Question

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

• A. $1:6$
• B. $8:1$
• C. $1:4$
• D. $1:64$

Answer 1

The correct option is D $1:64$

Inside a spherical drop of water, the excess pressure is given as,

$P=\frac{2T}{R}$

It is given that,

$P_1=4P_2$

$\frac{2T}{R_1}=4\times \frac{2T}{R_2}$

$R_2=4R_1$

The mass ratio is given as,

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

$\frac{m_1}{m_2}=\frac{{R_1}^3}{{R_2}^3}$

$\frac{m_1}{m_2}=\frac{1}{64}$

Thus, the mass ratio is $1:64$.