Question

A spherical solid ball of volume V is made of a material of density ρ. It is falling through a liquid of density ρ₂ (ρ₂ < ρ₁). Assume that the liquid applies a viscous force on the ball that is proportional to square of its speed V. i.e. Fviscous = – KV² (K > 0). The terminal speed of the ball is 

(A) [{Vgρ₁} / K}] 

(B) √[{Vgρ₁} / K}] 

(C) [{Vg(ρ₁ – ρ₁)} / K}] 

(D) √[{Vg(ρ₁ – ρ₂)} / K}]

Answer 1

Correct option: (D) √[{Vg(ρ₁ – ρ₂)} / K}]

Explanation:

Weight of spherical ball = mg

= V ∙ ρ₁ ∙ g ------ mass = density × volume

Buoyancy = V ∙ ρ₂ ∙ g

Given: Fviscous = – KV² 

The ball will acquire terminal speed in the state of equilibrium i.e. F_net = 0

∴ Vρ₂g – Vρ₁g + KV² = 0

∴ KV² = Vρ₁g – Vρ₂g

V² = [{Vg(ρ₁ – ρ₂)} / K]

V = √[{Vg(ρ₁ – ρ₂)} / K]