Question

A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $d$. If $T$ is the surface tension, then the diameter of the drop of the liquid is

• A. $\sqrt{\frac{6T}{g(2\rho - d)}}$
• B. $\sqrt{\frac{T}{g(2\rho - d)}}$
• C. $\sqrt{\frac{2T}{g(2\rho - d)}}$
• D. $\sqrt{\frac{12T}{g(2\rho - d)}}$

Hint:

When setting up buoyancy and weight balance equations, remember to track the density variables clearly! Weight tracks the drop's own density ($\rho$) across its entire volume, while buoyancy tracks the displaced fluid's density ($d$) only across its submerged fractional volume. Keeping this distinction sharp prevents algebra sign errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Set up the buoyancy-weight equilibrium for a liquid drop, carefully distinguishing the drop's own density from the surrounding fluid's density.

Step 2: Key Formula or Approach:
Weight of drop = ρ × V_total × g. Buoyant force = d × V_submerged × g. At equilibrium, these balance: ρV_total = dV_submerged.

Step 3: Detailed Explanation:
The drop's weight tracks its own density ρ across the entire volume. The buoyant upthrust tracks the displaced fluid's density d only across the submerged fraction. Maintaining this clear separation—ρ belongs to the drop, d belongs to the fluid—prevents sign errors and incorrect cancellations when solving for the submerged fraction or the density ratio. This bookkeeping clarity is essential for all buoyancy problems.

Step 4: Final Answer:
Weight uses ρ over total volume; buoyancy uses d over submerged volume.