Question

The excess pressure inside a spherical drop of water is four times that of another water drop. Then the mass ratio of the two drops is

• A. $1:16$
• B. $1:64$
• C. $1:4$
• D. $1:8$

Answer 1

The correct option is B $1:64$

We know that excess pressure inside a water droplet is given by
$\Delta P=\frac{2T}{R}$, where $T$ is Surface Tension and $R$ is the radius of the water drop.
$\Rightarrow \Delta P\propto \frac{1}{R}$
$[$ Surface Tension of water is constant $]$
So, $\frac{\Delta P_1}{\Delta P_2}=\frac{R_2}{R_1}$
$\Rightarrow \frac{4P}{P}=\frac{R_2}{R_1}$
$[$ given excess pressure inside one drop is four times that of excess pressure inside the other drop $]$
$\Rightarrow \frac{R_1}{R_2}=\frac{1}{4}$ ........(1)
Now, mass of a water drop, $M=\rho V$, where $\rho$ is density of water and $V$ is volume of water drop.
$\Rightarrow M=\frac{4}{3}\pi R^3\rho$
$\Rightarrow M\propto R^3$
$[$ density of water is constant $]$
So, $\frac{M_1}{M_2}={\left(\frac{R_1}{R_2}\right)}^3$
$\Rightarrow \frac{M_1}{M_2}={\left(\frac{1}{4}\right)}^3=\frac{1}{64}$ $[$ from (1) $]$