Question

A body of mass $M$ is attached to lower end of a metal wire, whose upper end is fixed. The elongation of 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 $\frac{1}{2}Mgl$.
• D. Heat produced is $\frac{1}{2}Mgl$

Answer 1

Answer: (A), (C), (D)

The elastic potential energy stored in the wire is $\frac{1}{2}Mgl$.
Heat produced is $\frac{1}{2}Mgl$
Loss in gravitational potential energy of $M$ is $Mgl$.

Since it moves by l, gravitational potential energy will be Mgl.

Elastic potential energy = $1/2\times Stress\times Strain\times Volume=1/2\times \frac{Mg}{A}\times \frac{l}{L}\times AL=\frac{Mgl}{2}$

Heat produced =Change in energy=Elastic energy stored-Gravitational energy lost=$\frac{-Mgl}{2}$. Thus heat lost=$\frac{Mgl}{2}$