A conical hole is drilled in a circular cylinder of height $ 12 \mathrm{~cm} $ and base radius $ 5 \mathrm{~cm} $. The height and the base radius of the cone are also the same. Find the whole surface and volume of the remaining cylinder.

Given:

A conical hole is drilled in a circular cylinder of height \( 12 \mathrm{~cm} \) and base radius \( 5 \mathrm{~cm} \).

The height and the base radius of the cone are also the same. 

To do:

We have to find the whole surface and volume of the remaining cylinder.

Solution:

Radius of the base of the cylinder $r= 5\ cm$

Height of the cylinder $h = 12\ cm$

This implies,

Volume of the cylinder $= \pi r^2 h$

$= \pi \times 5^2 \times 12$

$= 300 \pi\ cm^3$

Radius of the cone $r= 5\ cm$

Height of the cone $h = 12\ cm$

This implies,

Volume of the cone $= \frac{1}{3} \pi r^2 h$

Therefore,

Volume of the remaining cylinder $=$ Volume of the cylinder $-$ Volume of the cone

$=300 \pi -100 \pi$

$=200 \pi \mathrm{cm}^{3}$

Slant height of the cone $l=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(5)^{2}+(12)^{2}}$

$=\sqrt{25+144}$

$=\sqrt{169}$

$=13 \mathrm{~cm}$

Curved surface area of the cone $=\pi r l$

$=\pi \times 5 \times 13$

$=65 \pi \mathrm{cm}^{2}$

Total surface area of the cylinder $=2 \pi r h+\pi r^{2}$

$=2 \pi \times 5 \times 12+\pi \times 5^2$

$=120 \pi+25 \pi$

$=145 \pi \mathrm{cm}^{2}$

Whole surface area of the remaining cylinder $=65 \pi+145 \pi$

$=210 \pi \mathrm{cm}^{2}$

The whole surface and volume of the remaining cylinder are $210 \pi\ cm^2$ and $200 \pi\ cm^3$ respectively.

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Updated on: 10-Oct-2022

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