A metallic cylinder has radius 3 cm and height 5 cm.

Solution:

We have,

the base radius of the cylinder, $R=3 \mathrm{~cm}$,

the height of the cylinder, $H=5 \mathrm{~cm}$,

the base radius of the conical hole, $r=\frac{3}{2} \mathrm{~cm}$ and

the height of the conical hole, $h=\frac{8}{9} \mathrm{~cm}$

Now,

Volume of the cylinder, $V=\pi R^{2} H$

$=\pi \times 3^{2} \times 5$

$=45 \pi \mathrm{cm}^{3}$

Also,

Volume of the cone removed from the cylinder, $v=\frac{1}{3} \pi r^{2} h$

$=\frac{\pi}{3} \times\left(\frac{3}{2}\right)^{2} \times\left(\frac{8}{9}\right)$

$=\frac{2 \pi}{3} \mathrm{~cm}^{3}$

So, the volume of metal left in the cylinder, $V^{\prime}=V-v$

$=45 \pi-\frac{2 \pi}{3}$

$=\frac{133 \pi}{3} \mathrm{~cm}^{3}$

$\therefore$ The required ratio $=\frac{V}{v}$

$=\frac{\left(\frac{133 \pi}{3}\right)}{\left(\frac{2 \pi}{3}\right)}$

$=\frac{133}{2}$

$=133: 2$

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.