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(D) The energy stored per unit volume $(u)$ is given by the formula: $u = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} = \frac{1}{2} \frac{(\

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$\sqrt{2}: 1$

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(D) The energy stored per unit volume $(u)$ is given by the formula: $u = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} = \frac{1}{2} \frac{(	ext{Stress})^2}{Y}$.Since $	ext{Stress} = \frac{F}{A} = \frac{F}{\pi (d/2)^2} = \frac{4F}{\pi d^2}$,we have $u = \frac{1}{2Y} \left( \frac{4F}{\pi d^2} ight)^2$.Given that the load $(F)$,length,and Young's modulus $(Y)$ are the same for both wires,$u \propto \frac{1}{d^4}$.Therefore,$\frac{u_1}{u_2} = \left( \frac{d_2}{d_1} ight)^4$.Given $\frac{u_1}{u_2} = \frac{1}{4}$,we have $\frac{1}{4} = \left( \frac{d_2}{d_1} ight)^4$.Taking the fourth root on both sides,$\frac{d_2}{d_1} = \left( \frac{1}{4} ight)^{1/4} = \left( \frac{1}{2^2} ight)^{1/4} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$.Thus,$\frac{d_1}{d_2} = \sqrt{2} : 1$.

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is $1: 4$,the ratio of their diameters is

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