Q136 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics
A circular copper ring at $30^{\circ} \mathrm{C}$ has a hole with an area of $9.98 \mathrm{~cm}^2$. It is made to slip onto a steel rod of crosssectional area of $10 \mathrm{~cm}^2$, by raising the temperature of both ring and rod simultaneously by an amount $\Delta \mathrm{T}$. If the coefficient of linear expansion of copper and steel are $17 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and $11 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, then minimum value of $\Delta \mathrm{T}$ should be
- Correct answerA$167.6^{\circ} \mathrm{C}$
- B$133.3{ }^{\circ} \mathrm{C}$
- C$83.3{ }^{\circ} \mathrm{C}$
- D$249.9^{\circ} \mathrm{C}$
Solution
For slipping of ring on steel rod
$$
\begin{aligned}
& \mathrm{A}_{\text {ring }}=\mathrm{A}_{\mathrm{rod}} \\
& \Rightarrow \mathrm{A}_{\mathrm{o}}(1+\beta \Delta \mathrm{T})=\mathrm{A}_{\mathrm{o}}{ }^{\prime}\left(1+\beta^{\prime} \Delta \mathrm{T}\right) \\
& \Rightarrow\left(\beta \mathrm{A}_{\mathrm{o}}-\beta^{\prime} \mathrm{A}_{\mathrm{o}}\right) \Delta \mathrm{T}=\mathrm{A}_{\mathrm{o}}^{\prime}-\mathrm{A}_{\mathrm{o}} \\
& \Rightarrow \Delta \mathrm{T}=\frac{\mathrm{A}_{\mathrm{o}}^{\prime}-\mathrm{A}_{\mathrm{o}}}{\beta \mathrm{A}_{\mathrm{o}}-\beta^{\prime} \mathrm{A}_{\mathrm{o}}^{\prime}} \\
& =\frac{10-9.98}{2\left[17 \times 10^{-6} \times 9.98-11 \times 10^{-6} \times 10\right]}=167.6^{\circ} \mathrm{C}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.