JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Elasticity & SHM

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.

(A)3 times
(B) $3/2$ times
(C)4 times
(D)2 times✓ Correct

Answer: (D)

Solution

Young's modulus relates stress and strain:

Y = \frac{F / A}{ℓ ^{'} / L}

, giving

ℓ^{'} = \frac{F L}{A Y} = \frac{F L}{π r ^{2} Y}

For fixed

L

and

Y

, this means

ℓ^{'} \propto \frac{F}{r ^{2}}

. Halving both

F

and

r

\frac{ℓ _{new}^{'}}{ℓ ^{'}} = \frac{F /2}{( r /2 ) ^{2}} \times \frac{r ^{2}}{F} = \frac{F /2}{r ^{2} /4} \times \frac{r ^{2}}{F} = 2

The increase in length becomes 2 times the original.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Elasticity & SHM

Solution

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