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.

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Updated on: 10-Oct-2022

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