Correct option is (iv) $200 \Omega$
Resistance of wire $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{AL}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{~V}}$
$\Longrightarrow \mathrm{R} \propto \mathrm{L}^{2}$ (Since $\mathrm{V}$ and $\rho$ are constant)
Given: $\mathrm{L}^{\prime}=2 \mathrm{~L}$
$$
\begin{array}{c}
\Longrightarrow \frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{\left(\mathrm{L}^{\prime}\right)^{2}}{\mathrm{~L}^{2}}=\frac{4 \mathrm{~L}^{2}}{\mathrm{~L}^{2}}=4 \\
\Longrightarrow \mathrm{R}^{\prime}=4 \mathrm{R}=4 \times 50=200 \Omega
\end{array}
$$
The length of a metallic wire of $50 \Omega$ resistance is stretched two times its initial length. Its new resistance is
i) $25 \Omega$
ii) $50 \Omega$
iii) $100 \Omega$
iv) $200 \Omega$
i) $25 \Omega$
ii) $50 \Omega$
iii) $100 \Omega$
iv) $200 \Omega$
The length of a metallic wire of $50 \Omega$ resistance is stretched two times its initial length. Its new resistance is
i) $25 \Omega$
ii) $50 \Omega$
iii) $100 \Omega$
iv) $200 \Omega$
i) $25 \Omega$
ii) $50 \Omega$
iii) $100 \Omega$
iv) $200 \Omega$