TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

According to given situation,
Volume of big drop = Volume of n small drops
‌

4

3

πR3=n×‌

4

3

πr3⇒r3=‌

R3

n

or r=‌

R

n1/3

nr2=‌

nR2

n2/3

=n1/3R2...(i)
Surface area of big drop, A=4πR2
Surface area of n small drops,
A′=n×4πr2
=4π(n1/3R2−R2) [from Eq.(i)].
=4πR2(n1/3−1)
The work done for change in area,
This work is store in the form of energy.