A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways namely,

The dimensions of the rectangular sheet of paper are 30 cm × 18 cm.

Let V1 and V2 be the volumes of the cylinders formed by rolling the rectangular sheet of paper along its length (i.e. 30 cm) and breadth (i.e. 18 cm), respectively.

Suppose r1 and h1 be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its length, respectively.

$\therefore 2 \pi r_{1}=30 \Rightarrow r_{1}=\frac{30}{2 \pi} \mathrm{cm}$

h1 = 18 cm

$\therefore V_{1}=\pi r_{1}^{2} h_{1}=\pi\left(\frac{30}{2 \pi}\right)^{2} \times 18 \mathrm{~cm}^{3}$

Also, suppose r2 and h2 be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its breadth, respectively. 

$\therefore 2 \pi r_{2}=18 \Rightarrow r_{2}=\frac{18}{2 \pi} \mathrm{cm}$

$h_{2}=30 \mathrm{~cm}$

$\therefore V_{2}=\pi r_{2}^{2} h_{2}=\pi\left(\frac{18}{2 \pi}\right)^{2} \times 30 \mathrm{~cm}^{3}$

Now,

$\frac{V_{1}}{V_{2}}=\frac{\pi\left(\frac{30}{2 \pi}\right)^{2} \times 18}{\pi\left(\frac{18}{2 \pi}\right)^{2} \times 30}=\frac{5}{3}$

$\Rightarrow V_{1}: V_{2}=5: 3$

Thus, the ratio of the volumes of the two cylinders thus formed is 5 : 3.