Q. An iron sphere of mass $20 \times 10^{- 3}$kg falls through a viscous liquid with terminal velocity $0.5\ {ms}^{- 1}$. The terminal velocity $(\ in\ {ms}^{- 1})$ of another iron sphere of mass$\ 54 \times 10^{- 2}$ kg is
A) 4.5
B) 3.5
C) 2.5
D) 1.5
Explanation
(a) Terminal velocity $v \propto r^{2}$
$or\ \ \frac{v^{1}}{v^{2}} = \left( \frac{r_{1}}{r_{2}} \right)^{2} = \left\lbrack \left( \frac{r_{1}}{r_{2}} \right)^{3} \right\rbrack^{\frac{2}{3}}$
$\frac{v_{1}}{v_{2}} = \left( \frac{M_{1}}{M_{2}} \right)^{2/3}$
$\frac{0.5}{v_{2}} = \left( \frac{20 \times 10^{- 3}}{54 \times 10^{- 2}} \right)^{2/3}$
$\frac{0.5}{v_{2}} = \frac{1}{9}$ ⟹ $v_{2} = 4.5\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.