Question

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

Answer 1

Given:
Length of the edge of the metal block, x = 12 cm
Specific gravity of mercury, ${\rho }_{Hg}$= 13.6 gm/cc
It is given that $\frac{1}{5}$th of the cubical block is inside mercury initially.
Let ${\rho }_b$ be the density of the block in gm/cc.
$$\begin{gathered} \therefore (x{)}^3\times {\rho }_b\times g=(x{)}^2\times \left(\frac{x}{5}\right)\times {\rho }_{\mathrm{Hg}}\times g \\ \Rightarrow (12{)}^3\times {\rho }_b\times g=(12{)}^2\times \frac{12}{5}\times 13.6 \\ \Rightarrow {\rho }_b=\frac{13.6}{5}\mathrm{gm}/\mathrm{cc} \end{gathered}$$
Let y be the height of the water column after the water is poured.
∴ V_b = V_Hg + V_w = (12)³
Here,
V_Hg = Volume of the block inside mercury
V_w = Volume of the block inside water
$$\begin{gathered} \therefore (V_{\mathrm{b}}\times {\rho }_b\times g)=(V_{\mathrm{Hg}}\times {\rho }_{\mathrm{Hg}}\times g)+(V_{\mathrm{w}}\times {\rho }_{\mathrm{w}}\times g) \\ \Rightarrow (V_{\mathrm{Hg}}+V_w)\times \frac{13.6}{5}=V_{\mathrm{Hg}}\times 13.6+V_{\mathrm{w}}\times 1 \\ \Rightarrow (12{)}^3\times \frac{13.6}{5}=(12-y)\times (12{)}^2\times 13.6+\left(y\right)\times (12{)}^2\times 1 \\ \Rightarrow 12\times \frac{13.6}{5}=(12-y)\times 13.6+(y) \\ \Rightarrow 12.6y=13.6\left(12-\frac{12}{5}\right)=(13.6)\times (9.6) \\ \Rightarrow y=\frac{(9.6)\times (13.6)}{(12.6)}=10.4\mathrm{cm} \end{gathered}$$