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) $2.67 \times 10^{-3}\ °\text{C}^{-1}$
Loss of weight = mass of liquid displaced = volume $\times$ density.
At $30°$C: $V_{30} = \dfrac{20\ \text{g}}{1.5\ \text{g/cc}} = 13.33$ cc. At $40°$C: $V_{40} = \dfrac{18}{1.25} = 14.4$ cc.
$$\gamma = \frac{V_{40} - V_{30}}{V_{30}\,\Delta T} = \frac{1.067}{13.33 \times 10} = 8 \times 10^{-3}\ °\text{C}^{-1}, \qquad \alpha = \frac{\gamma}{3} \approx 2.67 \times 10^{-3}\ °\text{C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics