A wire of resistance x Omega is drawn out,so | vedclass

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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.When a wire is stretched,its volume $V = A 	imes l$ remains constant.Let the original

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$5$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.When a wire is stretched,its volume $V = A 	imes l$ remains constant.Let the original length be $l$ and area be $A$. The new length is $l' = 2l$.Since volume is constant,$A 	imes l = A' 	imes l' \implies A 	imes l = A' 	imes (2l) \implies A' = \frac{A}{2}$.The new resistance $R'$ is given by $R' = ho \frac{l'}{A'} = ho \frac{2l}{A/2} = 4 \left( ho \frac{l}{A} ight) = 4R$.Given $R' = 20 \ \Omega$ and $R = x$,we have $20 = 4x$.Therefore,$x = \frac{20}{4} = 5 \ \Omega$.

$A$ wire of resistance $x \ \Omega$ is drawn out,so that its length is increased to twice its original length,and its new resistance becomes $20 \ \Omega$. Then $x$ will be ........ $\Omega$.

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