First slide

Elastic Modulie

Question

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 2019]

Moderate

A

$M g L$
B

$\frac{1}{2} M g l$
C

$\frac{1}{2} M g L$
D

$M g l$

Solution

In stretching a wire, the work done against internal restoring force is stored as elastic potential energy in
the wire and is given by
$U = W = \frac{1}{2} \times Force (F) \times$ Elongation (/)
$= \frac{1}{2} F l = \frac{1}{2} \times M g \times l = \frac{1}{2} M g l$

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