A rectangular sheet of tin foil of size 30 cm × 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinder thus formed.

Suppose the sheet is rolled along length so the circumference of the base = 30 cm and height = 18 cm. Let radius be r₁.

or, 2πr₁ = 30

$\Rightarrow 2 \times \dfrac{22}{7} \times r$1 = 30 \\[1em] \Rightarrow r1 = \dfrac{30 \times 7}{2 \times 22} \\[1em] \Rightarrow r_1 = \dfrac{210}{44} = \dfrac{105}{22}.

Suppose the sheet is rolled along breadth so the circumference of the base = 18 cm and height = 30 cm. Let radius be r₂.

or, 2πr₂ = 18

$\Rightarrow 2 \times \dfrac{22}{7} \times r$2 = 18 \\[1em] \Rightarrow r2 = \dfrac{18 \times 7}{2 \times 22} \\[1em] \Rightarrow r_2 = \dfrac{126}{44} = \dfrac{63}{22}.

Volume of cylinder = πr²h

$\therefore \dfrac{\text{Vol. of 1st cylinder}}{\text{Vol. of 2nd cylinder}} = \dfrac{π \times \Big(\dfrac{105}{22}\Big)^2 \times 18}{π \times \Big(\dfrac{63}{22}\Big)^2 \times 30} \\[1em] = \dfrac{105 \times 105 \times 18 \times 22^2}{63 \times 63 \times 30 \times 22^2} \\[1em] = \dfrac{105 \times 105 \times 18}{63 \times 63 \times 30} \\[1em] = \dfrac{198450}{119070} \\[1em] = \dfrac{50}{30} \\[1em] = \dfrac{5}{3}.$

Hence, the ratio of the volume of two cylinders = 5 : 3.