# 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?

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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 $85\ cm^3$.

Updated on 10-Oct-2022 13:46:38

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