Question

A solid floats in water with 3/4 of its volume below the surface of water. The density of the solid will be ______________

(A) 75 kgm^−3 (B) 750 kgm^−3 (C) 75×10^2 kgm^−3 (D) 75×10^3 kgm^−3

Answer 1

You have the relation, $\frac{V_{immersed}} {V_{total}} = \frac{p_{object}}{p_{fluid}}$
Then, density of solid, p(object) = 750 kg/m³

Answer 2

weight of water displaced = weight of the solid

Let V be the volume of the solid.

3/4 * V * 1000 kg/m³ * g =  V * density * g

Density = 750 kg/m³