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(A) The resistance $R$ of a wire is given by $R = ho \frac{l}{A}$. Since the volume $V = l 	imes A$ remains constant during stretching,$A = \frac{V

Vedclass · Accepted Answer

$12$

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(A) The resistance $R$ of a wire is given by $R = ho \frac{l}{A}$. Since the volume $V = l 	imes A$ remains constant during stretching,$A = \frac{V}{l}$.Substituting this,we get $R = ho \frac{l^2}{V}$,which implies $R \propto l^2$.If the length increases by $10\%$,the new length $l' = l + 0.1l = 1.1l$.The new resistance $R'$ is given by $R' = R 	imes (\frac{l'}{l})^2$.$R' = 10 	imes (1.1)^2 = 10 	imes 1.21 = 12.1\, \Omega$.Note: For small changes,the approximation $\Delta R/R \approx 2 \Delta l/l$ gives a $20\%$ increase,but the exact calculation yields $12.1\, \Omega$.

$A$ wire has a resistance of $10\, \Omega$. If its length is increased by $10\%$ by stretching,what will be the new resistance of the wire in $\Omega$?

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