Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
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
Solution:
2141 Upvotes
Verified Answer
The correct answer is:
$1: 2$
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
$$
\begin{array}{l}
\left(\mathrm{F}_{1}\right)_{\text {rest }}=\left(\mathrm{F}_{2}\right)_{\text {ret }} \\
\Delta \mathrm{l} \propto \frac{1}{\mathrm{R}^{2}} \quad\left(\because \mathrm{A}=\pi \mathrm{R}^{2}\right) \\
\therefore \frac{\Delta \mathrm{l}_{1}}{\Delta \mathrm{l}_{2}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}} \times \frac{\mathrm{R}_{2}{ }^{2}}{\mathrm{R}_{1}{ }^{2}} \\
\frac{\Delta \mathrm{l}_{1}}{\Delta \mathrm{l}_{2}}=\frac{2 \mathrm{~L}}{\mathrm{~L}} \times \frac{\mathrm{R}_{1}{ }^{2}}{4 \mathrm{R}_{1}{ }^{2}} \\
\frac{\Delta \mathrm{l}_{1}}{\Delta \mathrm{l}_{2}}=\frac{1}{2} \\
\therefore \frac{\Delta \mathrm{l}_{2}}{\Delta \mathrm{l}_{1}}=\frac{2}{1}=2
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.