Delta \ell}{F \ell_B}$$
$$\dfrac{7}{4} = \dfrac{2}{1.5} \times \dfrac{2^2}{r^2_A} \Rightarrow r^2_A = \dfrac{4 \times 2 \times 2^2}{1.5 \times 7}$$
$$r_A = 1.7 \,mm$$
For an equal stretching force F, the Young's modulus $$ (Y_s) $$ for steel and rubber $$ (Y_r) $$ are related as
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$$ Y_a = Y_r $$
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$$ Y_s < Y_r $$
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$$ Y_s > Y_r $$
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$$ Y_s \ge Y_r $$
If $$F$$ is the force applied to an elastic bar to produce an extension of $$\Delta\ell$$. The, the energy lost in the process is
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$$F.\Delta \ell$$
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$$\dfrac{F.\Delta \ell}{2}$$
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$$zero$$
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$$\ 3\dfrac{F.\Delta\ell}{2}$$
Explanation
$$\begin{aligned}F_{S}=f & \\\Rightarrow \frac{A Y \Delta l}{l} &=F \\\text { Energy stored } &=\frac{A Y \Delta l^{2}}{2 l} \\&=\frac{A Y \Delta l}{\ell} \times \frac{\Delta l}{2} \\&=\frac{F \Delta l}{2}\end{aligned}$$
The relation between $$ \gamma, \mu $$ and K for a elastic material is
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$$ \dfrac {1}{\mu} = \dfrac {1}{3\gamma} + \dfrac {1}{9K} $$
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$$ \dfrac {1}{K} = \dfrac {1}{3\gamma} + \dfrac {1}{9 \mu} $$
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$$ \dfrac {1}{\gamma} = \dfrac {1}{3K} + \dfrac {1}{9\mu} $$
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$$ \dfrac {1}{\gamma} = \dfrac {1}{3\mu} + \dfrac {1}{9K} $$
Explanation
$$ \begin{aligned} \mu &=\text { Modulus of Rigiditg } \\ k &=\text { Bulk Modulus. } \\ \sigma &=\text { Normal stress } \end{aligned} $$
$$ \begin{aligned} y=3 k(1-2 \sigma) &-(1) \\ y=2 \eta(1+\alpha) &-(2) \\ \end{aligned} $$
$$ \begin{array}{l} \dfrac{y}{3 k}=1-2 \sigma -(3), \dfrac{x}{2 \mu}=1+\sigma -(4) \\ \text { Multiply eq 4 by 2. } \end{array} $$
$$ \dfrac{y}{3k}+\dfrac{y}{\mu}=3-(5) $$
Adding equ" (3) and eqn (5)
$$ \begin{aligned} \frac{y}{3 k}+\dfrac{y}{\mu} &=3 \\ \dfrac{1}{3 k}+\dfrac{1}{\mu} &=\dfrac{3}{y} \\ \dfrac{1}{y}=\dfrac{1}{9 k} &+\dfrac{1}{3\mu} \end{aligned} $$
An elastic string of length $$42sm$$ and cross-sectional area $$10^{-4}m^2$$ is attached between two pegs ar distance of $$6mm$$ as shown in the figure. A particle of mass m is kept at midpoint of string and stretched as shown in figure by $$20cm$$ and release. As the string its natural length, the particle attains a speed of $$20m/s$$. Then young modulus $$Y$$ of string is of order
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$$10^8$$
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$$10^{12}$$
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$$10^6$$
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$$10^4$$
Explanation
$$\dfrac{1}{2}\times Y\times {(\dfrac{\Delta l}{L}) ^2}= \dfrac{1}{2}mv^2$$$$Y\times \dfrac{(0.2)^2}{0.42}\times 10^{4}$$$$= 0.05 \times 400 =\dfrac{1}{2} mv^2$$$$Y=\dfrac{0.05\times 400 \times 0.42}{(0.2)^2\times 10^{4}}$$$$=2.1\times 10^{6 }N/m^2$$
A block of mass $$4kg$$ is suspended from the ceiling with the help of a steel wire of radius $$2mm$$ and negligible mass. Find the stress in the wire.$$(g=\pi^2$$)
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$$4.0\times 10^6N/m^2$$
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$$3.14\times 10^6N/m^3$$
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$$3\times 10^5N/m^3$$
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$$2.0\times 10^6N/m^2$$
Explanation
Stress=$$\dfrac{F}{A}=\dfrac{mg}{A}=\dfrac{4\times \pi^2}{\pi r^2}=\dfrac{4\pi}{r^2}$$=$$\dfrac{4\times 3.1}{4\times10^{-6}}$$
$$=3.14\times10^6N/m^2$$
Select the correct alternative $$(s)$$
a) Elastic forces are always conservative
b) Elastic forces are not always conservative
c) Elastic forces are conservative only when Hooke's law is obeyed
d) Elastic forces may be conservative even when Hooke's law is not obeyed
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$$a,d$$
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$$b,c$$
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$$b,d$$
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$$a,c$$
Explanation
$$ \begin{array}{l} \text { Elastic forces are not aluays conservative } \\ \text { gt is conservative only when Hooke's law is obeyed } \\ \text { because wol Hookes law is obeyed thpto elastic limit. } \\ \text { After elastic limit Elastic force is not conservative. } \\ \text { as it dcesn't regain its original shape. } \end{array} $$
A rubber cord of density d, Youngs modulus Y and length L is suspended vertically. If the cord extends by a length 0.5 L under its own weight, then L is
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$$\dfrac{Y}{2dg}$$
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$$\dfrac{Y}{dg}$$
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$$\dfrac{2Y}{dg}$$
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$$\dfrac{dg}{2Y}$$
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$$\dfrac{dg}{Y}$$
Explanation
Mass of the rod having length $$x$$
$$m'=\dfrac{M}{L}x$$
Now, change of the length in dx is $$\Delta L$$
$$\Delta L=\displaystyle\int^L_O\dfrac{m'gdx}{Ar}$$
$$=\dfrac{1}{Ar}\displaystyle\int^L_O\dfrac{M}{L}xg dx$$
$$=\dfrac{Mg}{ALY}\displaystyle\int^L_Oxdx=\dfrac{Mg}{ALY}\left[\dfrac{x^2}{2}\right]^L_O$$
$$=\dfrac{Mg}{ALY}\times \dfrac{L^2}{2}=\dfrac{MgL}{2AY}$$ .........$$(2)$$
$$\Delta L=0.5L$$ given
and, $$d=\dfrac{mass}{volume}=M=dv$$
so, from $$(1)$$
$$0.5L=\dfrac{dvgL}{2AY}=\dfrac{dALgL}{2AY}$$
$$0.5=\dfrac{dgL}{2Y}$$
$$\dfrac{1}{2}=\dfrac{dgL}{2Y}$$
$$L=\dfrac{Y}{dg}$$ $$\rightarrow B$$
So, the correct option is $$(B)$$
A fixed volume of iron is drawn into a wire of length $$L$$. The extension $$x$$ produced in the wire by a constant force $$F$$. $$F$$ is proportional to ______.
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$$\dfrac{1}{L^2}$$
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$$\dfrac{1}{L}$$
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$$L^2$$
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$$L$$
Explanation
The volume of the material remains the same as the fixed volume is drawn into a wire.
Fixed volume = $$V$$
The volume of the wire is given as:
$$V = A \times L$$
Here, $$A$$ is the area of cross-section of wire and $$L$$ is the length of wire.
$$\Delta l = x \rightarrow $$ extension
From the stress-strain relation:
$$\dfrac{F}{A} = Y \dfrac{\Delta l}{L}$$
$$\Delta l = \dfrac{F}{Y} \times \dfrac{L}{V} \times L$$
$$x = \dfrac{FL^2}{YV}$$
$$x \, \propto \, L^2$$
Option (c) correct.
The graph shows the extension of a sample of a type of rubber as different loads F are applied and then gradually removed.
What is the best estimate of the strain energy in the rubber when a load of $$80$$ N is applied?
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$$0.40$$ J
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$$0.64$$ J
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$$0.88$$ J
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$$1.3$$ J
To determine the Young modulus of a wire, several measurements are taken.
In which row can the measurement not be taken directly with the stated apparatus?
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measurement : area of cross-section of wire ; apparatus : micrometer screw gauge
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measurement : extension of wire ; apparatus : vernier scale
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measurement : mass of load applied to wire ; apparatus : electronic balance
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measurement : original length of wire ; apparatus : metre rule
Explanation
Measurement of area of any object is not a proper way of measuring .
Instead of measuring area , one should have to measure the diameter.
So, option A is correct answer.
The amount of water in slug containing by a cylindrical vessel of length $$10$$cm and cross-sectional radius $$5$$cm is?(The density of water is $$1000$$ $$kg/m^3$$):
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$$157\times 10^{-3}$$ slug
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$$53.76\times 10^{-3}$$ slug
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$$10.7$$ slug
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$$14.6$$ slug
Which of the following statement related to stress-strain relation is correct?
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Stress is linearly proportional to strain irrespective of the magnitude of the strain
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Stress is linearly proportional to strain above
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Stress is linearly proportional to strain for stress much smaller than at the yield point
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Stress-strain curve is same for all materials
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Stress is inversely proportional to strain
Explanation
Stress is linearly proportional to strain for stress much smaller than that at the yield point as per the Hook's law.
Stress $$\propto$$ Strain.
The bulk modulus for an incompressible liquid is?
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$$\infty$$
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$$0$$
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$$1$$
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$$2$$
Explanation
Bulk Modulus $$K= \dfrac{-dP}{ (dV/V)} =-\dfrac{VdP}{dV}$$
For incompressible fluid, $$dV = 0$$
On putting the value of $$dV$$
$$K=\dfrac{Vdp}{0}=\infty$$
Option A
Theoretically the value of Poisson's ratio $$\sigma$$ lies between:
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$$0 < \sigma < 1$$
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$$-1 < \sigma < 0.5$$
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$$0.2 < \sigma < 0.4$$
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$$-1 < \sigma < 1$$
Three blocks system is shown in the figure. Each has mass $$3$$ kg. String connected to P and Q are of equal cross-section and Young's modulus of $$0.005$$ $$cm^2$$ and $$2\times 10^{11}N/m^2$$ respectively, neglect friction. The longitudinal strain in A and B are.
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$$2.5\times 10^{-4}, 1\times 10^{-4}$$
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$$1\times 10^{-4}, 2\times 10^{-4}$$
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$$0.2\times 10^{-4}, 2\times 10^{-4}$$
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None of these
The pressure applied from all directions on a cube is P. How much its temperature should be raised to maintain the original volume? The volume elasticity of the cube is B and the coefficient of volume expansion is $$\gamma$$:
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$$\dfrac{P}{\gamma B}$$
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$$\dfrac{P\gamma}{B}$$
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$$\dfrac{PB}{\gamma}$$
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$$\dfrac{\gamma B}{P}$$
A one metre long steel wire of cross-sectional area $$1$$ $$mm^2$$ is extended by $$1$$ mm. If $$Y=2\times 10^{11}Nm^{-2}$$, then the work done is?
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$$0.1$$ J
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$$0.2$$ J
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$$0.3$$ J
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$$0.4$$ J
Young's modulus is defined for:
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Solid
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Liquid
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Gas
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All of these
The bulk modulus of water is $$2.1\times 10^9\ N/m^2$$. The pressure required to increase the density of water by $$0.1\%$$ is :
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$$2.1\times 10^3\ N/m^2$$
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$$2.1\times 10^6\ N/m^2$$
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$$2.1\times 10^5\ N/m^2$$
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$$2.1\times 10^2\ N/m^2$$
From the ceiling, a light rod of length $$200$$ cm is suspended horizontally with the help of two vetical wires of equal lenght as shown in figure. If one wire is made of brass and have cross-sectional area $$0.2$$ $$cm^2$$ and other of steel of $$0.1$$ $$cm^2$$ of cross-sectional area, then at what distance along rod a weight may be hung to produce equal stress in both the wires?
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$$\dfrac{4}{3}$$ m from steel wire
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$$\dfrac{4}{3}$$ m from brass wire
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$$1$$ m from steel wire
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$$\dfrac{1}{4}$$ m from brass wire
A long wire hangs vertically with its upper end clamped. A torque of $$8$$ Nm applied to the free end twisted it through $$45^o$$, the potential energy of the twisted wire is?
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$$\pi$$ joule
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$$\dfrac{\pi}{2}$$ joule
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$$\dfrac{\pi}{4}$$ joule
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$$\dfrac{\pi}{8}$$ joule
The dimensions of Poisson's ratio is?
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$$[M^0L^0T^0]$$
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$$[ML^{-1}T^{-2}]$$
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$$[ML^2T^{-4}]$$
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$$[ML^2T^{-3}]$$
The stress for one litre of a perfect gas, at a pressure of $$72$$ cm of Hg, when it is compressed isothermally to a volume of $$900$$ cc, is?
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$$9.88\times 10^3$$ $$N/m^2$$
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$$10.88\times 10^3$$ $$N/m^2$$
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$$1.088\times 10^3$$ $$N/m^2$$
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$$2\times 10^3$$ $$N/m^2$$
Choose the correct statements from the following
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Steel is more elastic than rubber
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The stretching of a coil spring is determined by the Young's modulus of the wire of the spring
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The frequency of a tuning fork is determined by shear modulus of the material of the fork
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When a material is subjected to a tensile (stretching) stress the restoring forces are caused by interatomic attraction
Explanation
Statement (a) is correct because Young's modulus of steel is greater than that of rubber.
Statement (b) is incorrect. If a spring is stretched, both the total length of the wire in the coil and the volume of the wire do not change. Only the shape of the coils of the wire undergoes a change. Hence, it is the shear modulus that determines the stretching of the coil.
Statement (c) is also incorrect. The bending moment of the prongs of a tuning fork is determined by Young's modulus of the material. Hence, the restoring force on the prongs depends on Young's modulus which determines the frequency of the fork.
Statement (d) is correct. when the material is not subjected to any stress, its atoms are in their normal (equilibrium) positions. When a tensile stress is applied, the separation $$R$$ between the atoms becomes greater than the equilibrium separation $$R_{0}$$. For $$R > R_{0}$$, the interatomic forces are attractive.
Which of the following are correct?
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The product of bulk modulus of elasticity and compressibility is $$1$$
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A rope $$1\ cm$$ in diameter breaks if the tension in it exceeds $$500\ N$$. The maximum tension that may be given to a similar rope of diameter $$2\ cm$$ is $$2000\ N$$
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Both a and b are correct
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Neither a nor b is correct
Explanation
Bulk modulus of elasticity is denoted by $$K$$. And bulk modulus of compressibility is denoted by $$\frac{1}{K}$$. So, their product is always equal to $$1$$.
Hence, option $$(A)$$ is correct.
For the rope,
Young's modulus, $$Y=\frac{Stress}{Strain}=\frac{\frac{T}{A}}{{\frac{\triangle l}{1}}}$$
$$\Rightarrow T= Y * \frac{\triangle l}{1}*\frac{\pi d^2}{4}$$
where $$d$$ is the diameter of the rope.
We can conclude,
$$T\propto d^2 $$ $$(1)$$
Let $$T_{1}, d_{1}$$ and $$T_{2},d_{2} $$ be the tension and diameter of the ropes respectively. Hence from equation $$(1)$$,
$$\frac{T_{2}}{T_{1}}=\frac{d_{2}^2}{d_{1}^2} $$
Given that $$T_{1}=500 N, d_{2}=2cm$$ and $$ d_{1}=1cm$$
$$\Rightarrow \frac{T_{2}}{500}=\frac{4}{1}$$
$$\Rightarrow T_{2}=2000 N$$
$$\therefore$$ option $$(B)$$ is also correct.
Since, both options $$(A)$$ and $$(B)$$ are correct, option $$(C)$$ is also correct.
If for a material, Y and B are Young's modulus and Bulk modulus then:
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$$Y < 3B$$
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$$Y=3B$$
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$$Y > 3B$$
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$$3Y=B$$
A given quantity of an ideal gas is at the pressure $$P$$ and the absolute temperature $$T$$. The isothermal bulk modules of the gas is :
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$$\dfrac {2}{3}P$$
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$$P$$
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$$\dfrac {3}{2}P$$
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$$2P$$
When water freezes inside a pipe, it expands by about $$9\%$$ due to expansion. What would be the pressure increase inside the pipe? The bulk modulus of ice is $$200\times 10^9\ N/ m^2$$.
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$$1.80\times 10^{10}\ N/ m^2$$
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$$3.60\times 10^8 \ N/ m^2$$
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$$9\times 10^7 \ N/ m^2$$
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$$7.2\times 10^8 \ N/ m^2$$
A rod of length $$1000$$ min and coefficient of linear expansion $$\alpha =10^{-4}$$ per degree is placed symmetrically between fixed walls separated by $$1001$$ mm. The Young's modulus of the rod is $$10^{11}$$ $$N/m^2$$. If the temperature is increased by $$20^oC$$, then the stress developed in the rod is (in $$N/m^2$$):
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$$10$$
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$$2\times 10^8$$
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$$10^8$$
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Cannot be calculated
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect
Explanation
Assertion is correct.
In reason part, elasticity of rod depends upon the mass, area of cross section and pressure exerted on the rod.
So, reason part is wrong.
A vertical bar of uniform section is fixed at both of its ends and a load $$W=5000\ N$$ is applied axially at an intermediate section as shown in figure.
Select the correct statements (s):
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Reaction at the bottom support is $$3000\ N$$
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Reaction at the bottom support is $$2000\ N$$
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Reaction at the top support is $$1000\ N$$
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Reaction at the top support is $$3000\ N$$
Explanation
We know from the formula of young's modulus
$$Y=\dfrac{F\times l}{A\times \Delta l}$$
$$\Rightarrow F=\dfrac{Y\times A\times \Delta l}{l}$$
As both the ends are fixed $$\Delta l=0$$
So the net force acting will be zero
$$F_1+F_2=0$$
And the force will be divided inversely to the value of length.
$$\dfrac{F_1}{F_2}=\dfrac{l_2}{l_1}=\dfrac{2}{3}$$
Solving these two equations we get, $$F_1=2000N$$ and$$ F_2=3000N$$
Select the correct information:
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Work done in stretching $$=\dfrac{1}{2Y}\times$$ (stress)$$^{2}$$
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For constant volume poisson's ration $$\sigma =0.5$$
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normal stress causes shear strain always
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Stress caused by heating is independent of length
Explanation
As we know,
$$\dfrac{W}{V}=\dfrac{1}{2Y} \times (stress)^2$$
and
For an incompressible material Poisson's ration σ=0.5
Stress caused by heating is independent of length as
$$stress=Y\alpha \Delta T$$
Normal stress always causes longitudinal deformation or strain.
In a wire of length $$L$$, the increase in its length is $$l$$ If the length is reduced to half, the increase in its length will be
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$$l$$
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$$2l$$
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$$\dfrac{l}{2}$$
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None of the above
Explanation
$$l\ \alpha\ L$$ i.e. if length is reduced to half then increase in length will be $$\dfrac{l}{2}$$
In $$CGS$$ system, the Young's modulus of a steel wire is $$2 \times 10^{12}$$. To double the length of a wire of unit cross section area, the force required is
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$$4 \times 10^{6} dynes$$
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$$2 \times 10^{12} dynes$$
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$$2 \times 10^{13} newtons$$
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$$2 \times 10^{8} dynes$$
Explanation
To double the length of wire,
$$Stress = $$\text{Young's modulus}$$
$$\dfrac{F}{A} = 2 \times 10^{12} \dfrac{dyne}{cm^{2}}$$
if $$A=1$$ then $$F = 2 \times 10^{12}dynes$$
To double the length of a iron wire having $$0.5 cm^{2}$$ area of cross- section, the required force will be ($$Y =10^{12} dyne / cm^{2}$$)
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$$1.0 \times 10^{-7}N$$
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$$1.0 \times 10^{7}N$$
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$$0.5 \times 10^{-7}N$$
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$$0.5 \times 10^{12} dyne$$
Explanation
If length of wire doubled then strain = $$1$$
$$Y = \text{stress} \implies F =Y \times A = 10^{12} \times 0.5 = 0.5 \times 10^{12} dyne$$
If Youngs modulus of iron is $$2 \times 10^{11} N/m^{2}$$ and the interatomic spacing between two molecules is $$3 \times 10^{-10} metre$$, the interatomic force constant is
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$$60N/m$$
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$$120N/m$$
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$$30N/m$$
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$$180N/m$$
Explanation
Interatomic force constant $$K=Y \times r_{0}$$
$$=2 \times 10^{11} \times 3 \times 10^{-10}$$
$$=60N/m$$
A copper and a steel wire of the same diameter are connected end to end A deforming force F is applied to this composite wire which causes a total elongation of $$1cm $$. The two wires will have:
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The same stress
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Different stress
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The same strain
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Different strain
Explanation
Internal force generated by the external force will be equal to $$F$$
$$\therefore stress = \dfrac{F}{A}$$
$$\because$$ area of cross section for both wire same and stretched by same force. So their stress are equal verifies option (a)
$$Stress = \dfrac{Stress}{Y}$$
As stress for both wires are same , so
$$\left(Strain \right)_{steel} \propto \dfrac{1}{Y_{steel}}$$
and $$\left(Strain\right)_{Al} \propto \dfrac{1}{Y_{Al}}$$
$$\dfrac{\left(Strain\right)}{\left(Strain\right)}_{Al} = \dfrac{Y_{Al}}{Y_{steel}}$$
$$Y_{Al}< Y_{S}$$ So $$\dfrac{Al}{steel}$$
or $$\left(strain \right)_{steel} < \left(Strain \right)_{Al}$$
Verifies option $$(d)$$.
The force required to stretch a steel wire of $$1 cm^{2}$$ cross-section to $$1.1$$ times its length would be ($$Y =2 \times 10^{11} Nm^{-2}$$
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$$2 \times 10^{6}N$$
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$$2 \times 10^{3}N$$
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$$2 \times 10^{-6}N$$
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$$2 \times 10^{-7}N$$
Explanation
$$F =A \times Y \times \text{Strain} = 1 \times 10^{-4} \times 2 \times 10^{11} \times 0.1 = 2\times 10^6N$$
A steel wire is stretched with a definite load. If the Young's modulus of the wire is $$Y$$. For decreasing the value of $$Y$$
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Radius is to be decreased
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Radius is to be increased
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Length is to be increased
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None of the above
Explanation
It is the specific property of a particular metal at a given temperature which can be changed only by temperature variations.
Which statement is true for a metal
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$$Y <\eta$$
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$$Y=\eta$$
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$$Y>\eta$$
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$$Y< \dfrac{1}{\eta}$$
Explanation
$$Y = 2\eta (1+\sigma)$$
The length of a rod is $$20 cm$$ and area of cross-section $$2 cm^2$$. The Young's modulus of the material of wire is $$1.4 \times 10^{11} N/m^2$$ . If the rod is compressed by $$5 kg-wt$$ along its length, then increase in the energy of the rod in joules will be
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$$8.57 \times 10^{-6}$$
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$$22.5 \times 10^{-4}$$
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$$9.8 \times 10^{-5}$$
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$$45.0 \times 10^(-5)$$
Explanation
Energy =$$ \dfrac{1}{2}Fl=\dfrac{1}{2}\times F\left( \dfrac{Fl}{AY}\right) = \dfrac{1}{2}\times \dfrac{F^{2}l}{AY}= \dfrac{\left( 50\right) ^{2}\times 20\times 10^{-2}}{2\times 10^{-4} \times 1\cdot 4\times 10^{11}} = 8.57 \times 10^{-6}$$
The spring balance does not read properly after its long use, because
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The elasticity of spring increases
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The elasticity decreases
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Its plastic power decreases
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Its plastic power increases
Explanation
Due to elastic fatigue its elasticity decreases
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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