A spherical solid ball of volume V is made of a material of density $\rho_{1}$. It is falling through a liquid of density $\rho_{2}\left(\rho_{2} 0)$. The terminal speed of the ball is

(A) $\sqrt{\frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}}$

(C) $\sqrt{\frac{V g \rho_{1}}{k}}$

(D) $\frac{\mathrm{Vg}\left(\rho_{1}-\rho_{2}\right)}{\mathrm{k}}$

JEE Main · 2008 · 27 Apr 2008 · Offline · Fluid Mechanics

A spherical solid ball of volume V is made of a material of density

ρ_{1}

. It is falling through a liquid of density

ρ_{2} (ρ_{2} < ρ_{1})

. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed

v

, i.e.,

F_{viscous} = - k v^{2} (k > 0)

. The terminal speed of the ball is

Answer: (A)

At the terminal speed the ball moves steadily, so the three forces on it balance. Gravity pulls down with

ρ_{1} V g

, while buoyancy

ρ_{2} V g

and the viscous drag

k v^{2}

both act upward. The balance is

ρ_{1} V g = ρ_{2} V g + k v_{T}^{2}

, so

k v_{T}^{2} = (ρ_{1} - ρ_{2}) V g

, and solving for the terminal speed gives

v_{T} = \frac{V g ( ρ _{1} - ρ _{2} )}{k}

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option