CGPET 2019 Solved Paper

Given, density of liquid drop =ρ
Density of floating half immersed in liquid =ρ0
Surface tension of liquid =s
Now, balancing the forces acting on the drops floating
∴w=Fb+Fs
where, w is the weight of the liquid drop, Fb is the Buoyant force acting on the drop and FS is the surface tension acting on drop.
Then,
w=Fb+Fs
or

4

3

πr3ρg=

2

3

πr3ρ0g+s×2πr
or

4

3

πr3ρg−

2

3

πr3ρ0g=s×2πr
or r3[

4

3

πρg−

2

3

πρ0g]=s×2πr
or r2

[4πρg−2πρ0g]

3

=s×2π
or r2

2πg[2p−ρ0]

3

=s×2π
or r2

g(2p−p0)

3

=s
or r2=

3s

g(2p−p0)

or r=√

3s

g(2ρ−ρ0)

.
So, the radius of drop is r=√

3s

g(2ρ−ρ0)

.