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(D) The elongation $\Delta L$ in a string is given by the formula $\Delta L = \frac{FL}{AY}$, where $F$ is the tension, $L$ is the length, $A$ is the

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$6$

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(D) The elongation $\Delta L$ in a string is given by the formula $\Delta L = \frac{FL}{AY}$, where $F$ is the tension, $L$ is the length, $A$ is the cross-sectional area, and $Y$ is the Young's modulus.For string $A$, the total mass supported is $M + 2M = 3M$, so the tension $F_A = 3Mg$.For string $B$, the mass supported is $2M$, so the tension $F_B = 2Mg$.Given: $A_A = A_B = A$, $L_A/L_B = 2$, and $Y_A/Y_B = 0.5$.The elongation in $A$ is $\Delta L_A = \frac{F_A L_A}{A Y_A} = \frac{3Mg L_A}{A Y_A}$.The elongation in $B$ is $\Delta L_B = \frac{F_B L_B}{A Y_B} = \frac{2Mg L_B}{A Y_B}$.The ratio of elongations is $\frac{\Delta L_A}{\Delta L_B} = \left( \frac{3Mg}{2Mg} \right) \cdot \left( \frac{L_A}{L_B} \right) \cdot \left( \frac{Y_B}{Y_A} \right)$.Substituting the values: $\frac{\Delta L_A}{\Delta L_B} = \left( \frac{3}{2} \right) \cdot (2) \cdot \left( \frac{1}{0.5} \right) = 3 \cdot 2 = 6$.

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