JAMB Physics 2019 Paper

Concept:Linear expansivity, $\alpha$, is the fractional change in length per degree change in temperature, given by $\alpha = \dfrac{L_2 - L_1}{L_1(T_2 - T_1)}$.Explanation:From the data: $L_1 = 100 \, \text{cm}$, $T_1 = 200^\circ\text{C}$, $L_2 = 99.4 \, \text{cm}$, and $\alpha = 2 \times 10^{-5} \, {}^{\circ}\text{C}^{-1}$.Since $L_2 < L_1$, the rod contracted, meaning the final temperature $T_2$ is below $200^\circ\text{C}$.Substitute the values into the expansivity formula:$$2 \times 10^{-5} = \frac{99.4 - 100}{100\,(T_2 - 200)}$$Simplify the numerator:$$2 \times 10^{-5} = \frac{-0.6}{100\,(T_2 - 200)}$$Cross-multiply to solve for $T_2$:$$2 \times 10^{-5} \times 100\,(T_2 - 200) = -0.6$$$$0.002\,(T_2 - 200) = -0.6$$Divide both sides by $0.002$:$$T_2 - 200 = -300$$$$T_2 = 200 - 300 = -100^\circ\text{C}$$Answer:The required temperature is $-100^\circ\text{C}$, so the correct option is D.