Question

A block of density (D) floats in a liquid with its one third volume immersed .the density of the liquid is ??????????????????????????????????????????????????????

Answer 1

Question: a block of density $D$ floats in a liquid with its one third volumes immersed. What is the density of the liquid?

Solution: the block floats in a liquid, so it means that this block is in equilibrium state: the sum of all forces acting on it is zero. There are two forces acting on this block: downward gravitational force $F_{g} = mg$ and upward buoyant force. The buoyant force we find from equation

$$F _ {b} = \rho g V,$$

Here $\rho$ is the density of a liquid and $V$ — volume of immersed part of the body. Since $F_{g} = F_{b}$, we become:

$$\rho g V _ {\text {immersed}} = m g$$ $$\rho g \frac {V}{3} = D V g,$$

Because the mass of the body is $m = D \cdot V$. From last equation we obtain density of the liquid:

$$\rho = 3 D.$$

Answer: $\rho = 3D$.