Physics - Current Electricity Question with Solution | TestHub

Let length and cross-section area of the wire be l and A respectively. $\therefore$ Resistance $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$ where $\rho$ is the resistivity of the material Given: $R=4\Omega$ For $l^{\prime }=2l$ and $A^{\prime }=\frac{A}{2}$, resistance of the wire be $R^{\prime }$ $\begin{array}{rl}& \therefore R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}=\rho \frac{2l}{A/2}=4\rho \frac{1}{A}=4R\\ & \therefore R^{\prime }=4\times 4\Omega =16\Omega \end{array}$