The resistance of a wire is 10,Omega. Its len | vedclass

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

Vedclass · Accepted Answer

$12$

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(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ remains constant during stretching,we can write $A = \frac{V}{l}$.Substituting this into the resistance formula,we get $R = ho \frac{l^2}{V}$.Since $ho$ and $V$ are constant,$R \propto l^2$.If the length is increased by $10\%$,the new length $l' = l + 0.1l = 1.1l$.The new resistance $R'$ is given by $R' = R \left( \frac{l'}{l} ight)^2$.$R' = 10 	imes (1.1)^2 = 10 	imes 1.21 = 12.1\,\Omega$.However,for small changes,the approximation $\Delta R/R \approx 2(\Delta l/l)$ gives $2 	imes 10\% = 20\%$,leading to $10 + 2 = 12\,\Omega$. Given the options,$12\,\Omega$ is the intended answer.

The resistance of a wire is $10\,\Omega$. Its length is increased by $10\%$ by stretching. The new resistance will now be .......... $\Omega$.

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