Mechanical Properties of Solids - Result Question 8
9. 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 :
[2019]
(a) $Mg l$
(b) $MgL$
(c) $\frac{1}{2} Mg l$
(d) $\frac{1}{2} MgL$
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Solution:
- (c) Here,
$Kx_0=Mg$
where $K=$ force constant
$\Delta E=\frac{1}{2} Kx_0^{2}$
$=\frac{1}{2} \frac{Mg}{x_0} \times x_0^{2}$
$=\frac{1}{2} Mgx_0$
Stored elastic potential energy in extended wire, $=\frac{1}{2} Mg \ell[.$ here $.x_0=\ell]$
