A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer.

Answer :

Given : Diameter of  hemispherical bowl  =  36 cm  , So radius of hemispherical bowl, r = diameter/2  =  18 cm

And we know Volume of hemisphere  = $\frac{2 \mathrm{\pi } {\mathrm{r}}^3}{3}$ , So

Volume of hemispherical bowl  = $\frac{2}{3}{\mathrm{\pi r}}^3 = \frac{2}{3}\mathrm{\pi }{\left(18\right)}^3 {\mathrm{cm}}^3$
And  ,now, the liquid is transferred into 72 bottles and 10% liquid is wasted in this transfer.
So,
$\mathrm{volume} \mathrm{of} \mathrm{liquid} \mathrm{that} \mathrm{is} \mathrm{transferred} = \frac{2}{3}\mathrm{\pi }{\left(18\right)}^3 - \frac{10}{100}\times \frac{2}{3}\mathrm{\pi }{\left(18\right)}^3 = \frac{9}{10}\times \frac{2}{3}\mathrm{\pi }{\left(18\right)}^3 {\mathrm{cm}}^3$

Given diameter of each cylindrical bottle  =  6 cm , So radius , R =  3 cm
And let the height of each bottle = h cm
We know volume of cylinder  = ${\mathrm{\pi R}}^2 \mathrm{h}$ , So we get
$$\begin{gathered} 72 \left( \mathrm{\pi } {\mathrm{R}}^2 \mathrm{h}\right) = \frac{9}{10}\times \frac{2}{3}\mathrm{\pi }{\left(18\right)}^3 \\ \Rightarrow 72 \times \mathrm{\pi } \times {\left(3\right)}^2 \times \mathrm{h} = \frac{9}{10}\times \frac{2}{3}\mathrm{\pi }{\left(18\right)}^3 \\ \Rightarrow \mathrm{h} = 5.40 \mathrm{cm} \end{gathered}$$
So,
Height of each cylindrical bottle =  5.40 cm                                ( Ans )