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A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is \( 2 \mathrm{~cm} \) and the diameter of the base is \( 4 \mathrm{~cm} \). If a right circular cylinder circumscribes the toy, find how much more space it will cover.
Given:
A solid toy is in the form of a hemisphere surmounted by a right circular cone.
Height of the cone is \( 2 \mathrm{~cm} \) and the diameter of the base is \( 4 \mathrm{~cm} \).
A right circular cylinder circumscribes the toy.
To do:
We have to find how much more space it will cover.
Solution:
Height of the cone $h = 2\ cm$
Diameter of the base $= 4\ cm$
This implies,
Radius of the cone $r=\frac{4}{2}$
$=2 \mathrm{~cm}$
Therefore,
Volume of the toy $=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$
$=\frac{1}{3} \pi r^{2}(h+2 r)$
$=\frac{1}{3} \pi(2)^{2}(2+2 \times 2)$
$=\frac{1}{3} \pi \times 4(2+4)$
$=\frac{1}{3} \pi \times 4 \times 6$
$=8 \pi \mathrm{cm}^{3}$
Volume of the cylinder which circumscribes the toy $= \pi r^2 h$
$ = \pi (2)^2 \times 4$
$= 16 \pi\ cm^3$
Amount of space it covers more $=$ Difference of the volumes
$= 16\pi – 8\pi$
$= 8 \pi\ cm^3$
Hence, it covers $8 \pi\ cm^3$ of more space.