First slide

Viscosity

Question

Two equal drops are falling through air with a steady velocity of 15 cm/s. If the drops coalesce, the new terminal velocity will be

Easy

A

15/(2)cm/s
B

$5 \times (4)^{2 / 3} c m / s$
C

$15 \times (4)^{1 / 3} cm / s$
D

$5 / (4)^{2 / 3} c m / s$

Solution

When the two drops coalesce,
$r^{'} = (2)^{1 / 3} r We know that V \propto r^{2} So, terminal velocity = 15 \times {(2^{1 / 3})}^{2} = 15 \times (4)^{1 / 3} cm / s$

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