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A solid spherical ball of volume V and density ρ1 is allowed to fall in a liquid of density ρ2, with ρ2 < ρ1. 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 velocity is
- √(Vg(σ1 - σ2))/k
- Vgσ1/k
- √(Vgσ1)/k
- Vg(σ1 - σ2)/k
Correct answer: √(Vg(σ1 - σ2))/k
Solution
At terminal velocity the net downward force equals drag: V*g*(rho1 - rho2) = k*v^2, so v = sqrt(V*g*(rho1 - rho2)/k). The stored option omits the square root.
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