Properties Of Solids And Liquids Question 222
Question: 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 [MP PMT 1990]
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
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} $
therefore $ R^{3}=2r^{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}} $
therefore $ v _{2}={{2}^{2/3}}\times v _{1}={{2}^{2/3}}\times (5)=5\times {{(4)}^{1/3}}m/s $
BETA
