Question

A soft drink is available in two packs - (i) a tin can with a rectangular base of length $5 c m$ and width $4 c m,$ having a height of $15 c m$ and (ii) a plastic cylinder with a circular base of diameter $7 c m$ and height $10 c m .$ Which container has greater capacity and by how much? (Assume $π = \frac{22}{7}$ )

Accepted Answer

Given that, a soft drink is available in two packs -

(i) a tin can with a rectangular base of length $5 c m$ and width $4 c m,$ having a height of $15 c m$ and

(ii) a plastic cylinder with a circular base of diameter $7 c m$ and height $10 c m .$

The length $(l)$ of tin can $= 5 c m$

Breadth $(b)$ of tin can $= 4 c m$

Height $(h)$ of tin can $= 15 c m$

Capacity of the can $= l \times b \times h$ $= 300 {cm}^{3}$

Radius $(r)$ of circular end of plastic cylinder $= \frac{7}{2} = 3.5 c m$

Height $(H)$ of plastic cylinder $= 10 c m$

Capacity of the plastic cylinder $= π r^{2} H$

$= [\frac{22}{7} \times {(3.5)}^{2} \times 10] {cm}^{3} = [\frac{22}{7} \times 3.5 \times 3.5 \times 10] {cm}^{3} = [\frac{\overset{11}{22}}{7} \times \frac{\overset{5}{35}}{\underset{2}{10}} \times \frac{35}{10} \times 10] {cm}^{3}$
$= 11 \times 35 c m^{3}$
$= 385 c m^{3}$

Therefore, the plastic cylinder has greater than the tin can.

The difference between two containers $=$ Capacity of the plastic cylinder $-$ Capacity of the tin can
$= (385 - 300) {cm}^{3}$
$= 85 c m^{3}$
Hence, the plastic cylinder has greater capacity by $85 c m^{3}$ than the tin can.

EXERCISE

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