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 at final equilibrium state is $l$ .

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The elastic potential energy stored in the wire is $M g l$

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

Loss in gravitational potential energy of $M$ is $M g l$

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

answer is A, C, D.

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Detailed Solution

When a block attached to a spring is released from its unstretched position, it will go down to stretch the spring by $x$ such that
$\frac{1}{2} k x^{2} = m g x \Rightarrow x = \frac{2 m g}{k}$
Ultimately, due to air friction, etc., the block will settle down in its equilibrium position, stretching the spring by
$x_{0} = \frac{m g}{k} = (= \frac{x}{2})$

Thus, energy equal to $\frac{m g}{k} \frac{x}{2}$ is lost forever. This energy is mainly lost an heat in overcoming viscosity of air.
In case of the wire (which is just like a very stiff spring), similar thing happens. We may not be able to observe oscillations as in the case of a spring due to small amplitude. Ultimately, the system settles in equilibrium where half the loss in gravitational Pe of the block is lost as heat.

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