In this page we have *Important questions on Elasticity for JEE Main/Advanced and CBSE* . Hope you like them and do not forget to like , social share
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## Multiple Choice questions

**Question -1** A wire whose cross-section area is A

_{1} is stretched by L

_{1} by a certain weight. How far will a wire of same material and same length and cross-section area A

_{2} stretch if same weight is applied to it

Solution
We know that

$\Delta L= \frac {FL}{AY}$

Or

$A\Delta L = \frac {FL}{Y}$

As F,L and Y are same So ,

$A\Delta L =constant$

$A_1 L_1 = A_2 L_2$

Or

$L_2 =\frac {A_1 L_1}{A_2}$

**Question 2:**
A wire is stretched by x mm when a load F is hanged on it.If the same wire goes over a pulley and two weight F each are hung at the two end’s the what will be the elongation in the wire

a) x

b) 2x

c) x/2

d) 0

Solution
When wire is stretched by F

$Y =\frac {F/A}{x/L}$ --- (1)

When the wire goes over the pulley, the length on each side will be L/2 .And each side will experience the same elongation .Let it be y

Then

$Y=\frac {F/A}{y/5L}$ ---(2)

From equation 1 and 2

We get

y=x/2

Total elongation=2y=x

__Linked Type Comprehensions__
One end of a uniform wire of length L and μ mass per unit length is attached rigidly to a point in a ceiling. A Mass M is suspended from its lower end. Area of cross-section of the wire is A.

**Question 3** Find the stress at a distance x from the ceiling point

Solution
Weight of wire below the point =(L-x)μg +Mg

Stress =F/A

So $S= \frac {(L-x)\mu g + Mg}{A}$

**Question 4**:

Find the elongation of the wire. If Young Modulus of wire is Y

Solution
Let us consider a small element of length dx at distance x from ceiling point.

Let dy be the elongation of that element due dx

We know from previous question that

Stress will be

$S= \frac {(L-x)\mu g + Mg}{A}$

Strain =dy/dx

Now

$Y = \frac {Stress}{Strain}$

$Y=\frac {[(L-x) \mu g + Mg]dx}{Ady}$

Or

$dx= \frac {[(L-x) \mu g + Mg]dx}{AY}$

Total elongation of the wire can be found by integrating from 0 to L

__Multiple Choice questions__
**Question 5**: A block of gelatin is 60 mm by 60 mm by 20 mm when unstressed. A force of .245 N is applied tangentially to the upper surface causing a 5mm displacement relative to the lower surface. The block is placed such that 60X60 comes on the lower and upper surface. Find the shearing stress, shearing strain and shear modulus

a) (68.1 N/m

^{2},.25,272.4 N/m

^{2})

b) (68 N/m

^{2},.25,272 N/m

^{2})

c) (67 N/m

^{2},.26,270.4 N/m

^{2})

d) (68.5 N/m

^{2},.27,272.4 N/m

^{2})

Solution
Shear stress=F/A=.245/36*10^{-4} ==68.1 N/m^{2}

Shear strain= tanθ= d/h=5/20=.25

Shear modulus (S) =shear stress/shear strain=272.4 N/m^{2}

**Question 6**
A wire of radius r is stretched without sag and tension between two point separated by distance 2L.A weight is hanged at the middle of the wire which displaced the point by a distance d. Young modulus of elasticity is Y. Find the tension in the wire

Solution
Length of wire after weight is hanged

$=2\sqrt {L^2 + d^2} $

Length before hanging=2L

Change in length

$=2\sqrt {L^2 + d^2} -2L $

So

Strain = $ \frac {2\sqrt {L^2 + d^2} -2L }{2L}$

Or

$strain = \sqrt {1+ (\frac {d}{L})^2} -1$

Now

$Y =\frac {stress}{strain}$

Let F be the tension then

**Question 7**:

If in previous question d<<<<L, what will be the value of Tension

Solution
$strain = \sqrt {1+ (\frac {d}{L})^2} -1$

$=(1+ \frac {d^2}{L^2} )^ 1/2 -1 $

Expanding bionomically and as d<<<L ,neglecting higher terms

$=(1+ \frac {d^2}{2L^2} -1 $

$=\frac {d^2}{2L^2}$

Then tension

$F= Y \pi r^2 \frac {d^2}{2L^2}$

$F=\frac {Y \pi r^2 d^2}{2L^2}$

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