Consider a wire of length L with a resistance | vedclass

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(B) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The resistance of a wire is given by $R = ho \frac{L}{A}$,where $ho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.Since the volume $V = A 	imes L$ remains constant during elongation,we can write $A = \frac{V}{L}$.Substituting this into the resistance formula,we get $R = ho \frac{L}{V/L} = ho \frac{L^2}{V}$.Since $ho$ and $V$ are constant,$R \propto L^2$.Let the initial resistance be $R_1 = 5 \Omega$ and initial length be $L_1 = L$.The new length is $L_2 = 3L$.Therefore,the new resistance $R_2$ is given by:$\frac{R_2}{R_1} = \left( \frac{L_2}{L_1} ight)^2 = \left( \frac{3L}{L} ight)^2 = 3^2 = 9$.$R_2 = 9 	imes R_1 = 9 	imes 5 \Omega = 45 \Omega$.

Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force,the wire is elongated such that its length becomes $3 L$. Assuming the resistivity and density of the material remain unchanged,the resistance of the elongated wire is (in $Omega$)

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