Question

A cubical metal block of edge 12 cm 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 = 13.6.

Answer 1

Given, x= 12 cm

= Length of the edge of the block

$P_{Hg}$ = 13.6 gm/cc

Given that initially $\frac{1}{5}$ of block is inside mercury.

Let Pb $\to$ density of block in gm/cc.

$\therefore (x{)}^3\times {\rho }_b\times g=(x{)}^2\times \left(\frac{x}{5}\right)\times \rho Hg\times g$

$\Rightarrow (12{)}^3\times {\rho }_b\times g=(12{)}^2\times \frac{12}{5}\times 13.6$

${\rho }_b=\frac{13.6}{5}$ gm/cc

After water poured, let x =height of water column.

$V_b=V_{Hg}+V_w=(12{)}^3$

Where $V_{H}g\text{and}{V}_w$ are volumes of block inside mercury and water respctively.

$\therefore (V_b\times {\rho }_b\times g)=(V_{H}g\times {\rho }_{Hg}\times g)+(V_w\times {\rho }_w\times g)$

$\Rightarrow (V_{Hg}\times V_m)\times \frac{13.6}{5}$

= $V_{Hg}\times 13.6+V_w\times 1$

$\Rightarrow (12{)}^3\times \frac{13.6}{5}=(12-x)\times (12{)}^2\times 13.6+\left(x\right)\times (12{)}^2\times 1$

$\Rightarrow 12\times \frac{13.6}{5}=(12-x)\times 13.6+\left(x\right)$

$\Rightarrow 12.6x=13.6(12-\frac{12}{5})$

= (13.6) $\times$ (9.6)

$\Rightarrow x=\frac{\left(9.6\right)\times \left(13.6\right)}{\left(12.6\right)}=10.4$cm