SATHEE: Properties Of Solids And Liquids Question 222

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) $5 \times {(4)}^{1 / 3} c m$ per sec

D) $5 \times \sqrt{2} c m$ 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 $\frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} + \frac{4}{3} π r^{3}$

therefore $R^{3} = 2 r^{3} \Rightarrow R = 2^{1 / 3} r$ .

If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $v \propto r^{2}$

$\frac{v_{2}}{v_{1}} = {(\frac{R}{r})}^{2} = {(\frac{2^{1 / 3} r}{r})}^{2}$

therefore $v_{2} = 2^{2 / 3} \times v_{1} = 2^{2 / 3} \times (5) = 5 \times {(4)}^{1 / 3} m / s$

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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:

Answer:

Solution:

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