The work done on a wire of volume 2 ,cm^3 is | vedclass

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(D) The energy density (work done per unit volume) in a stretched wire is given by the formula: $u = \frac{1}{2} Y \varepsilon^2$, where $Y$ is the Yo

Vedclass · Accepted Answer

$0.02$

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(D) The energy density (work done per unit volume) in a stretched wire is given by the formula: $u = \frac{1}{2} Y \varepsilon^2$, where $Y$ is the Young's modulus and $\varepsilon$ is the strain.Total work done $W = u 	imes V = \frac{1}{2} Y \varepsilon^2 V$.Given: $W = 16 	imes 10^2 \,J$, $V = 2 \,cm^3 = 2 	imes 10^{-6} \,m^3$, $Y = 4 	imes 10^{12} \,N/m^2$.Rearranging for strain $\varepsilon$: $\varepsilon = \sqrt{\frac{2W}{YV}}$.Substituting the values: $\varepsilon = \sqrt{\frac{2 	imes 16 	imes 10^2}{4 	imes 10^{12} 	imes 2 	imes 10^{-6}}} = \sqrt{\frac{3200}{8 	imes 10^6}} = \sqrt{400 	imes 10^{-6}} = 20 	imes 10^{-3} = 0.02$.Thus, the strain produced is $0.02$.

The work done on a wire of volume $2 \,cm^3$ is $16 \times 10^2 \,J$. If the Young's modulus of the material of the wire is $4 \times 10^{12} \,N/m^2$, then the strain produced in the wire is:

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