Question

Volume of an ice block is $1600c{m}^3$ and its density is $\frac{0.8g}{c{m}^3}$. How much part of it remains above the surface of water, when it is kept in water?

• A. 0.2
• B. 0.8
• C. 0.4
• D. 0.5

Answer 1

The correct option is A

0.2

According to the principle of floatation,
Weight of floating body = Weight displaced by the liquid
$Or{W}_{ice}=W_{water}$
$Or{M}_{ice}\times g=M_{water}\times g$
$Or{\rho }_{ice}\times V_{ice}={\rho }_{water}\times V_{water}$
$Or\frac{{\rho }_{ice}}{{\rho }_{water}}=\frac{V_{water}}{V_{ice}}$
Or $\frac{0.8}{1}$ = Part of ice immersed in water
Part of ice immersed in water = $\frac{8}{10}$
We know that, Part of solid immersed in water = $\frac{V_{liquid}}{V_{solid}}$
Part of solid outside the water = $1-\frac{V_{liquid}}{V_{solid}}$
Part of ice outside the water = $1-\frac{8}{10}=\frac{10-8}{10}=\frac{2}{10}=0.2$
Thus, ${0.2}^{th}$ part of volume of ice will remain above the surface of water.