Two rectangular sheets of paper each 30 cm × 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders, thus formed, is :

1. 2 : 1

2. 3 : 2

3. 4 : 3

4. 5 : 3

For cylinder 1, rolled along its length:

Height of cylinder, h = 18 cm

Radius of cylinder be r cm

Circumference of base = 2πr = 30

$\Rightarrow 2 \times π \times \text{r} = 30 \\[1em] \Rightarrow 2π \times \text{r} = 30 \\[1em] \Rightarrow \text{r} = \dfrac{30}{2π} \\[1em] \Rightarrow \text{r} = \dfrac{15}{π} \text{ cm.}$

Volume of cylinder 1 = πr²h

$= π \times \Big(\dfrac{15}{π}\Big)^2 \times 18 \\[1em] = π \times \Big(\dfrac{225}{π^2}\Big) \times 18 \\[1em] = \dfrac{4050}{π} \text{ cm}^3$

For cylinder 2, rolled along its breadth:

Height of cylinder, H = 30 cm

Radius of cylinder be R cm

Circumference of base = 2πR = 18

$\Rightarrow 2 \times π \times \text{R} = 18 \\[1em] \Rightarrow 2π \times \text{R} = 18 \\[1em] \Rightarrow \text{R} = \dfrac{18}{2π} \\[1em] \Rightarrow \text{R} = \dfrac{9}{π} \text{ cm.}$

Volume of cylinder 2 = πR²H

$= π \times \Big(\dfrac{9}{π}\Big)^2 \times 30 \\[1em] = π \times \Big(\dfrac{81}{π^2}\Big) \times 30 \\[1em] = \dfrac{2430}{π} \text{ cm}^3$

Ratio of the volumes of the two cylinders:

$= \dfrac{\text{Volume of cylinder 1}}{\text{Volume of cylinder 2}} \\[1em] = \dfrac{\dfrac{4050}{π}}{\dfrac{2430}{π}} \\[1em] = \dfrac{4050}{π} \times \dfrac{π}{2430} \\[1em] = \dfrac{5}{3}$

Hence, option 4 is the correct option.