A soft drink is available in two packs:
(i) a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 15 cm, and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
(i) Length of tin can, l = 5 cm
Breadth of tin can, b = 4 cm
Height of tin can, h = 15 cm
∴ Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm3
(ii) Radius of plastic cylinder, $r=\frac{7}{2} \mathrm{~cm}$
Height of plastic cylinder, h = 10 cm
$\therefore$ Volume of soft drink in plastic cylinder $=\pi r^{2} h=\frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \times 10=385 \mathrm{~cm}^{3}$
So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.
Difference in the capacities of the two packs = 385 − 300 = 85 cm3
Thus, the capacity of the plastic cylinder pack is 85 cm3 more than the capacity of the tin can pack.
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