Question

A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to

• A. $R^{2}$
• B. $R$
• C. $1/R$
• D. $I/R^2$

Answer 1

Answer: (A)

$R^{2}$

Answer 2

Since sphere is moving with constant velocity, there is no acceleration in it. When the sphere of radius $R$ is falling in a liquid of density $\sigma$ and coefficient of viscosity $\eta$ it attains a terminal velocity $v$, under two forces (i) Effective force acting downward $=V(\rho-\sigma) g=\frac{4}{3} \pi R^{3}(\rho-\sigma) g$ where $\rho$ is density of sphere. (ii) Viscous forces acting upwards $=6 \pi \eta R v$ Since the sphere is moving with a constant velocity $v$, there is no acceleration in it, the net force acting on it must be zero. That is $6 \pi \eta R v =\frac{4}{3} \pi R^{3}(\rho-\sigma) g$ $\Rightarrow v =\frac{2}{9} \frac{R^{2}(\rho-\sigma) g}{\eta}$ $\Rightarrow v \propto R^{2}$ Thus, terminal velocity is proportional to the square of its radius.