Q. Eight drops of a density $\rho$ and each of radius $a$ are falling through air with a constant velocity $375cm\ s^{- 1}$. When the eight drops coalesce to from a single drop the terminal velocity of the new drop will be
A) $1.5 \times 10^{- 2}\ {ms}^{- 1}$
B) $2.4 \times 10^{- 2}\ {ms}^{- 1}$
C) $0.75 \times 10^{- 2}\ {ms}^{- 1}$
D) $15 \times 10^{- 2}\ {ms}^{- 1}$
Explanation
(d) Terminal velocity of single drop,
$v = 3.75{cms}^{- 1} = 3.75 \times 10^{- 2}\ {ms}^{- 1}$
Terminal velocity of the big drop
$\ V = n^{\frac{2}{3}}v$
$= {(8)}^{2/3} \times 3.75 \times 10^{- 2}$
$= 4 \times 3.75 \times 10^{- 2} = 15 \times 10^{- 2}{ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.