Question

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

• A. $1:16$
• B. $1:27$
• C. $1:9$
• D. $1:3$

Answer 1

The correct option is B $1:27$

We know that excess pressure inside a spherical water drop is given by,
$\Delta P=\frac{2T}{R}$, where, $T$ is surface tension and $R$ is radius of the sphere.
Given,
$\Delta P_1=3\Delta P_2$
$\Rightarrow \frac{2T}{R_1}=3\times \frac{2T}{R_2}$
$\Rightarrow \frac{R_1}{R_2}=\frac{1}{3}$ ...........(1)
Now, mass,
$m=\rho V$, where, $\rho$ is density and $V$ is volume.
So,
$\frac{m_1}{m_2}=\frac{V_1}{V_2}$
$\Rightarrow \frac{m_1}{m_2}=\frac{\frac{4}{3}\pi R_1^3}{\frac{4}{3}\pi R_2^3}$
$\Rightarrow \frac{m_1}{m_2}={\left(\frac{R_1}{R_2}\right)}^3$
$\Rightarrow \frac{m_1}{m_2}={\left(\frac{1}{3}\right)}^3=\frac{1}{27}=1:27$
[from (1)]