A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular

Given data is as follows:

Dimensions of the rectangular sheet of paper = 30 cm × 18 cm

We have to find the ratio of the volumes of the cylinders formed by rolling the sheet along its length and along its breadth.

Let V1 be the volume of the cylinder which is formed by rolling the sheet along its length.

When the sheet is rolled along its length, the length of the sheet forms the perimeter of the cylinder. Therefore, we have,

2πr1 = 30

r1 = 15/π

The width of the sheet will be equal to the height of the cylinder. Therefore,

h1 = 18 cm

Therefore, $\mathrm{V}_{1}=\pi \mathrm{r}_{1}^{2} \mathrm{~h}_{1}$

$=\pi \times 15 / \pi \times 15 / \pi \times 18$

$v_{1}=225 / \pi \times 18 \mathrm{~cm}^{3}$

Let V2 be the volume of the cylinder formed by rolling the sheet along its width.

When the sheet is rolled along its width, the width of the sheet forms the perimeter of the base of the cylinder. Therefore, we have,

2πr2 = 18

r2 = 9/π

The length of the sheet will be equal to the height of the cylinder. Therefore,

h2 = 30 cm

Therefore, $\mathrm{V}_{2}=\pi \mathrm{r}_{2}^{2} \mathrm{~h}_{2}$

$=\pi \times \frac{9}{\pi} \times \frac{9}{\pi} \times 30$

$V_{2}=\frac{81 \times 30}{\pi}$

Now that we have the volumes of two cylinders, we have,

$\frac{V_{1}}{V_{2}}=\frac{225 \times 18}{81 \times 30}$

$\frac{V_{1}}{V_{2}}=\frac{5}{3}$

Therefore, the ratio of the volumes of the two cylinders is 5: 3.