Question

Eight identical drops of water falling through air with uniform velocity of $10 \,cm/s$ combine to form a single drop of big size, then terminal velocity of the big drop will be

• A. $80 \,cm/s$
• B. $10 \,cm/s$
• C. $30 \,cm/s$
• D. $40 \,cm/s$

Answer 1

Answer: (D)

Given, terminal velocity of each small drop
$v_{1}=10 cm / s$
When 8 small drops are combined to form bigger drop, then volume will be same. If $r_{1}$ be radius of small drop and $r_{2}$ be radius of bigger drop, then $v_{2}=v_{1}$
$\frac{4}{3} \pi r_{2}^{3}=8 \times \frac{4}{3} \pi r_{1}^{3} \Rightarrow r_{2}=2 r_{1}$
$\therefore $ Terminal velocity of the drop
$v=\frac{2}{9} \frac{r^{2}(\rho-\sigma)}{9 \eta}$
i.e., $v \propto r^{2}$
$\therefore \frac{v_{1}}{v_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{r_{1}^{2}}{\left(2 r_{1}\right)^{2}} \Rightarrow \frac{v_{1}}{v_{2}}=\frac{1}{4}$
$v_{2}=4 v_{1}=4 \times 10=40 cm / s$