Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the

Q: Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$ If the two drops coalesce, the terminal velocity would be

c (c)If two drops of same radius r coalesce then radius of new drop is given by R $\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3} + \frac{4}{3}\pi {r^3}$==> ${R^3} = 2{r^3} \Rightarrow R = {2^{1/3}}r$ If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $v \propto {r^2}$ $\frac{{{v_2}}}{{{v_1}}} = {\left( {\frac{R}{r}} \right)^2} = {\left( {\frac{{{2^{1/3}}r}}{r}} \right)^2}$ ==> ${v_2} = {2^{2/3}} \times {v_1} = {2^{2/3}} \times (5) = 5 \times {(4)^{1/3}}m/s$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$ If the two drops coalesce, the terminal velocity would be

A$10 cm$ per sec
B$2.5 cm$ per sec
C$5 \times {(4)^{1/3}}cm$ per sec
D$5 \times \sqrt 2 \,cm$ per sec

Diffcult

STD 11 Science
NEET
STD 11 - 9.1. fluid mechanics
Physics

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