EAMCET Engineering 2009 Solved Paper

Let now radius of big drop is R.
Then,
‌

4

3

πR3‌=‌

4

3

×πr3⋅8
R‌=2r
where r is radius of small drops. Now, terminal velocity of drop in liquid.
ve=‌

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 Eq. (ii), we get
‌

6

v′

=‌

‌ 2 9 r2 η (ρ−σ)g

2 9 ‌ R2 η (ρ−σ)g

or ‌

6

v′

=‌

r2

(2r)2

or v′=24‌cm‌ s−1