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(D) The strain energy per unit volume $u$ is given by the formula $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.Since $	ext{stress} = \

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$81$

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(D) The strain energy per unit volume $u$ is given by the formula $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.Since $	ext{stress} = \frac{F}{A}$ and $	ext{strain} = \frac{	ext{stress}}{Y}$,we have $u = \frac{1}{2} 	imes \frac{	ext{stress}^2}{Y} = \frac{1}{2} 	imes \frac{F^2}{A^2 Y}$.Given that the wires have the same length,material ($Y$ is constant),and are stretched by the same force $F$,the energy density $u$ is inversely proportional to the square of the area of cross-section $A^2$.Since $A = \pi r^2$,we have $A \propto d^2$ (where $d$ is the diameter),so $u \propto \frac{1}{(d^2)^2} = \frac{1}{d^4}$.Therefore,the ratio of strain energy per unit volume for the smaller diameter wire $(d_S)$ to the larger diameter wire $(d_L)$ is $\frac{u_S}{u_L} = \left( \frac{d_L}{d_S} ight)^4$.Given the ratio of diameters $d_S : d_L = 1 : 3$,we have $\frac{u_S}{u_L} = \left( \frac{3}{1} ight)^4 = 81 : 1$.

Two wires having the same length and material are stretched by the same force. Their diameters are in the ratio $1:3$. The ratio of strain energy per unit volume for these two wires (smaller to larger diameter),when stretched,is (in $:1$)

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