A solid whose volume does not change with temperature floats in a liquid. At temperatures $t_1$ and $t_2$ of the liquid, the fractions of the volume of the solid that remain submerged are $f_1$ and $f_2$ respectively. The coefficient of absolute expansion of the liquid is
Answer: (B) $\dfrac{f_1 - f_2}{f_2t_1 - f_1t_2}$
Concept. A floating body's weight equals the buoyant force, so (submerged fraction) $\times$ (liquid density) is constant. The density changes as $\rho = \dfrac{\rho_0}{1 + \gamma t}$.
Step 1. Equate.
$$\frac{f_1}{1 + \gamma t_1} = \frac{f_2}{1 + \gamma t_2} \;\Rightarrow\; f_1 + f_1\gamma t_2 = f_2 + f_2\gamma t_1$$
Step 2. Solve.
$$\gamma = \frac{f_1 - f_2}{f_2t_1 - f_1t_2}$$
Answer. $\dfrac{f_1 - f_2}{f_2t_1 - f_1t_2}$ (Option B).
Solution by Sreeraj P, M.Sc Physics