Physics - Fluid Question with Solution | TestHub

For the cylinder to be in equilibrium, the total downward gravitational force (weight) must be balanced by the total upward buoyant force exerted by the two liquids.

Let the cross-sectional area of the cylinder be $A$ and its total length be $l$.
The weight of the cylinder is given by:
$W=\text{Volume}\times \text{density}\times g=(A\cdot l)\cdot d\cdot g$

The cylinder is submerged in two liquids. The length of the cylinder in the denser liquid (lower liquid) is $\frac{l}{3}$.
Therefore, the length of the cylinder in the lighter liquid (upper liquid) is $l-\frac{l}{3}=\frac{2l}{3}$.

The buoyant force is the sum of the buoyant forces from each liquid:
Buoyant force from the upper liquid ($F_{B1}$) = Volume in upper liquid $\times$ density of upper liquid $\times g$
$F_{B1}=\left(A\cdot \frac{2l}{3}\right)\cdot \rho \cdot g$

Buoyant force from the lower liquid ($F_{B2}$) = Volume in lower liquid $\times$ density of lower liquid $\times g$
$F_{B2}=\left(A\cdot \frac{l}{3}\right)\cdot (3\rho )\cdot g$

Equating the weight of the cylinder to the total buoyant force:
$W=F_{B1}+F_{B2}$
$A\cdot l\cdot d\cdot g=\left(A\cdot \frac{2l}{3}\right)\cdot \rho \cdot g+\left(A\cdot \frac{l}{3}\right)\cdot (3\rho )\cdot g$

Dividing both sides by $A\cdot l\cdot g$:
$d=\frac{2\rho }{3}+\frac{3\rho }{3}$
$d=\frac{5\rho }{3}$

Answer: $\frac{5\rho }{3}$