Question

A ball of mass m and radius r is released in the viscous liquid. The value of its terminal velocity is proportional to :

• A. $\left(\frac{1}{r}\right)$only
• B. $\frac{m}{r}$
• C. $(\frac{m}{r}{)}^{1/2}$
• D. $m$ only.

Answer 1

Answer: (B)

$\frac{m}{r}$

Lets consider $m=$mass of a ball

$r=$ radius of a ball

$v=$ terminal velocity

By stocks law,

$Viscousforce+Buoyancyforce=weight$

$6\pi \eta rv+\frac{4}{3}\pi r^3\rho g=mg$

$6\pi \eta vr=mg-\frac{4}{3}\pi r^3\rho g$

$v=\frac{mg}{6\pi \eta r}-\frac{4\pi r^3\rho g}{18\pi \eta r}$. . . . . .(1)

If the density $\rho$ is negligible,

Equation (1) become,

$v=\frac{mg}{6\pi \eta r}$

$v\propto \frac{m}{r}$

The correct option is B.