In this page we have Elasticity Problems for JEE Main/Advanced And NEET . Hope you like them and do not forget to like , social share
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Subjective Questions
Question 1
When a 100 Kg mass is suspended from wire of length 1 meter and and area 1 cm2, it length increases.
(a) Find the stress in the wire
(b) Find the increase in length of the wire
(c) Find the elastic potential energy stored in the wire
(d) Find the tension in the wire in loaded position Solution
Given g=10 m/sec2 and Young Modulus of the material of the wire=2X1011 N/m2
Stress =Load/Area
$=\frac {mg}{A}=\frac {100 \times 10}{10^{-4}} =10^7 \ N/m^2$
Now
Young Modulus=Stress/Strain
So strain= Stress/Young Modulus
$=5 \times 10^{-5}$
Now we know that
Strain = Increase in length/Initial length
=> Increase in Length=strain * Initial length
$=5 \times 10^{-5}$ m
Elastic Potential energy
$ =\frac{1}{2}stress \times strain \times Volume$
$=5 \times 10^{-2} \ J$
The tensions in the load position will the weight of the mass
So T=mg
=100*10=1000 N
Question 2
A thin metallic rod of Length L and radius of cross-section r rotates with the angular velocity in a horizontal plane about a vertical axis passing through one of its end.
Density of rod is ρ and Young Modulus of elasticity is Y. Breaking stress of the rod is S
Find out the following
(a) Find the tension in the rod as a function of distance from the fixed end
(b) Find the tension at the fixed end, midpoint and end point of the rod
(c) Find the maximum angular velocity with which it can rotate with out breaking
(d) Find the total elongation in the rod Solution
Let us consider a small element dx at a distance x from the rotating end.
Now mass of the rod =$L\pi r^2\rho$
Mass per unit length of the rod=$\pi r^2\rho$
Mass of the small element=$\pi r^2dx\rho$
Centripetal force acting on the element
$dF=\pi r^2\rho x\omega^2dx$
The Tension in the element would be due to the centripetal force of the outer portion i.e from x=x to x=L
$F=\int_{x}^{L}{\pi r^2\rho x\omega^2dx}$
$F=\frac{1}{2}\pi r^2\rho\omega^2(L^2-x^2)$
Tension at fixed point
x=0
$F=\frac{\rho\pi r^2\omega^2L^2}{2}$
At x=L/2
$F=\frac{3\rho\pi r^2\omega^2L}{8}$
So tension is maximum at fixed end and decrease towards outer end.
So
$S\pi r^2>\frac{\rho\pi r^2\omega^2L^2}{2}$
Or $\sqrt{\frac{2S}{\rho L^2}}<\omega$
Let dy be the elongation at the element of length dx
Then
$\frac{dy}{dx}=\frac{1}{Y}\frac{F(x)}{A}$
Substituting the value of F(x) from previous part
$dy=\frac{\pi r^2\rho\omega^2(L^2-x^2)dx}{2Y\pi r^2}$
$y=\int_{0}^{L}{\frac{\rho\omega^2}{2Y}(L^2-x^2)dx=\frac{\rho\omega^2L^3}{3Y}}$
Question 3
An elastic string has a mass M suspended at its lower end,the upper being fixed to a support. When the mass is pulled down over a short distance and let go, then it executes SHM with Time period given by
$T= 2 \pi \sqrt {\frac {l_1}{g}}$
Where l1 is the elongation of the string.Now the mass m is added to the mass M, then it is found that Time period of oscillation (T2) such that
$\frac{T_1}{T_2}=\frac{5}{4}$
Find the ratio m:M
Given Y is the Young modulus of elasticity. L be the original length of string. Solution
Let $l_2$ be the elongation in second case then
$T_1=2\pi\sqrt{\frac{l_1}{g}}$ ---(1)
$T_2=2\pi\sqrt{\frac{l_2}{g}}$ ---(2)
So that
$\frac{T_1^2}{T_2^2}=\frac{l_1}{l_2}$
Or
$\frac{l_1}{l_2}=\frac{16}{25}$
Now
For first case
$Y=\frac{MgL}{Al_1} $ --(3)
Second case
$Y=\frac{(M+m)gL}{Al_2}$ --(4)
From 3 and 4
$\frac{l_1}{l_2}=\frac{M}{M+m}$
Or
$\frac{M}{M+m}=\frac{16}{25}$
$\frac{M+m}{M}=\frac{25}{16}$
Or
$\frac{M+m-M}{M}=\frac{25-16}{19}$
Or
$\frac{m}{M}=\frac{9}{16}$
Question 4
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 temperature of each rod increases by $\Delta T$. Find change in the angle ABC. [Coeff. of linear expansion for Cu is $\alpha _1$ ,Coeff. of linear expansion for Al is $alpha _2$ ] Solution
Let $l_1$ = AB, $l_2$ = AC, $l_3$ = BC
$cos \theta = \frac {l_3^2 + l_1^2 -l_2^2}{2l_3L_1}$
or
$2l_3L_1 cos \theta=l_3^2 + l_1^2 -l_2^2$
Differentiating
$2(l_3dL_1 + l_1dl_3 ) cos \theta - 2l_3L_1 sin \theta d\theta=2l_3dl_3 + 2l_1dl_1 - 2l_2dl_2$
Now
$dl_1 = l_1 \alpha _1 \Delta T$
$dl_2 = l_2 \alpha _1 \Delta T$
$dl_3 = l_3 \alpha _2 \Delta T$
and $l_1 = l_2 =l_3 =l$
Substituting these values
$2(l^2 \alpha _1 \Delta T + l^2 \alpha _2 \Delta T)cos \theta -2 l^2 sin \theta d\theta=2l^2 \alpha _1 \Delta T + 2l^2 \alpha _1 \Delta T -2 l^2 \alpha _2 \Delta T$
$sin \theta d\theta = 2 \alpha _1 \Delta T (1 - cos \theta) - \alpha _2 \Delta T$
Substituting $\theta = 60^0$
$d \theta = \frac {2 (\alpha _1 - \alpha _2) \Delta T}{\sqrt 3}$
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