A soft drink is available in two packs - (i) a tin can with a rectangular base of length $ 5 \mathrm{~cm} $ and width $ 4 \mathrm{~cm} $, having a height of $ 15 \mathrm{~cm} $ and (ii) a plastic cylinder with circular base of diameter $ 7 \mathrm{~cm} $ and height $ 10 \mathrm{~cm} $. Which container has greater capacity and by how much?
Given:
A soft drink is available in two packs -
(i) a tin can with a rectangular base of length $5\ cm$ and width $4\ cm$, having a height of $15\ cm$ and
(ii) a plastic cylinder with circular base of diameter $7\ cm$ and height $10\ cm$.
To do:
We have to find the container that has greater capacity and by how much.
Solution:
In the first case,
The length of the base of the tin can $= 5\ cm$
Width of the tin can $= 4\ cm$
Height of the tin can $= 15\ cm$
Therefore,
Volume of the soft drink $= lbh$
$= 5 \times 4 \times 15$
$= 300\ cm^3$
In the second case,
Diameter of the base of the plastic cylinder $= 7\ cm$
Radius of the plastic cylinder $=\frac{7}{2} \mathrm{~cm}$
Height of the plastic cylinder $=10 \mathrm{~cm}$
Therefore,
Volume of the soft drink $=\pi r^{2} h$
$=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 10$
$=385 \mathrm{~cm}^{2}$
The soft drink in the second container is greater by $385\ cm^3 - 380\ cm^3 = 85\ cm^3$.
Hence, the plastic cylinder has a greater capacity of $85\ cm^3$.
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