The excess pressure inside a first spherical drop of water is three times that of second spherical drop of water. Then the ratio of mass of first spherical drop to that of second spherical drop is

Excess pressure inside the \(1^{\text {st }}\) spherical drop is given by,
\(\mathrm{P}_1 =\frac{2 \mathrm{~T}}{\mathrm{r}_1} \)
\( \text {Similarly, for } 2^{\text {nd }} \text {drop } \)
\( \mathrm{P}_2 =\frac{2 \mathrm{~T}}{\mathrm{r}_2} \)
\( \mathrm{P}_1 =3 \mathrm{P}_2 .......(Given)\)
\( \therefore \frac{2 \mathrm{~T}}{\mathrm{r}_1} =3\left(\frac{2 \mathrm{~T}}{\mathrm{r}_2}\right) \)
\( \therefore \frac{\mathrm{r}_1}{\mathrm{r}_2} =\frac{1}{3}\)
Now, \(\frac{m_1}{m_2}=\frac{V_1 \rho_1}{V_2 \rho_2}\)
As both the drops are of water, \(\rho_1=\rho_2\)
\(\therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2} \)
\( \therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{4 / 3 \pi_1^3}{4 / 3 \mathrm{r}_2^3} \)
\( \therefore \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{27}\)