AP EAPCET 23-Aug-2021 Shift 2 Paper

Let radius of big drop =R
Radius of small drop =T
Number of small drops, n=1000
In this process, the volume remains same
Vf=Vi
⇒

4

3

πR3=n(

4

3

πr3)
⇒R3=1000r3⇒R=10r ...(i)
Initial surface area,A1=n4πr2
Final surface area, A2=4πR2
Now, increase in surface area,
ΔA=A2−A1=4π[R2−nr2]
We know that, surface energy,
SE=TA [where, T= surface tension of liquid ]
According to question.

SEf

SEi

=

TA2

TA1

=

4πR2

n(4πr2)

=

R2

1000r2

[From Eq. (i)]
=

(10r)2

1000r2

=

100r2

1000r2

=

1

10

or 1:10