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 \[U=\frac{1}{2}Mgl\]