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(C) Let the thin wire have length $L_1 = L$ and radius $r_1 = R$. Its area is $A_1 = \pi R^2$.Let the thick wire have length $L_2 = 2L$ and radius $r_

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(C) Let the thin wire have length $L_1 = L$ and radius $r_1 = R$. Its area is $A_1 = \pi R^2$.Let the thick wire have length $L_2 = 2L$ and radius $r_2 = 2R$. Its area is $A_2 = \pi (2R)^2 = 4\pi R^2 = 4A_1$.Since the wires are in series,the same force $F$ acts on both wires.Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta L/L}$,so $\Delta L = \frac{FL}{AY}$.For the thin wire: $\Delta L_1 = \frac{FL_1}{A_1 Y} = \frac{FL}{\pi R^2 Y}$.For the thick wire: $\Delta L_2 = \frac{FL_2}{A_2 Y} = \frac{F(2L)}{(4\pi R^2) Y} = \frac{FL}{2\pi R^2 Y}$.The ratio of elongation in the thin wire to that in the thick wire is $\frac{\Delta L_1}{\Delta L_2} = \frac{FL/\pi R^2 Y}{FL/2\pi R^2 Y} = 2$.

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