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A wire of 4m in length suspended vertically stretches by 1mm when mass of 25Kg is attached to the lower end.What is the elastic potential energy gained by the wire?$g=10m/s^2$

$\begin{array}{1 1}(A)\;.10J\\(B)\;.12J\\(C)\;.25J\\(D)\;.20J\end{array} $

1 Answer

Elastic potential energy =$\large\frac{1}{2}$$\times$ Force$\times \Delta l$
$\Rightarrow \large\frac{1}{2}$$\times 25\times 10\times 1\times 10^{-3}$
$\Rightarrow$ .125J
Hence (B) is the correct answer.
answered May 8, 2014 by sreemathi.v
 

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