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. Elastic potential energy stored in the wire is `(Mgl)/2`C. Elastic potential energy stored in the wire is `Mgl`D. Elastic potential energy stored in the wire is `(Mgl)/3`

A body of mass `M` is attached to the lower end of a metal wire, whose

1.	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. Elastic potential energy stored in the wire is `(Mgl)/2`C. Elastic potential energy stored in the wire is `Mgl`D. Elastic potential energy stored in the wire is `(Mgl)/3`
Answer» Correct Answer - A::B Loss in gravitational potential energy of `M` is `Mgl` as `M` falls down by `l`. Elastic potential energy stored in wire is `U=1/2xx"Stress"xx"Strain"xx"Volume"` `=1/2 1/2 xx(Mg)/Axxl/LxxAL=(Mgl)/2` Now work done by gravityy force is not equal to elastic potential energy stored in wire. This is due to the fact that some work has been done against air friction etc., which increases the internal energy of wire.

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