Question:
A large drop of oil (density 0.8 g/cm³ and viscosity \( \eta_0 \)) floats up through a column of another liquid (density 1.2 g/cm³ and viscosity \( \eta_L \)). Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on:
A large drop of oil (density 0.8 g/cm³ and viscosity \( \eta_0 \)) floats up through a column of another liquid (density 1.2 g/cm³ and viscosity \( \eta_L \)). Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on:
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The velocity of an oil drop rising in a liquid depends on the viscosity of the liquid, not the oil.
Updated On: Mar 25, 2026
- \( \eta_0 \) only
- \( \eta_L \) only
- both on \( \eta_0 \) and \( \eta_L \)
- neither \( \eta_0 \) nor \( \eta_L \)
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The Correct Option is B
Solution and Explanation
Step 1: Stokes' law.
The velocity of a drop rising in a liquid is given by Stokes' law: \[ v = \frac{2r^2 (\rho_{\text{oil}} - \rho_{\text{liquid}}) g}{9 \eta_L} \] This shows that the velocity depends on the viscosity of the liquid \( \eta_L \), not the oil's viscosity.
Thus, the correct answer is (2).
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