Q106 / 180 · NEET 2013 (Karnataka) · Physics
In the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are $p, q$ and $s$ respectively, then the corresponding ratio of increase in their length would be:
- A$\frac{7 q}{\left(7 s p^2\right)}$
- Correct answerB$\frac{7 q}{\left(5 s p^2\right)}$
- C$\frac{2 q}{(5 s p)}$
- D$\frac{7 q}{(5 s p)}$
Solution
We know that
$$
\text { Young's modulus } \begin{aligned}
\gamma & =\frac{\mathrm{FL}}{\Delta 4 \mathrm{~L}} \\
& =\frac{4 \mathrm{FL}}{\pi \mathrm{D}^2 4 \mathrm{~L}} \\
\Rightarrow \quad \Delta \mathrm{L} & =\frac{4 \mathrm{FL}}{\pi \mathrm{D}^2 \mathrm{~L}}
\end{aligned}
$$
According to questions.
$$
\begin{aligned}
& \frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}}=\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{F}_{\mathrm{C}}} \cdot \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}} \cdot \frac{\mathrm{D}_{\mathrm{C}}^2 \mathrm{~L}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{S}}^2 \mathrm{~L}_{\mathrm{C}}} \\
& \text { Here } \mathrm{F}_{\mathrm{S}}=(5 \mathrm{~m}+2 \mathrm{~m}) \mathrm{g}=7 \mathrm{mg} \\
& \mathrm{F}_{\mathrm{C}}=5 \mathrm{mg},=?, \frac{\mathrm{D}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{C}}}=P, \\
& \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{S}
\end{aligned}
$$
Putting in eq. (ii)
$$
\begin{aligned}
\frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}} & =\frac{7 \mathrm{mg}}{5 \mathrm{mg}} \times q+\left(\frac{1}{\mathrm{P}}\right)^2\left(\frac{1}{\mathrm{~S}}\right) \\
& =\frac{7 q}{5 p^2 \mathrm{~S}}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.