Question

A solid floats with (1/4)th of its volume above the surface of water. Calculate the density of the solid

Answer 1

Volume of solid = V
Volume of water displaced = 3V/4

Weight of solid = Buoyant Force
mg = F
Vd’g = (3V/4)dg
d’ = 3d/4
d’ = (3 × 1 g/cm³) / 4
d’ = 0.75 g/cm³

Density of solid is 0.75 g/cm³

Answer 2

✬ Density = 750 kg m-³ ✬

Explanation:

Given:

• A solid floats with (1/4)th of its volume above the surface of water.

To Find:

• What is the density of the solid?

Solution: Let V be the volume and p be the density of the solid.

Weight of the body = V p g

• Volume of solid body inside the water:-

$= \: V \: - \frac{V}{4} \\ \\ = > \frac{3V}{4}$

Weight of water displaced by the immersed portion of the solid =

$\frac{3V}{4} \times {10}^{3} \times g$

Using the principal of floatation,

$Vpg \: = \frac{3V}{4} \times {10}^{3} \times g \\ \\ or \: \: p = \frac{3}{4} \times {10}^{3} \\ \\ →750 \: kg \: m ^{ - 3}$