JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Elasticity & SHM

Two wires A and B are made of same material having ratio of lengths

\frac{L _{A}}{L _{B}} = \frac{1}{3}

and their diameters ratio

\frac{d _{A}}{d _{B}} = 2

. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

Answer: (B)

Solution

The elongation under a stretching force is

Δ L = \frac{F L}{A Y}

with

A = \frac{π d ^{2}}{4}

. For the same force and the same material,

\frac{Δ L _{A}}{Δ L _{B}} = \frac{L _{A}}{L _{B}} (\frac{d _{B}}{d _{A}})^{2} .

Using

L_{A} / L_{B} = 1/3

and

d_{A} / d_{B} = 2

(so

d_{B} / d_{A} = 1/2

\frac{Δ L _{A}}{Δ L _{B}} = \frac{1}{3} \times (\frac{1}{2})^{2} = \frac{1}{12} .

The ratio of their elongations is

1 : 12

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option