Question

When a body of mass $M$ is attached to lower end of a wire having length $l$ and upper end is fixed, then elongation of the wire is $l$. In this situation, mark out the correct statement(s).

• A. Loss in gravitational potential energy of mass $M$ is $Mgl$.
• B. Elastic potential energy stored in the wire is $\frac{Mgl}{2}.$
• C. Elastic potential energy stored in the wire is $Mgl$.
• D. Elastic potential energy stored in the wire is $\frac{Mgl}{3}.$

Answer 1

Answer: (A), (B)

The correct options are
A Loss in gravitational potential energy of mass $M$ is $Mgl$.
B Elastic potential energy stored in the wire is $\frac{Mgl}{2}.$

Loss in gravitational potential energy of mass $M$ is $Mgl$ as it falls down by $l$.
Elastic potential energy stored in the wire is,
$U=\frac{1}{2}\times averagestress\times strain\times volume$
$=\frac{1}{2}\times \frac{1}{2}\times \frac{Mg}{A}\times \frac{l}{L}\times AL=\frac{Mgl}{2}$
Now, we can see that work done by gravity force is not equal to elastic potential energy stored in wire. This is due to the fact that some work has been done against air friction etc., which increases the internal energy of wire.