Two drops, of the same radius, are falling through air with a steady velocity of 5 cm s-1 If the two drops coalesce, the terminal velocity would be

Question

Two drops, of the same radius, are falling through air with a steady velocity of 5 cm s-1 If the two drops coalesce, the terminal velocity would be (A) 10 cm s-1 (B) 2.5 cm s-1 (C) 5 × (4)1/3 cm s-1 (D) 5 × √3 cm s-1

Tardigrade Admin · Accepted Answer

If

R

is radius of bigger drop formed, then

\frac{4}{3} π R^{3} = 2 \times \frac{4}{3} π​ r^{3} or R = 2^{1 / 3} r

As,

v_{0} \propto r^{2}

∴

\frac{{(v_{0})}_{1}}{v_{0}} = \frac{R^{2}}{r^{2}} = \frac{(2^{1 / 3} r)^{2}}{r^{2}} = 2^{2 / 3}

or

{(v_{0})}_{1} = v_{0} 2^{2 / 3}

5 \times {(4)}^{1 / 3}

= (5) \times {(2)}^{2 / 3}

= 5 \times {(4)}^{1 / 3} {cm s}^{-1}

Q. Two drops, of the same radius, are falling through air with a steady velocity of $5 c m s^{- 1}$ If the two drops coalesce, the terminal velocity would be

$10 c m s^{- 1}$

$2.5 c m s^{- 1}$

$5 \times (4)^{1/3} c m s^{- 1}$

$5 \times 3 c m s^{- 1}$

Q. Two drops, of the same radius, are falling through air with a steady velocity of 5cms−1 If the two drops coalesce, the terminal velocity would be

10cms−1

2.5cms−1

5×(4)1/3cms−1

5×3​cms−1

If R is radius of bigger drop formed, then 43πR3=2×43π​r3orR=21/3r As, v 0 ∝ r 2 ∴ ( v 0 ) 1 v0=R2r2=(21/3r)2r2=22/3 or ( v 0 ) 1 = v 0 2 2/3 5× ( 4 ) 1/3 =( 5 )× ( 2 ) 2/3 =5×(4)1/3 cm s-1

Q. Two drops, of the same radius, are falling through air with a steady velocity of $5 c m s^{- 1}$ If the two drops coalesce, the terminal velocity would be

$10 c m s^{- 1}$

$2.5 c m s^{- 1}$

$5 \times (4)^{1/3} c m s^{- 1}$

$5 \times 3 c m s^{- 1}$