JEE Main · 2004 · 24 Apr 2004 · Offline · Elasticity & SHM

A wire fixed at the upper end stretches by length by applying a force

F

. The work done in stretching is

Answer: (D)

Solution

As the wire stretches, the pull grows steadily from zero up to the final value

F

over the extension

ℓ

, so the average force is

\frac{F}{2}

and the work stored is

W = \frac{1}{2} F ℓ

. The same follows from the elastic energy

\frac{1}{2} k ℓ^{2}

once you write

F = k ℓ

, which again gives

\frac{1}{2} F ℓ

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option