# A 1kg lead(density=11200kg/m^3) block floats in mercury(13600kg/m^3). Determine the submerged volume of the block,the % of the block submerged and whether the object will submerge fully in water(1k kg/m^3)?

Aug 10, 2015

Volume submerged $= 7.35 \times {10}^{- 5} {m}^{3}$
Percentage submerged = 82.36%
The object will be completely submerged in water.

#### Explanation:

$1 {m}^{3}$ needs to displace $11200$ Kgm of mercury in order to float.

$11200$ Kgm of mercury has a volume of
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{11200}{13600} {m}^{3} = \frac{14}{17} {m}^{3}$

For each ${m}^{3}$ of lead floating in mercury
$\textcolor{w h i t e}{\text{XXXX}}$14/17 m^3 = 82.36% will be submerged

$1$ kgm of mercury has a volume of
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{1}{11200} {m}^{3}$

82.36%# of $\frac{1}{11200} {m}^{3}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= 7.35 \times {10}^{- 5} {m}^{3}$

Since the density of water is less than the density of lead, it is impossible for water to support the block of lead.