One end of thick horizontal copper wire of length '2L' and radius '2R' is welded to an end of another thin horizontal copper wire of length 'L' and radius 'R'. When they are stretched by applying same force at two ends, the ratio of the elongation in the thick wire to that in thin wire is

Correct option is (\(\mathrm{C}\) )
We have change in length
\(\triangle \mathrm{l}=\frac{\mathrm{F} . \mathrm{l}}{\mathrm{YA}}\)
Since the two rods are in series
\(
\begin{array}{l}
\left(\mathrm{F}_{1}\right)_{\text {rest }}=\left(\mathrm{F}_{2}\right)_{\text {ret }} \\
\Delta \mathrm{l} \propto \frac{1}{\mathrm{R}^{2}} \quad\left(\because \mathrm{A}=\pi \mathrm{R}^{2}\right) \\
\therefore \frac{\Delta \mathrm{l}_{1}}{\Delta \mathrm{l}_{2}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}} \times \frac{\mathrm{R}_{2}{ }^{2}}{\mathrm{R}_{1}{ }^{2}} \\
\frac{\Delta \mathrm{l}_{1}}{\Delta \mathrm{l}_{2}}=\frac{2 \mathrm{~L}}{\mathrm{~L}} \times \frac{\mathrm{R}_{1}{ }^{2}}{4 \mathrm{R}_{1}{ }^{2}} \\
\frac{\Delta \mathrm{l}_{1}}{\Delta \mathrm{l}_{2}}=\frac{1}{2} \\
\therefore \frac{\Delta \mathrm{l}_{2}}{\Delta \mathrm{l}_{1}}=\frac{2}{1}=2
\end{array}
\)