Question

A wire fixed at one end is stretched by constant force $F$. If elongation in wire is $l$, then energy stored in wire will be

• A. $F.l$
• B. $2F.l$
• C. $\frac{F.l}{2}$
• D. $\frac{F.l}{4}$

Answer 1

The correct option is C $\frac{F.l}{2}$

$K=\frac{F}{l}$
At equilibrium, $U=\frac{1}{2}.K.l^2=\frac{1}{2}Kl.l=\frac{F.l}{2}$
$U=\frac{Fl}{2}$