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(D) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.Since the volume of the wire remains constant during stretchi

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$16$

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(D) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.Since the volume of the wire remains constant during stretching,$V = A_1 L_1 = A_2 L_2$.Therefore,$L_1 \pi r_1^2 = L_2 \pi r_2^2$,which implies $\frac{L_2}{L_1} = \left( \frac{r_1}{r_2} \right)^2$.The ratio of resistances is $\frac{R_2}{R_1} = \frac{L_2}{L_1} \cdot \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^2 \cdot \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^4$.Given $r_2 = \frac{r_1}{2}$,we have $\frac{r_1}{r_2} = 2$.Substituting this into the ratio,$\frac{R_2}{R} = (2)^4 = 16$.Thus,the new resistance is $R' = 16\, R$.

$A$ wire of resistance $R$ is stretched until its radius is halved. What will be the new resistance (in $, R$)?

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