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(A) Young's modulus is given by $Y = \frac{Fl}{A\Delta l}$.Since the volume $V = A 	imes L$ is the same for both wires,and the cross-sectional areas

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$9F$

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(A) Young's modulus is given by $Y = \frac{Fl}{A\Delta l}$.Since the volume $V = A 	imes L$ is the same for both wires,and the cross-sectional areas are $A_1 = A$ and $A_2 = 3A$,their lengths must be $L_1 = 3l$ and $L_2 = l$ respectively.For the first wire:$\Delta l = \frac{F \cdot (3l)}{A \cdot Y} = \frac{3Fl}{AY} \quad ...(i)$For the second wire,let the required force be $F'$. The extension is the same $\Delta l$:$\Delta l = \frac{F' \cdot l}{(3A) \cdot Y} = \frac{F'l}{3AY} \quad ...(ii)$Equating $(i)$ and $(ii)$:$\frac{3Fl}{AY} = \frac{F'l}{3AY}$$3F = \frac{F'}{3}$$F' = 9F$

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