A metallic cylinder has radius $3 \mathrm{~cm}$ and height $5 \mathrm{~cm}$. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of $\frac{3}{2} \mathrm{~cm}$ and its depth is $\frac{8}{9} \mathrm{~cm}$. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.
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.
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