Question

A solid body floats in water with 1/7th of its

volume above the surface of water. What fraction of

its volume will be projected if it floats in a liquid of

density 1250kgm-3

​

Answer 1

Given

• When a solid body is immersed in water,it floats with 1/7th part above water.

• Then it is immersed in a liquid of density 1250 kg/m³

To Find

• Fraction of volume of the body outside the liquid given.

Concept

• When a solid body is immersed in a liquid,a force acts on it which is given by $\rho V'g$ . This force is called buoyant force. In short,this force equals mass of liquid displaced.

Calculations

Let the volume of the solid material be V and mass be m.

Now, when it is immersed in water, Buoyant force(in upward direction) on it equals its weight( in downward direction).So,

mg = $\rho \dfrac{6V}{7} g$

$\implies \: m\:=\: 1000 \times \dfrac{6V}{7}$ .......(i)

Now, Consider the free body diagram in the second liquid.

mg = $\rho V'g$ Where V' is the volume inside liquid.

$\implies \: m\:=\: 1250 \times V'$ .....(ii)

From (i) and (ii)

$1000 \times \dfrac{6V}{7} \:=\: \: 1250 \times V' \\ \\ \\ \implies V' \:=\: 1000 \times \dfrac{6V}{7 \times 1250} \\ \\ \\ \implies V' \:=\: 0.685 \:V$

Thus, Volume of solid outside the liquid = V - 0.685 V = 0.315 V.

Thus, 0.315 (315/100) part of solid remains outside the liquid.

Answer 2

Hope it's helpful for you