A cylindrical vessel of diameter $ 14 \mathrm{~cm} $ and height $ 42 \mathrm{~cm} $ is fixed symmetrically inside a similar vessel of diameter $ 16 \mathrm{~cm} $ and height $ 42 \mathrm{~cm} $. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required?

Given:

A cylindrical vessel of diameter \( 14 \mathrm{~cm} \) and height \( 42 \mathrm{~cm} \) is fixed symmetrically inside a similar vessel of diameter \( 16 \mathrm{~cm} \) and height \( 42 \mathrm{~cm} \).

The total space between the two vessels is filled with cork dust for heat insulation purposes. 

To do:

We have to find the volume of cork dust required.

Solution:

Diameter of inner cylinder $= 14\ cm$

Thios implies,

Radius of inner cylinder $r =\frac{14}{2}$

$= 7\ cm$
Diameter of outer cylinder $= 16\ cm$

This implies,

Radius of outer cylinder $R =\frac{16}{2}$

$= 8\ cm$
Height of the cylinders $h = 42\ cm$

Therefore,

Space between the two cylinders $=$ Volume of outer cylinder $-$ Volume of inner cylinder

Volume of cork dust required $=\pi \mathrm{R}^{2} h-\pi r^{2} h$

$=\pi h(\mathrm{R}^{2}-r^{2})$

$=\frac{22}{7} \times 42(8^{2}-7^{2})$

$=22 \times 6 \times(64-49)$

$=22 \times 6 \times 15$

$=1980 \mathrm{~cm}^{3}$

Hence, $1980 \mathrm{~cm}^{3}$ of cork dust will be required.

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Updated on: 10-Oct-2022

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