A uniform metal wire of length $l$ has $10 \Omega$ resistance. Now this wire is stretched to a lengt

Question

A uniform metal wire of length $l$ has $10 \Omega$ resistance. Now this wire is stretched to a length $2 l$ and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is

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(a) : The volume of wire remains same even after stretching the wire, i.e., $V^{\prime}=V ; A^{\prime} l^{\prime}=A l$ \ $ A^{\prime}=\frac{A l}{l^{\prime}}=\frac{A l}{2 l}=\frac{A}{2} $ \ $ \left(\because l^{\prime}=2 light) $ and Resistance of wire is $R=\frac{ho L}{A}=10 \Omega$ After stretching, resistance of circle becomes \ $ R^{\prime}=ho \frac{(2 l)(2)}{A}=4 R=40 \Omega $ \ As the wire is bent at its mid point, to form a circle, thus, the effective resistance of semi-circle is $=\frac{R^{\prime}}{2}=\frac{40}{2}=20 \Omega$ ∴ Equivalent resistance across diameter of circle, \ $ \begin{aligned} & =20 \Omega \| 20 \Omega \ & \frac{1}{R_{e q}}=\frac{1}{20}+\frac{1}{20} \ 	herefore \quad & R_{e q}=10 \Omega \end{aligned} $