Question

A body of mass $M$ is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of the 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 correct options are
A Heat produced is $\frac{1}{2}Mgl$
C Loss in gravitational potential energy of $M$ is $Mgl$
D The elastic potential energy stored in the wire is $\frac{1}{2}Mgl$

We know that gravitational potential energy is $V_G=Mgh$ , where $h=$ distance from equilibrium position.
so here $h=l$, loss in gravitational P.E. is $V_G=Mgl$

Elastic potential energy is $V_e=\frac{1}{2}Fl=\frac{1}{2}Mgl$

As the energy is conserved so loss in P.E. is equal to the sum of elastic potential energy stored and heat energy.

Heat energy = $Mgl-\frac{1}{2}Mg=\frac{1}{2}Mgl$