Question

Two non-mixing liquids of densities $1000{\text{kg/m}}^3$ and $2000{\text{kg/m}}^3$ having height of $1\text{m}$ each are taken in a container, as shown in the figure. When a solid cylinder of length $0.5\text{m}$ and density $\rho$ is inserted in this container, it is observed that the cylinder floats with its axis being vertical and a length $0.2\text{m}$ of it in the denser liquid. The density $\rho$ of the solid cyclinder will be equal to

• A. $1500{\text{kg/m}}^3$
• B. $1300{\text{kg/m}}^3$
• C. $1600{\text{kg/m}}^3$
• D. $1400{\text{kg/m}}^3$

Answer 1

The correct option is D $1400{\text{kg/m}}^3$

Given,
Density of liquid in the upper half of container ${\rho }_1=1000{\text{kg/m}}^3$
Density of liquid in the lower half of container ${\rho }_2=2000{\text{kg/m}}^3$


From the figure given in the question, we can deduce that, at equilibrium, force of buoyancy due to both liquids should balance the weight of the cylinder (as shown in figure-2).

Let $F_{b1}$ and $F_{b2}$ be the force of buoyancy due to liquids of density ${\rho }_1$ and ${\rho }_2$.
$\therefore F_{b1}=m_1\times g={\rho }_1\times V_1\times g$
$=1000\times g\times [0.3\times A]=300gA.......\left(1\right)$
Similarly, force of buoyancy due to liquid of density ${\rho }_2$
$F_{b_2}=2000\times g\times [0.2\times A]=400gA.........\left(2\right)$
For equilibrium (floating) of cylinder (From figure-2)
$mg=F_{b_1}+F_{b_2}$
By using $\left(1\right)\text{and}\left(2\right)$, we get
$\Rightarrow 0.5A\rho g=300Ag+400Ag$
$\Rightarrow 0.5\rho =700$
$\Rightarrow \rho =1400{\text{kg/m}}^3$
Hence, option (d) is the correct answer.