VITEEE 2009 Solved Paper

Let radius of big drop = R.
∴

4

3

πR3=

4

3

×πr3.8
R = 2r
Here, r = radius of small drops.
Now, terminal velocity of drop in liquid
vT=

2

9

×

r2

η

(ρ−σ)g
where η is coefficient of viscosity and ρ is density of drop, σ is density of liquid.
Terminal speed drop is 6cm s‌−1
∴ 6=

2

9

×

r2

η

(ρ−σ)g...(i)
Let terminal velocity becomes v' after coalesce, then
v'=

2

9

R2

η

(ρ−σ)g ...(ii)
Dividing Eq. (i) by (ii), we get

6

v'

=

2 9 r2 η (ρ−σ)g

2 9 R2 η (ρ−σ)g

⇒

6

v'

=

r2

(2r)2

or v' = 24cm s‌−1