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An $8\,m$ long copper wire and $4\,m$ long steel wire, each of cross section $0.5\,cm^2$ are fastened end to end and stretched by $500\,N$ force. The elastic potential energy of the system is (Youngs mod $: Y_{cu}= 1\times 10^{11}\,N/m^2,$ $Y_{steel} = 2\times 10^{11}\,N/m^2$ ) :
$1/4\,\,J$
$1/3\,\,J$
$1/2\,\,J$
$1\,\,J$
Solution
Work $=$ Energy stored
$=\frac{1}{2} \times(\text { Force }) \times(\Delta l)$
$\Delta l_{1}=\frac{\mathrm{FL}_{1}}{\mathrm{Y}_{\mathrm{cu}} \mathrm{A}}=0.8 \mathrm{mm} \quad \& \mathrm{\Delta l}_{2}=\frac{\mathrm{FL}_{2}}{\mathrm{Y}_{\mathrm{steel}} \mathrm{A}}=0.2 \mathrm{mm}$
Net $\Delta l=\Delta l_{1}+\Delta l_{2}=1 \mathrm{mm}=10^{-3} \mathrm{m}$
Elastic potential energy $=\frac{1}{2} \times 500 \times 10^{-3}=?$
$=\frac{1}{4} J$
