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
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Total weight is balanced by the sum of buoyant forces from all displaced liquids.
Step 1: Understanding the Question:
This is a problem of buoyancy and equilibrium in a layered fluid. Step 2: Key Formula or Approach:
Upthrust \( U = \text{Weight of fluid displaced} \).
In equilibrium, Weight of cylinder \( W = \text{Total Upthrust} \). Step 3: Detailed Explanation:
Let the top liquid have density \( \rho \) and the bottom denser liquid have density \( 3\rho \).
Length of cylinder in top liquid \( = 3l/4 \).
Length of cylinder in bottom liquid \( = l/4 \).
Total upthrust \( U = \text{Upthrust from top liquid} + \text{Upthrust from bottom liquid} \).
\[ U = (A \times 3l/4) \rho g + (A \times l/4) 3\rho g \]
\[ U = \frac{3}{4} A l \rho g + \frac{3}{4} A l \rho g = \frac{6}{4} A l \rho g = \frac{3}{2} A l \rho g \]
Weight of cylinder \( W = (A \times l) d g \).
Equating \( W \) and \( U \):
\[ A l d g = \frac{3}{2} A l \rho g \implies d = \frac{3}{2} \rho \]
Step 4: Final Answer:
The density of the cylinder is \( \frac{3}{2} \rho \).
