CGPET 2021 Solved Paper

According to problem, n small drop of radius r coalesce to form a big liquid drop of radius R. Since, the volume remain same.
i.e.

4

3

πR3=n×

4

3

πr3
∴ R=r×n

1

3

.......(i)
During the formation of big drop, surface area decreases and hence, the energy released.
Now, change in surface area,
∆A=n×4πr2−4πR2
Therefore, energy released,
∆E=T×∆A=T×[n×4πr2−4πR2]
=4πTR2[

nr2

R2−1

]
Using eq. (i), we get
⇒ ∆E=4πTR2(n

1

3

−1)
=4πTR2(

R

r

−1)=4πTR3[

1

r

−

1

R

]
=(

3

3

.4πR3)T[

1

r

−

1

R

]=3VT[

1

r

−

1

R

]