Question

A solid cylinder of length \(l\) and cross-sectional area \(\frac{a}{5}\) is immersed such that it floats with its axis vertical at the liquid-liquid interface with length \(l / 4\) in the denser liquid as shown in figure. The lower density liquid ( \(\rho\) ) is open to atmosphere having pressure \(\mathrm{P}_0\). The density d of solid cylinder is

• A. \(\frac{1}{2}\rho\)
• B. \(\frac{3}{2}\rho\)
• C. \(\frac{3}{4}\rho\)
• D. \(\rho\)

Hint:

The weight of the cylinder equals the sum of the buoyant forces from the two liquids; use the immersed lengths in each.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
From the figure, the upper liquid has density \(\rho\) and the lower liquid has density \(3\rho\).
The cylinder floats in equilibrium, so weight \(=\) total buoyant force.

Step 2: Immersed Lengths
Length in the denser liquid \(=\dfrac{l}{4}\), so length in the lighter liquid \(=\dfrac{3l}{4}\). Let the cross-sectional area be \(S\).

Step 3: Balance of Forces
\(S\,l\,d\,g=S\left(\dfrac{3l}{4}\right)\rho g+S\left(\dfrac{l}{4}\right)(3\rho)g\)

Step 4: Calculation
\(d=\dfrac{3\rho}{4}+\dfrac{3\rho}{4}=\dfrac{3}{2}\rho\)
The atmospheric pressure and the area cancel out.

Final Answer:
The density of the cylinder is 3/2 times rho. \[ \boxed{\text{(B) }\dfrac{3}{2}\rho} \]