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(D) Let the length of both rods be $L$, density be $\rho$, and Young's modulus be $Y$. Since the masses are equal, $m_1 = m_2 \implies \rho A_1 L = \r

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$\frac{\pi^2}{4}$

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(D) Let the length of both rods be $L$, density be $ho$, and Young's modulus be $Y$. Since the masses are equal, $m_1 = m_2 \implies ho A_1 L = ho A_2 L \implies A_1 = A_2$. For the circular rod, $A_1 = \pi r^2 = \pi (d/2)^2 = \pi (0.5)^2 = \frac{\pi}{4} \ cm^2$. For the square rod, $A_2 = s^2 = 1^2 = 1 \ cm^2$. Wait, the problem states they have equal mass and equal length (implied by rods), so $A_1 = A_2$. However, the dimensions given ($d=1 \ cm$ and $s=1 \ cm$) lead to $A_1 = \frac{\pi}{4}$ and $A_2 = 1$. This implies the lengths must be different to keep mass equal. Let $L_1$ and $L_2$ be the lengths. $m = ho A_1 L_1 = ho A_2 L_2 \implies L_1 A_1 = L_2 A_2$. Elongation $\Delta L = \frac{FL}{AY}$. Ratio $\frac{\Delta L_1}{\Delta L_2} = \frac{L_1}{A_1} 	imes \frac{A_2}{L_2} = \frac{L_1}{L_2} 	imes \frac{A_2}{A_1}$. Since $\frac{L_1}{L_2} = \frac{A_2}{A_1}$, the ratio is $(\frac{A_2}{A_1})^2$. $A_1 = \frac{\pi}{4}$, $A_2 = 1$. Ratio $= (\frac{1}{\pi/4})^2 = (\frac{4}{\pi})^2 = \frac{16}{\pi^2}$.

$A$ steel rod with a circular cross-section of diameter $1 \ cm$ and another steel rod with a square cross-section of side $1 \ cm$ have equal mass. If the two rods are subjected to the same tension, the ratio of the elongations of the two rods is

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