The excess pressure inside a spherical drop of water A is four times that of another drop B. Then the ratio of mass of drop \(A\) to that of drop \(B\) is

Excess pressure inside \(1^{\text {st }}\) spherical drop (A) is given by, \(\dot{P}_A=\frac{2 T}{r_1}\)
For \(2^{\text {nd }} \operatorname{drop}(B), P_B=\frac{2 T}{r_2}\)
\(\therefore \quad-\frac{2 \mathrm{~T}}{\mathrm{r}_1}=4\left(\frac{2 \mathrm{~T}}{\mathrm{r}_2}\right) \ldots\) (given, \(\mathrm{P}_{\mathrm{A}}=4 \mathrm{P}_{\mathrm{B}}\) )
\(\therefore \quad \frac{r_1}{r_2}=\frac{1}{4}\)
Now, \(\frac{m_1}{m_2}=\frac{V_1 \rho_1}{V_2 \rho_2}\)
As both are drops of water, \(\rho_1=\rho_2\)
\(\therefore \quad \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\frac{4}{3} \pi \mathrm{r}_1^3}{\frac{4}{3} \pi \pi_2^3}=\left(\frac{1}{4}\right)^3=\frac{1}{64}\)