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ExamsJEE MainPhysics

A solid spherical ball of volume V is dropped into a liquid of density ρ2. The ball is made of a material whose density is ρ1, with ρ1 > ρ2. The liquid exerts a resistive force on the ball proportional to the square of its speed, given by F = -kv², where k > 0. The ball’s terminal speed is

  1. √(Vg(ρ1-ρ2))/k
  2. Vgρ1/k
  3. √(Vgρ1)/k
  4. Vg(ρ1-ρ2)/k

Correct answer: √(Vg(ρ1-ρ2))/k

Solution

At terminal speed the net downward force (rho1-rho2)*V*g equals the drag k*v^2, so v = sqrt(V*g*(rho1-rho2)/k). The buoyancy-corrected square-root form (option with (rho1-rho2)) is correct, not the buoyancy-free expression.

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