Question

Body A of density $\left({\rho }_A\right)$ is floating in the water. When another body B having density $({\rho }_B=1500kg{m}^{-3})$ and same dimensions is placed on body A, then the system just sinks in the water. What fraction of body A was originally immersed in the water?
The density of water is ${\rho }_w=1000kg{m}^{-3}$.

• A. 0.5

Answer 1

The correct option is A 0.5
Volume of body A and B are equal,
$V_A=V_B=V$

When the system just sinks, the average density of the system is equal to the density of water.
${\rho }_w=\frac{\text{Total mass}}{\text{Total volume}}$
${\rho }_w=\frac{m_A+m_A}{V_A+V_B}$
${\rho }_w=\frac{{\rho }_A{V}_A+{\rho }_B{V}_B}{V_A+V_B}$
$1000=\frac{V\times {\rho }_A+1500\times V}{V+V}$
$1000=\frac{{\rho }_A+1500}{2}$
${\rho }_A=500kg{m}^{-3}$
The density of A is equal to half of the density of water.

When the system was floating, weight of the body is equal to the weight of the water displaced. Let fraction of A sunk in water be $n$.
$\left(W_A\right)g=F_b$
${\rho }_{A}Vg={\rho }_w\times \left(nV\right)\times g$
$n=\frac{{\rho }_A}{{\rho }_w}$
$n=\frac{1}{2}$