Question

A light metal wire length $L$ is suspended vertically from rigid support. When a body of mass $M$ is attached to the lower end of the wire, the elongation of the wire is $\Delta L$ in its equilibrium position. Then,

• A. the loss in gravitational potential energy of mass M is $Mg\Delta L$
• B. the elastic potential energy stored in the wire is $Mg\Delta L$
• C. the elastic potential energy stored in the wire is $\frac{1}{2}Mg\Delta L$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A the elastic potential energy stored in the wire is $\frac{1}{2}Mg\Delta L$
C the loss in gravitational potential energy of mass M is $Mg\Delta L$

When the mass is attached, the wire elongates by $\Delta L$ in the downward direction.
So, the mass comes down, hence loses P.E
Decrease in $P.E=Mg\Delta L$
In the same process, wire gains elastic potential Energy,
Elastic $P.E=Mg\frac{\Delta L}{2}$