Answer to Question #97686 in Physics for Adedayo

Question

A body weighs 0.5 kg in air ,0.3kg in water and 0.2 kg in a liquid .What is the relative density of the liquid

Answer 1

Let the weights of the body in the air, water and liquid be $m_1 = 0.5 kg, m_2 = 0.3 kg, m_3 = 0.2 kg$ respectively.

The net force acting on the body in liquid/gas is $mg - \rho g V_{body}$, where $\rho$ is the density of the liquid/gas.

Hence, the weights, expressed in terms of gravitational force and buoyant force are $m_2 = m_1 - \rho_w V_{body}$, $m_3 = m_1 - \rho_{liq} V_{body}$.

Dividing last two equations, obtain $\frac{\rho_{liq}}{\rho_{w}} = \frac{m_1 - m_2}{m_1 - m_3} = \frac{0.3 kg}{0.2 kg}$ , from where $\rho_{liq} = 1.5 \rho_{w}$, so the liquid is 1.5 times denser than water.