A wire of resistance 5 Omega is stretched so | vedclass

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(A) The resistance $R$ of a wire is given by $R = \rho \frac{\ell}{A}$.Since the volume $V = A \ell$ remains constant when the wire is stretched, $A =

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$45 \Omega$

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(A) The resistance $R$ of a wire is given by $R = ho \frac{\ell}{A}$.Since the volume $V = A \ell$ remains constant when the wire is stretched, $A = \frac{V}{\ell}$.Substituting this into the resistance formula, we get $R = ho \frac{\ell^2}{V}$.Since $ho$ and $V$ are constant, $R \propto \ell^2$.Given the new length $\ell_2 = 3 \ell_1$, the new resistance $R_2$ is related to the original resistance $R_1$ by the ratio $\frac{R_2}{R_1} = \left(\frac{\ell_2}{\ell_1}ight)^2$.Substituting the values: $\frac{R_2}{5 \Omega} = (3)^2 = 9$.Therefore, $R_2 = 5 \Omega 	imes 9 = 45 \Omega$.

$A$ wire of resistance $5 \Omega$ is stretched so that its new length is $3$ times its original length. What is the resistance of the new wire?

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