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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?$g=10m/s^2$

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

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.