Question

A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of this wire is l.

• A. Loss in gravitational potential energy of M is Mgl
• B. The elastic potential energy stored in the wire is Mgl.
• C. The elastic potential energy stored in the wire is Mgl.
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A

Loss in gravitational potential energy of M is Mgl

C

The elastic potential energy stored in the wire is Mgl.

$v=\frac{1}{2}\times stress\times strain\times volume$

= $\frac{1}{2}\times \frac{Mg}{A}\times \frac{l}{L}\times AL$

= $\frac{1}{2}Mgl$

And we know change in gravitational potential energy will be Mgl.