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MCQEasyJEE 2023Viscosity & Stoke's Law

JEE Physics 2023 Question with Solution

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

  • A

    F=Mg(1+ρoρ)F = Mg\left(1 + \frac{\rho_o}{\rho}\right)

  • B

    F=Mg(1+ρρo)F = Mg\left(1 + \frac{\rho}{\rho_o}\right)

  • C

    F=Mg(1−ρoρ)F = Mg\left(1 - \frac{\rho_o}{\rho}\right)

  • D

    F=Mg(1+ρρo)F = Mg\left(1 + \rho\rho_o\right)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A small ball of mass MM and density ρ\rho falls in a viscous liquid of density ρo\rho_o and eventually moves with constant velocity.

Find: The viscous force on the ball.

When the ball moves with constant velocity, its acceleration is zero, so the net force on it is zero. Therefore, the downward weight is balanced by the upward buoyant force and the upward viscous force.

F=mg−FbF = mg - F_b

Using the relation shown in the solution for buoyant force,

F=mg−ρo(mgρ)F = mg - \rho_o\left(\frac{mg}{\rho}\right)

So,

F=mg(1−ρoρ)F = mg\left(\frac{1-\rho_o}{\rho}\right)

This corresponds to the intended option

F=Mg(1−ρoρ)F = Mg\left(1 - \frac{\rho_o}{\rho}\right)

Therefore, the viscous force on the ball is given by F=Mg(1−ρoρ)F = Mg\left(1 - \frac{\rho_o}{\rho}\right), so the correct option is C.

Force Balance at Terminal Velocity

Given: The ball falls with constant velocity in the liquid.

Find: The viscous force acting on the ball.

At constant velocity, the ball has reached terminal velocity. Hence,

net force=0\text{net force} = 0

The forces acting are:

  • weight downward = MgMg
  • buoyant force upward = FbF_b
  • viscous force upward = FF

So the balance is

Mg=F+FbMg = F + F_b

Therefore,

F=Mg−FbF = Mg - F_b

From the extracted solution,

Fb=ρo(Mgρ)F_b = \rho_o\left(\frac{Mg}{\rho}\right)

Substituting,

F=Mg−ρo(Mgρ)F = Mg - \rho_o\left(\frac{Mg}{\rho}\right)F=Mg(1−ρoρ)F = Mg\left(1 - \frac{\rho_o}{\rho}\right)

Therefore, the correct option is C.

Common mistakes

  • Taking the viscous force equal to MgMg ignores the buoyant force. This is wrong because at terminal velocity the weight is balanced by both buoyancy and viscous drag. First subtract buoyancy, then find the viscous force.

  • Using the wrong sign for buoyant force leads to Mg(1+ρoρ)Mg\left(1 + \frac{\rho_o}{\rho}\right). This is incorrect because buoyancy acts upward and reduces the required viscous force. Use F=Mg−FbF = Mg - F_b.

  • Confusing ρ\rho of the ball with ρo\rho_o of the liquid gives the wrong ratio. The buoyant force depends on the liquid density, not the ball density alone. Keep the symbols identified correctly before substitution.

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