A

Step 1: Define the parameters involved. Let:

• Volume of the cube = $V = a^3$, where $a$ is the side length of the cube.
• Coefficient of linear expansion of the cube = $\alpha_s$.
• Coefficient of volume expansion of the liquid = $\gamma_L$.
• Temperature change = $\Delta T$.

Step 2: Calculate the change in volume of the cube due to the temperature increase. Since the cube experiences linear expansion, its new volume $V'$ can be calculated using the formula: $$ V' = V(1 + 3\alpha_s \Delta T) $$ Step 3: Since the cube is floating, we need to consider the change in volume of the liquid as well. The volume of the liquid displaced by the cube remains the same, which is equal to the submerged volume of the cube. Step 4: For the liquid, its change in volume $V_L$ can be expressed as: $$ V_L = V_{liquid}(1 + \gamma_L \Delta T) $$ Step 5: For the cube to remain at the same depth in the liquid, the increase in the volume due to heating of the cube must equal the increase in the volume of the displaced liquid. Thus, we have: $$ V(1 + 3\alpha_s \Delta T) = V_L(1 + \gamma_L \Delta T) $$ Step 6: As the cube is submerged in liquid, we let $V_L$ be equal to the volume of the cube originally submerged, which we can denote as $V_{submerged}$. Therefore: $$ V(1 + 3\alpha_s \Delta T) = V_{submerged}(1 + \gamma_L \Delta T) $$ Step 7: As the cube is floating, it displaces an amount of liquid equal to its submerged volume. Since $V = a^3$ and for the cube submerged, we require that the relationship remains constant, we can derive: $$ 1 + 3\alpha_s \Delta T = 1 + \gamma_L \Delta T $$ Step 8: Subtracting 1 from both sides gives: $$ 3\alpha_s \Delta T = \gamma_L \Delta T $$ Step 9: Canceling $\Delta T$ (where $\Delta T \neq 0$) leads to the relationship: $$ 3\alpha_s = \gamma_L $$ Step 10: Therefore, the relation between the coefficients of expansion is: $$ \alpha_s = \frac{\gamma_L}{3} $$
Hence, we find that the relationship between the coefficient of linear expansion of the cube and the coefficient of volume expansion of the liquid is $\alpha_s = \frac{\gamma_L}{3}$. This concludes the proof.