question_answer

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes\[(L+l)\]. The elastic potential energy stored in the extended wire is-               [NEET 5-5-2019]

A)  \[\frac{1}{2}Mgl\]

B) \[\frac{1}{2}MgL\]

C) \[Mgl\]              

D) \[MgL\]

Correct Answer: A

Solution :

Loss in gravitational P.E. =\[Mgl\] Elastic potential energy \[U=\frac{1}{2}Mgl\]