Q73 / 90 · JEE Main 2023 (06 Apr Shift 1) · Physics

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?

  • A
    F=Mg1+ρ0ρ
  • B
    F=Mg1+ρρ0
  • Correct answer
    C
    F=Mg1-ρ0ρ
  • D
    F=Mg1±ρρ0

Solution

The buoyant force is given by 

B=Vρ0g

where V is the volume of the ball. The weight of the ball

W=Mg=Vρg, is acting in the opposite direction of the buoyant force . Hence, at terminal velocity

Mg-B=Fv

⇒B+Fv=Mg

⇒Fv=Vρg-Vρ0g⇒Fv=ρVg(1-ρ0ρ)⇒Fv=Mg(1-ρ0ρ)

Hence, the correct answer is option: 'C'.

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