A 15cm long test tube having diameter of 3.6 cm has water upto a height of 6 cm. 20spherical drops ofoil of radius 9mm are dropped into it. what lenght of the test tube remains empty ?

Length of test tube = 15 cm
diameter of test tube = 3.6 cm 
then radius of test tube (r) = 1.8 cm
height of the water in the test tube (h) = 6 cm
Therefore, Volume of water inside test tube = volume of cylinderical test tube of height 6 cm  = ${\mathrm{\pi r}}^2\mathrm{h}$

$$\begin{gathered} = \mathrm{\pi } \times 1.8 \times 1.8 \times 6 \\ = \mathrm{\pi } \times \frac{18}{10} \times \frac{18}{10} \times 6 \\ = \mathrm{\pi } \times \frac{9}{5} \times \frac{9}{5} \times 6 \\ = \frac{486}{25}\mathrm{\pi } {\mathrm{cm}}^3 \\ \mathrm{radius} \mathrm{of} \mathrm{one} \mathrm{spherical} \mathrm{drop} \mathrm{of} \mathrm{oil} \left(\mathrm{R}\right) = 9 \mathrm{mm} =\frac{9}{10}\mathrm{cm} \\ \mathrm{Therefore} \mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{spherical} \mathrm{drop} = \frac{4}{3}{\mathrm{\pi R}}^3 \\ =\frac{4}{3}\mathrm{\pi } \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \\ =\frac{243\mathrm{\pi }}{250}{\mathrm{cm}}^3 \\ \mathrm{Therefore}, \mathrm{volume} \mathrm{of} 20 \mathrm{spherical} \mathrm{drops} = \frac{20 \times 243\mathrm{\pi }}{250} \\ =\frac{486\mathrm{\pi }}{25}{\mathrm{cm}}^3 \\ \mathrm{Total} \mathrm{volume} \mathrm{of} \mathrm{water} \mathrm{and} \mathrm{oil} \mathrm{drops} = \frac{486\mathrm{\pi }}{25}+\frac{486\mathrm{\pi }}{25} \\ =\frac{972\mathrm{\pi }}{25}{\mathrm{cm}}^3 \\ \mathrm{Let} \mathrm{H} \mathrm{be} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{mixture} \mathrm{of} \mathrm{water} \mathrm{and} \mathrm{oil} \mathrm{in} \mathrm{the} \mathrm{test} \mathrm{tube} \\ \mathrm{then}, {\mathrm{\pi r}}^2\mathrm{H} =\frac{972\mathrm{\pi }}{25} \\ \mathrm{\pi } \times 1.8 \times 1.8 \times \mathrm{H} =\frac{972\mathrm{\pi }}{25} \\ \Rightarrow \frac{18}{10} \times \frac{18}{10} \times \mathrm{H} =\frac{972}{25} \\ \Rightarrow \frac{81}{25} \times \mathrm{H} =\frac{972}{25} \\ \Rightarrow \mathrm{H} =\frac{972}{25} \times \frac{25}{81} =\frac{972}{81} = 12 \mathrm{cm} \\ \mathrm{The} \mathrm{length} \mathrm{of} \mathrm{test} \mathrm{tube} \mathrm{that} \mathrm{remains} \mathrm{empty} = 15 - 12 =3 \mathrm{cm}. \end{gathered}$$