Question

A wooden cube of side 10 m floats in water with 3 m of its vertical edge lying above the surface of water. The mass of water displaced by the cube is : (Take g = 10 $m{s}^{-2}$ and density of water = 1000 $kg{m}^{-3}$)

• A. 7000 kg
• B. 700 kg
• C. 70000 kg
• D. 700000 kg

Answer 1

The correct option is D

700000 kg

Given, side of the cube = 10 m, volume of the cube = $10\times 10\times 10$ = 1000 $m^3$

Now, length of side of cube immersed in water = 10-3 = 7 m

Volume of water displaced by the submerged part = $10\times 10\times 7$ = 700 $m^3$

Density of water displaced = $\frac{\text{Mass of water displaced}}{\text{Volume of water displaced}}$

1000 = $\frac{\text{Mass of water displaced}}{700}$

Mass of water displaced = 700000 kg