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

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

Answer 1

Answer: (C)

$5 \times ( 4 ) ^ { 1 / 3 } \mathrm {~cm}$ per sec

Answer 2

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

$\frac { v _ { 2 } } { v _ { 1 } } = \left( \frac { R } { r } \right) ^ { 2 } = \left( \frac { 2 ^ { 1 / 3 } r } { r } \right) ^ { 2 }$

⇒ $v _ { 2 } = 2 ^ { 2 / 3 } \times v _ { 1 } = 2 ^ { 2 / 3 } \times ( 5 ) = 5 \times ( 4 ) ^ { 1 / 3 } \mathrm {~m} / \mathrm { s }$