Question

In a cylindrical glass container a solid silica cylinder is placed vertically at its bottom and remaining space is filled with mercury up to the top level of the silica cylinder as shown in the figure . Assume that the volume of silica remains unchanged due to variation in temperature, the coefficient of cubical expansion of mercury is $\gamma$ and the coefficient of linear expansion of glass is $\alpha$. If the top surface of silica and mercury level remains at the same level with the variation in temperature then the ratio of volume of mercury and volume of silica is equal to

Answer 1

Hint: Kinetic energy is the free motion input to a mass to make the identical momentum change or the momentum change that releases the identical amount of free passage of mass onto another mass. Momentum is the result of the product of the mass of the body and its velocity. We will change wavelengths in terms of kinetic energy and mass; then we will get the ratio of wavelengths by putting the mass of electron and proton.

Complete step-by-step solution:
Given that the system of cylindrical glass container, with say volume $v_g$ initially contains silica of volume $v_s$ and liquid mercury of volume $v_m$.
Then, we have
$v_g=v_s+v_m$
When the system is heated, and the temperature is increased by and let the increase in volume of the glass cylinder be marked as $\Delta V_g$ and if $\Delta V_m$ and $\Delta V_s$ be the increase in volume of the mercury and silica, then
$\Delta V_g=\Delta V_m+\Delta V_s$
Given that $\Delta V_s=0$
Then we have, $\Delta V_g=\Delta V_m$,
From thermal expansion, we know that $\Delta V=V\times \alpha \times \Delta T$, where $\alpha$ is the coefficient of thermal expansion and $V$ is the initial volume.
Also given that the coefficient of cubical expansion of mercury is $\gamma$ and the coefficient of linear expansion of glass is $\alpha$ and $\alpha_v=3\alpha_l$, where$\alpha_v$ and $\alpha_l$ are the coefficient of volume expansion and the coefficient of linear expansion respectively.
Substituting, we have
$\implies \Delta T\times 3\alpha \times v_g=\Delta T\times \gamma \times v_m$
$\implies 3\alpha \times v_g=\gamma\times v _m$
$\implies 3\alpha(v_s+v_m)=\gamma \times v_m$
$\implies 3\alpha\left(\dfrac{v_s}{v_m}+1\right)=\gamma$
$\implies \left(\dfrac{v_s}{v_m}+1\right)=\dfrac{\gamma }{3\alpha}$
$\implies \dfrac{v_s}{v_m}=\dfrac{\gamma }{3\alpha}-1$
Thus the ratio of volume of mercury and volume of silica is equal to $\dfrac{v_s}{v_m}=\dfrac{\gamma }{3\alpha}-1$

Note: Thermal expansion can change the shape, area, volume and/or the density of the given substance. And $\alpha_l$ which is the coefficient of linear expansion. However, other expansions can also be written with respect to $\alpha_l$ as $\alpha_v=3\alpha_l$ and $\alpha_a=2\alpha_l$ are the coefficient of volume expansion and coefficient of area expansion respectively mainly for isotropic solids.