Question

An iceberg is floating in sea water. The density of ice is 0.92 gm/cm³ and that of sea water is 1.03g/cm³. What percentage of the iceberg will be below the surface of water?

• A. (a) 3%
• A. (b) 11%
• A. (c) 89%
• A. (d) 92%

Answer 1

(c)

For the floatation of ice-berg, Weight of ice

= upthrust due to displaced water

$V\rho g = V_{in}\sigma g$

⇒ $V_{in} = \left( \frac{\rho}{\sigma} \right)V = \left( \frac{0.92}{1.03} \right)V = 0.89V$

$\therefore\mspace{6mu}\frac{V_{in}}{V} = 0.89$ or 89%.