Question

Eight rain drops of radius one mm each falling down with a terminal velocity of 5 cm s^-1 coalesce to form a bigger drop. Calculate the terminal velocity of the bigger drop.

Answer 1

Radius of each small drop,

r = 1 mm = 0.1 cm

Terminal velocity of each small drop,

v_T = 5 cm s^-1

or, V_T = 2/9{r²(ρ - σ)g}/{η}  

or, 5 = 2/9{(0.1)².(ρ - σ)g}/{η} ...(i)

Now, volume of bigger drop = Volume of 8 drops

⇒ 4/3πR³ = 8 x 4/3πR³

or, R = 2r = 2 x 0.1 cm = 0.2 cm,

where R is the radius of the bigger drop

Let v_T' be the terminal velocity of the bigger drop, then,

v_T' = 2/9({2(0.2)² (ρ - σ)g}/{η})   ...(ii)

Dividing (ii) by (i), we get

v_T'/V_T = (0.2)²/(0.1)² = 4

or v_T' = 4 V_T = (4 x 5)cm s^-1 = 20 cm s^-1