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(B) The Young's modulus $Y$ is given by the formula: $Y = \frac{F \cdot l}{A \cdot \Delta l}$.Since volume $V = A \cdot l$,we can write $l = \frac{V}{

Vedclass · Accepted Answer

$16$

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(B) The Young's modulus $Y$ is given by the formula: $Y = \frac{F \cdot l}{A \cdot \Delta l}$.Since volume $V = A \cdot l$,we can write $l = \frac{V}{A}$.Substituting this into the formula: $Y = \frac{F \cdot (V/A)}{A \cdot \Delta l} = \frac{F \cdot V}{A^2 \cdot \Delta l}$.Rearranging for extension $\Delta l$: $\Delta l = \frac{F \cdot V}{Y \cdot A^2}$.Since $F, V,$ and $Y$ are constant for both rods,$\Delta l \propto \frac{1}{A^2}$.Since the cross-section is circular,area $A = \pi r^2 = \pi (d/2)^2$,so $A \propto d^2$.Therefore,$\Delta l \propto \frac{1}{(d^2)^2} = \frac{1}{d^4}$.Given $d_1 = \frac{1}{2} d_2$,or $d_2 = 2 d_1$.The ratio of extensions is $\frac{\Delta l_1}{\Delta l_2} = \left( \frac{d_2}{d_1} \right)^4 = \left( \frac{2 d_1}{d_1} \right)^4 = 2^4 = 16$.Thus,the ratio is $16: 1$.

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