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(A) Initial resistance of the wire,$R_1 = 2 R$. Let the initial length be $L_1$ and initial area of cross-section be $A_1$.When the wire is stretched

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$6$

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(A) Initial resistance of the wire,$R_1 = 2 R$. Let the initial length be $L_1$ and initial area of cross-section be $A_1$.When the wire is stretched to double its length,the new length $L_2 = 2 L_1$. Since the volume of the wire remains constant,$A_1 L_1 = A_2 L_2$.Substituting $L_2 = 2 L_1$,we get $A_1 L_1 = A_2 (2 L_1)$,which implies $A_2 = A_1 / 2$.The new resistance $R_2$ is given by $R_2 = ho \frac{L_2}{A_2}$.Substituting the values,$R_2 = ho \frac{2 L_1}{A_1 / 2} = 4 \left( ho \frac{L_1}{A_1} ight) = 4 R_1$.Since $R_1 = 2 R$,the new resistance $R_2 = 4 	imes (2 R) = 8 R$.The increase in resistance is $\Delta R = R_2 - R_1 = 8 R - 2 R = 6 R$.

$A$ wire of resistance $2 R$ is stretched such that its length is doubled. Then the increase in its resistance is (in $R$)

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