Question

A solid cylinder of length \( l \) and cross-sectional area \( a \) is immersed such that it floats with its axis vertical at the liquid-liquid interface with length \( l/4 \) in the denser liquid (\(\rho\)) as shown in figure. The lower density liquid (\( \rho/3 \)) is open to atmosphere having pressure \( P_0 \). 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:

Total weight is balanced by the sum of buoyant forces from all displaced liquids.

Answer 1

The Correct Option is A

Step 1: Principle of Floatation
Weight of cylinder = Upthrust from Liquid 1 + Upthrust from Liquid 2.
Step 2: Weight and Upthrust Calculation
Weight $= (a \cdot l) \cdot d \cdot g$
Upthrust (Denser liquid, $\rho$) $= (a \cdot l/4) \cdot \rho \cdot g$
Upthrust (Lighter liquid, $\rho/3$) $= (a \cdot 3l/4) \cdot (\rho/3) \cdot g$
Step 3: Balancing Equations
$a \cdot l \cdot d \cdot g = \frac{a \cdot l \cdot \rho \cdot g}{4} + \frac{a \cdot l \cdot \rho \cdot g}{4}$
$d = \frac{\rho}{4} + \frac{\rho}{4} = \frac{\rho}{2}$
Step 4: Conclusion
The density of the cylinder is $\frac{1}{2}\rho$.
Final Answer:(A)