Q128 / 150 · MHT CET 2025 (26 Apr Shift 1) · Physics
The difference in length between two rods $A$ and $B$ is 60 cm at all temperatures. If $\propto_A=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and $\propto_B=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, then the length of $\operatorname{rod} A$ and $\operatorname{rod} B$ at $0^{\circ} C$ is respectively
- A$\ell_{\mathrm{A}}=120 \mathrm{~cm}, \ell_{\mathrm{B}}=60 \mathrm{~cm}$.
- Correct answerB$\ell_{\mathrm{A}}=180 \mathrm{~cm}, \ell_{\mathrm{B}}=120 \mathrm{~cm}$.
- C$\ell_{\mathrm{A}}=240 \mathrm{~cm}, \ell_{\mathrm{B}}=180 \mathrm{~cm}$.
- D$\ell_{\mathrm{A}}=270 \mathrm{~cm}, \ell_{\mathrm{B}}=210 \mathrm{~cm}$.
Solution
The constant length difference of 60 cm at all temperatures implies the thermal expansion terms must cancel. At $T = 0^\circ\text{C}$, $L_{A0} - L_{B0} = 60$.
For arbitrary temperature $T$, the difference is $L_{A0}(1 + \alpha_A T) - L_{B0}(1 + \alpha_B T) = 60$. Substituting the known constant difference yields $(L_{A0}\alpha_A - L_{B0}\alpha_B)T = 0$ for all $T$, requiring $L_{A0}\alpha_A = L_{B0}\alpha_B$.
Given $\alpha_A = 18 \times 10^{-6}/^\circ\text{C}$ and $\alpha_B = 27 \times 10^{-6}/^\circ\text{C}$, we obtain $L_{A0}/L_{B0} = \alpha_B/\alpha_A = 3/2$.
Substituting $L_{A0} = (3/2)L_{B0}$ into the length difference equation gives $(1/2)L_{B0} = 60$, hence $L_{B0} = 120$ cm and $L_{A0} = 180$ cm.
The solution corresponds to option B.
Hence, the correct answer is option: 'B'.