Question

A cubical block of coefficient of linear expansion $\alpha _{s}$ is submerged partially inside a liquid of coefficient of volume expansion $\gamma _{\ell }.$ On increasing the temperature of the system by $\Delta T,$ the height of the cube inside the liquid remains unchanged. Find $\frac{\gamma _{\ell }}{\alpha _{s}}=?$

Answer 1

Let $L$ be the side of the cube at initial temperature and d the depth of cube submerged. Then according to the law of floatation, the weight of solid is the same as the weight of the liquid displaced,
$\Rightarrow mg=L^{2}d\rho g....\left(1\right)$
When the temperature is increased, the weight remains the same, the length of the side of the cube increases, the density of liquid decrease and the depth remains unchanged (as given)
$m g=\left(L^{\prime}\right)^{2} d \rho^{\prime} g$
$\rho^{\prime}=\frac{\rho}{(1+\gamma \Delta T)}$ and $L^{\prime}=L(1+\alpha \Delta T)$
$\Rightarrow\left(1+2 \alpha_{s} \Delta T\right)\left(1-\gamma_{l} \Delta T\right)=1 \Rightarrow \frac{\gamma_{l}}{\alpha_{s}}=2$