Elasticity | JEE Main previous year questions with Solutions

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Topic Question Set

Q 1 :
A wire of length $L$ and radius $r$ is clamped at one end. If its other end is pulled by a force $F$ , its length increases by $l$ . If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become [2024]

3 times
4 times
$\frac{3}{2}$ times
2 times

1

2

3

4

(D) Young's modulus, $Y = \frac{s t r e s s}{s t r a i n} = \frac{F / A}{∆ l / L} = \frac{F L}{A ∆ l}$

In case I : $Y = \frac{F L}{{πr}^{2} l}$ ...(i)

In case II : $Y = \frac{(F / 2) L}{π {(\frac{r}{2})}^{2} ∆ l} = \frac{2 F L}{π r^{2} ∆ l}$ ...(ii)

From (i) and (ii), $\frac{F L}{π r^{2} l} = \frac{2 F L}{π r^{2} ∆ l} \Rightarrow ∆ l = 2 l$

Q 2 :
Young's modulus of material of a wire of length 'L' and cross-sectional area A is Y. If the length of the wire is doubled and cross-sectional area is halved then Young's modulus will be [2024]

$Y / 4$
$4 Y$
$Y$
$2 Y$

1

2

3

4

(C) Young's modulus depends on the material not on length and cross-sectional area. So young's modulus remains same.

Q 3 :
With rise in temperature, the Young's modulus of elasticity [2024]

changes erratically
decreases
increases
remains unchanged

1

2

3

4

(B) $Y = \frac{F / A}{∆ l / l} = \frac{F / A}{α ∆ T}$ as $∆ T$ increases, $Y$ decreases

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