JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Elasticity & SHM

A wire of length L is hanging from a fixed support. The length changes to

L_{1}

and

L_{2}

when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to:

Answer: (C)

Solution

By Hooke's law, the extension is proportional to the applied force:

\frac{F _{1}}{F _{2}} = \frac{Δ L _{1}}{Δ L _{2}}

. With

F_{1} = 10

N and

F_{2} = 20

N, we have

\frac{10}{20} = \frac{L _{1} - L}{L _{2} - L}

. Simplifying:

\frac{1}{2} = \frac{L _{1} - L}{L _{2} - L}

, so

L_{2} - L = 2 (L_{1} - L)

. Solving for L:

L_{2} - L = 2 L_{1} - 2 L

, thus

L = 2 L_{1} - L_{2}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option