Question

A cubical metal block of edge $12cm$ floats in mercury with one fifth of the height inside the mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury is $13.6$.

Answer 1

Hint: First, we will have to calculate the density of the block through Newton's third law of motion as we equate the forces acting on the block and the mercury. When the water is poured, the height of the water column increases, and then we again equate the forces according to Newton's third law of motion.

Complete step by step answer:
The length of the cubical metal block is 12cm. Thus supposing that x is the length of the cubical metal block, then x=12cm. The specific gravity of mercury is given as 13.6 and as the density of water is \[1\dfrac{{gm}}{{cc}}\], so the density of the mercury would be \[13.6\,gm/cc\]. Now, let the density of the block be \[{\rho _b}\]. Let the density of mercury be\[{\rho _m}\]. We know that according to Newton’s third law, the force applied on the metal block is equal to the force applied on the mercury.
 \[
{{x}^{3}}{{\rho }_{b}}g={{x}^{2}}\centerdot \left( \dfrac{x}{5} \right)\centerdot {{\rho }_{m}}g \\
\Rightarrow {{12}^{3}}{{\rho }_{b}}={{12}^{2}}\left( \dfrac{12}{5} \right)13.6 \\
\therefore {{\rho }_{b}}=\dfrac{13.6}{5}\dfrac{gm}{cc} \\
\]
When the water is poured, and then let us assume the height of the water column which is increased is denoted by x. Then we have;
\[{V_b} = {V_m} + {V_w} = {\left( {12} \right)^3}\]
Here, ${V_b}$ is the volume of the block, Vm is the volume of mercury and Vw is the volume of water. Thus when comparing the forces acting upon the materials, we have;
\[
{{\rho }_{b}}{{V}_{b}}g={{\rho }_{m}}{{V}_{m}}g+{{\rho }_{w}}{{V}_{w}}g \\
\Rightarrow ({{V}_{m}}+{{V}_{w}}){{\rho }_{b}}={{\rho }_{m}}{{V}_{m}}+{{\rho }_{w}}{{V}_{w}} \\
\Rightarrow ({{V}_{m}}+{{V}_{w}})\centerdot \dfrac{13.6}{5}={{V}_{m}}\times 13.6+{{V}_{w}}\times 1 \\
\Rightarrow {{(12)}^{3}}\centerdot \dfrac{13.6}{5}=(12-x)\centerdot {{(12)}^{2}}\times 13.6+{{(x)}^{2}}\centerdot {{(12)}^{2}} \\
\therefore x=10.4cm \\
\]
Thus, the height of the water column to be poured is 10.4cm.

Note:We assume that there is very less frictional force between the block and the mercury, that is that the viscosity of mercury is negligible, and the only force which is being applied on the block is the buoyancy force and the gravitational force which is why there is no other force term which balances out in the equation.