A wire of resistance 4 is stretched to twice its original length....

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}=2 l) and (A^{\prime}=\frac{A}{2}), resistance of the wire be (R^{\prime}) $$\begin{aligned} & \therefore R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{A / 2}=4 \rho \frac{1}{A}=4 R \\ & \therefore R^{\prime}=4 \times 4 \Omega=16 \Omega \end{aligned}$$)$