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A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed.Initially the body is supported at its bottom . The support is slowly lowered and finally removed . The elongation of the wire is l.

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Loss in gravitational potential energy of M is M g l

The elastic potential energy stored in the wire is $Mgl$

The elastic potential energy stored in the wire is $\frac{1}{2} Mgl$

Heat produced is $\frac{1}{2} M g l$

answer is A, C, D.

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Detailed Solution

Half of energy is lost in heat and the rest half is stored as elastic potential energy.

$l = \frac{M g L}{A Y} \dots . . (i) U = \frac{1}{2} K l^{2} = \frac{1}{2} (\frac{Y A}{L}) l^{2} . . . . . . (i i)$

From Eqs. (i) and (ii), we can prove that

$U = \frac{1}{2} Mgl$

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