A piece of metal weighs $45$ g in air and $25$ g in a liquid of density $1.5 \times 10^3\ \text{kg m}^{-3}$ kept at $30°$C. When the temperature of the liquid is raised to $40°$C, the metal piece weighs $27$ g. The density of the liquid at $40°$C is $1.25 \times 10^3\ \text{kg m}^{-3}$. The coefficient of linear expansion of the metal is
Answer: (C) $2.67 \times 10^{-3}\ °\text{C}^{-1}$
Concept. The loss of weight equals the mass of liquid displaced, which gives the metal's volume at each temperature; then $\gamma = \dfrac{\Delta V}{V\Delta T}$ and $\alpha = \dfrac{\gamma}{3}$.
Step 1. Volume at $30°$C. $V_{30} = \dfrac{45 - 25}{1.5} = 13.33$ cc.
Step 2. Volume at $40°$C. $V_{40} = \dfrac{45 - 27}{1.25} = 14.4$ cc.
Step 3. Coefficients.
$$\gamma = \frac{14.4 - 13.33}{13.33 \times 10} = 8 \times 10^{-3}\ °\text{C}^{-1},\qquad \alpha = \frac{\gamma}{3} \approx 2.67 \times 10^{-3}\ °\text{C}^{-1}$$
Answer. $2.67 \times 10^{-3}\ °\text{C}^{-1}$ (Option C).
Solution by Sreeraj P, M.Sc Physics