Question

The R.D. of ice is 0.92 and that of sea water is 1.025. Find the total volume of an iceberg which floats with its volume 800 cm³ above water.

Answer 1

Relative density of Ice = 0.92
Relative density of sea water = 1.025
Let the total volume of iceberg = X cm³.
The volume of the iceberg above water = 800 cm³.
The volume of the iceberg is submerged in the water = (X - 800) cm³.
Fraction of iceberg submerged = (X- 800)/X
Now we know that the fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of ice / Density of sea water) = fraction submerged
0.92/1.025 = (X-800)/X
0.8975 X = X - 800
X - 0.8975 X = 800
0.1025 X = 800
X = 800/0.1025 = 7804.8 cm³.
Total volume of iceberg = 7804.8 cm³.