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. $8:1$
• C. $1:4$
• D. $1:64$

Answer 1

Answer: (D)

Excess pressure inside a spherical drop of water
$p=\frac{2 T}{R}$
$Given, \, p_{1}=4p_{2}$
$\frac{2 T}{R_{1}}=4\times \frac{2 T}{R_{2 \, }}$
$or \, \, \, R_{2}=4R_{1}$
$Now, \, \, \frac{m_{1}}{m_{2}}=\frac{4 \pi R_{1}^{3} d_{1}}{4 \pi R_{2}^{3} d_{2}}$
$or \, \, \, \frac{m_{1}}{m_{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{m_{1}}{m_{2 \, }}=\frac{1}{64}$