Question

The excess pressure inside a spherical drop of water is $4$ times that of another drop. Then their respective mass ratio is

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

Answer 1

The correct option is B $1:64$

Excess pressure inside a liquid drop $\left(\Delta P\right)=\frac{2T}{R}$

Let the excess pressure in drop $1$ is $\Delta P_1$ and drop $2$ is $\Delta P_2$ and their radii are $R_1$ and $R_2$ respectively.
Given,
$\Delta P_1=4\times \Delta P_2$
$\Rightarrow \frac{2T}{R_1}=4\times \frac{2T}{R_2}$
$\Rightarrow R_2=4R_1$
$\Rightarrow \frac{R_1}{R_2}=\frac{1}{4}$

Mass ratio $=\frac{\text{mass of droplet 1}\left(m_1\right)}{\text{mass of droplet 2}\left(m_2\right)}$
Here, $m_1=\frac{4}{3}\pi R_1^3\rho$;
$m_2=\frac{4}{3}\pi R_2^3\rho$
$\therefore \frac{m_1}{m_2}=\frac{R_1^3}{R_2^3}={\left(\frac{R_1}{R_2}\right)}^3={\left(\frac{1}{4}\right)}^3=\frac{1}{64}$