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(D) Given,initial resistance,$R_{1} = 3 \Omega$.Let the original length of the wire be $l$. When the wire is stretched to twice its length,the new len

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$12$

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(D) Given,initial resistance,$R_{1} = 3 \Omega$.Let the original length of the wire be $l$. When the wire is stretched to twice its length,the new length becomes $l^{\prime} = 2l$.Since the volume of the wire remains constant during stretching,$V = A 	imes l = A^{\prime} 	imes l^{\prime}$.Therefore,$A^{\prime} = \frac{A 	imes l}{l^{\prime}} = \frac{A 	imes l}{2l} = \frac{A}{2}$.The resistance of a wire is given by $R = ho \frac{l}{A}$.The new resistance $R_{2}$ is $R_{2} = ho \frac{l^{\prime}}{A^{\prime}} = ho \frac{2l}{A/2} = 4 \left( ho \frac{l}{A} ight) = 4 R_{1}$.Substituting the value of $R_{1}$,we get $R_{2} = 4 	imes 3 \Omega = 12 \Omega$.

$A$ wire of resistance $3 \Omega$ is stretched to twice its original length. The resistance of the new wire will be (in $\Omega$)

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