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(C) Given: $l_1: l_2 = 1: 2$,$d_1: d_2 = 3: 2$,$F_1: F_2 = 3: 1$.The elastic potential energy stored per unit volume $(U)$ is given by the formula $U

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$16: 9$

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(C) Given: $l_1: l_2 = 1: 2$,$d_1: d_2 = 3: 2$,$F_1: F_2 = 3: 1$.The elastic potential energy stored per unit volume $(U)$ is given by the formula $U = \frac{\sigma^2}{2Y}$,where $\sigma$ is the stress and $Y$ is Young's modulus.Since stress $\sigma = \frac{F}{A} = \frac{F}{\pi (d/2)^2} = \frac{4F}{\pi d^2}$,we have $\sigma \propto \frac{F}{d^2}$.Therefore,$U \propto \sigma^2 \propto \left(\frac{F}{d^2}ight)^2$.The ratio of energies is $\frac{U_1}{U_2} = \left(\frac{F_1}{F_2}ight)^2 	imes \left(\frac{d_2}{d_1}ight)^4$.Substituting the given values: $\frac{U_1}{U_2} = \left(\frac{3}{1}ight)^2 	imes \left(\frac{2}{3}ight)^4 = 9 	imes \frac{16}{81} = \frac{16}{9}$.Thus,the ratio is $16: 9$.

Two copper wires $A$ and $B$ of lengths in the ratio $1: 2$ and diameters in the ratio $3: 2$ are stretched by forces in the ratio $3: 1$. The ratio of the elastic potential energies stored per unit volume in the wires $A$ and $B$ is

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