JEE Main 2023 · 6 April · morning · Single correct · +4 / −1

A small ball of mass M and density ρ is dropped in a viscous liquid of density ρ₀. After some time, the ball falls with a constant velocity. What is the viscous force on the ball ?

  1. (a) F=Mg(1-ρ_O/ρ)
  2. (b) F=Mg(1+ρ/Pₒ)
  3. (c) F=Mg(1+ρₒ/ρ)
  4. (d) F=M g(1 ± ρ ρ₀)
Answer and solution

Answer (a) F=Mg(1-ρ_O/ρ)

Solution

Physics sir · 4 steps

Constant velocity means zero net force, so viscous drag is just weight minus buoyancy.

  1. Forces on the ball. Weight Mg acts down. Buoyancy acts up and equals the weight of liquid displaced. Viscous drag F also acts up, opposing the downward motion.
  2. Find the buoyancy. The ball's volume is V = M/ρ. The displaced liquid has mass Vρ₀ = Mρ₀/ρ, so the buoyant force is Mg ρ₀/ρ.
  3. Balance at terminal velocity. At constant velocity the acceleration is zero, so the upward forces add up to the weight: F + Mg ρ₀/ρ = Mg.

    F = Mg - Mgρ₀/ρ = Mg(1 - ρ₀/ρ)

  4. So. The viscous force is Mg(1 - ρ₀/ρ), which is option (a).

The idea to keep. At terminal velocity, write the force balance: weight equals buoyancy plus drag. Buoyancy is always (volume) times (liquid density) times g, with volume = M/ρ.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Fluid Mechanics

Asked about 90 times in JEE Main since 2020. This question is from 2023.