Question

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

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

Answer 1

The correct option is B $1:64$

let, Surface Tension = T

excess pressure of these two spherical drop

$P_1$ and $P_2$ and Radii $R_1$ and $R_2$.

let $P_1=9P_2$

we know, excess pressune inside a

liquid drop $=\frac{2T}{R}$

$\begin{array}{cc}\frac{2T}{R_1}& =9\times \frac{2T}{R_2}\\ \text{or}{R}_2& =9R_1\end{array}$

$\frac{m_1}{m_2}=\frac{4/3\pi R_1^3\times \rho }{4/3\pi R_2^3\times \rho }$

$\begin{array}{cc}\frac{m_1}{m_2}& ={\left(\frac{1}{9}\right)}^3\\ \text{or}& \frac{m_1}{m_2}=\frac{1}{64}\end{array}$

Correct option is (B).