Physics - Current Electricity Question with Solution | TestHub

Resistance per unit length $=\rho =\frac{R}{2\pi r}$ Lengths of sections $APB$ and $AQB$ are $r\theta$ and $r(2\pi -\theta )$ Resistances of sections $APB$ and $AQB$ are $\begin{array}{rl}& R_1=\rho r\theta =\frac{R}{2\pi r}r\theta =\frac{R\theta }{2\pi }\\ & \text{ and }{R}_2=\frac{R}{2\pi r}r(2\pi -\theta )=\frac{R(2\pi -\theta )}{2\pi }\end{array}$ As $R_1$ and $R_2$ are in parallel between $A$ and $B$, their equivalent resistance is $\begin{array}{rl}& R_{\mathrm{eq}}=\frac{R_1{R}_2}{R_1+R_2}\\ & =\frac{\frac{R\theta }{2\pi }\cdot \frac{R(2\pi -\theta )}{2\pi }}{\frac{R\theta }{2\pi }+\frac{R(2\pi -\theta )}{2\pi }}=\frac{\frac{R^2\theta (2\pi -\theta )}{4{\pi }^2}}{\frac{R}{2\pi }[\theta +2\pi -\theta ]}=\frac{R(2\pi -\theta )\theta }{4{\pi }^2}\\ & \end{array}$