Question

A water drop is divided into 8 equal droplets. The pressure difference between inner and outer side of big drop is
A. same as for smaller droplet
B. 1/2 of that for smaller droplet
C. 1/4 of that for smaller droplet
D. twice that for smaller droplet

Answer 1

We have to consider the radius of the smaller droplet and the bigger drop as an unknown variable and then find the relation between the two radius. We can do it by comparing the volumes of both the bigger drop and the droplets formed. After that we know that by the pressure difference in the question it means the excess pressure in the drop. So we have to find a relation between the pressure difference for the bigger drop and the smaller drop.

Formula used: $\dfrac{4}{3}\pi {r^3}$
$P = \dfrac{{2T}}{R}$

Complete step by step solution:
So a big water drop is divided into 8 smaller droplets,
So we will find the relation between the radius of the smaller droplets and the radius of the big droplet,
Now, let the radius for the bigger drop be ‘R’ and for the smaller drop be ‘r’.
We know that the volume is the same as there was no loss of water, the same bigger droplet was divided into smaller droplets.
We know that droplets are more or less like a small sphere.
We now comparing the volumes,
Volume of bigger drop = volume of smaller droplets
$\dfrac{4}{3}\pi {R^3} = 8 \times \dfrac{4}{3}\pi {r^3}$ (we are multiplying the volume of smaller droplets with 8 as there are 8 droplets made from the bigger drop)
\eqalign{ & {R^3} = 8{r^3} \cr & \Rightarrow R = 2r \cr} ,
So we can say from the above relation that the radius of the bigger drop is twice the radius of the smaller drop.
Now, the pressure difference between the inner and outer side of the drop means the excess pressure only. Because the excess pressure will get accumulated between the inner and the outer side of the big drop. This excess pressure can be mathematically given as:
${P_{{\text{excess}}}} = \dfrac{{2T}}{R}$ ,
Now, excess pressure for the bigger drop will be given by,
${P_b} = \dfrac{{2T}}{R}$
And, the excess pressure for the smaller drop will be given by,
\eqalign{ & {P_s} = \dfrac{{2T}}{r} = \dfrac{{2T \times 2}}{R} \cr & \Rightarrow {P_s} = \dfrac{{4T}}{R}{\text{ }}\left[ {\because R = 2r} \right] \cr}
We can say that, the pressure in the bigger drop is,
$\therefore {P_b} = \dfrac{1}{2} \times {P_s}$,

So, the correct answer is “Option B”.

Note: In the equation $P = \dfrac{{2T}}{R}$, ‘T’ is the surface tension for the drops or droplets. Students have to analyse the problem properly. We are calculating the volume while finding a relation between the radius of the smaller and the bigger drop as dividing the volume of each single droplet must be less than the bigger drop but when all the droplets are considered together the volume is the same.