ICSE Class 10 Maths Chapter 20 Cylinder Cone and Sphere

Unit 4: Mensuration

20 Cylinder, Cone and Sphere

(Surface Area and Volume)

20.1 Cylinder

A solid which has uniform circular cross-section, is called a cylinder (or a circular cylinder).

Let r be the radius of circular cross-section and h the height of the cylinder; then

1. Area of cross-section = \(\pi r^2\).

2. Perimeter (circumference) of cross-section = \(2\pi r\)

3. Curved surface area = Perimeter of cross-section \(\times\) height = \(2\pi rh\).

4. Total surface area = Curved surface area + 2 (Area of cross-section) = \(2\pi rh + 2(\pi r^2) = 2\pi r(h + r)\).

5. Volume = Area of cross-section \(\times\) height (or, length) = \(\pi r^2h\).

20.2 Hollow Cylinder

Let R be the external radius of a hollow cylinder, r its internal radius and h its height or length; then

1. Thickness of its wall = R - r

2. Area of cross-section = \(\pi R^2 - \pi r^2 = \pi(R^2 - r^2)\)

3. External curved surface = \(2\pi Rh\)

4. Internal curved surface = \(2\pi rh\)

5. Total surface area = External curved surface area + Internal curved surface area + 2 (Area of cross-section) = \(2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2)\)

6. Volume of material = External volume - Internal volume = \(\pi R^2h - \pi r^2h = \pi(R^2 - r^2)h\).

Example 1

The area of the curved surface of a cylinder is 4,400 cm² and the circumference of its base is 110 cm. Find:

(i) the height of the cylinder,

(ii) the volume of the cylinder.

Solution

(i) Given, \(2\pi rh = 4400\) cm² and \(2\pi r = 110\) cm

\(\therefore \frac{2\pi rh}{2\pi r} = \frac{4400}{110}\) cm \(\Rightarrow h = 40\) cm Ans.

(ii) Since, \(2\pi r = 110\)

\(\therefore r = \frac{110}{2\pi} = \frac{110 \times 7}{2 \times 22}\) cm = \(\frac{35}{2}\) cm

and, volume = \(\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 40\) cm³ [Since, volume = \(\pi r^2h\)]

= 38,500 cm³ Ans.

Teacher's Note

Understanding cylinder dimensions helps in real-world applications like calculating water tank capacity or paint needed for cylindrical pipes in construction.

Example 2

The barrel of a fountain-pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

Answer correct to the nearest 100 words

Solution

One-fifth of a litre = \(\frac{1}{5} \times 1000\) cm³ = 200 cm³ [- 1 litre = 1000 cm³]

Volume of ink in the barrel = \(\pi r^2h\)

= \(\frac{22}{7} \times (0.25)^2 \times 7\) cm³ [Since, \(r = \frac{5}{2}\) mm = 0.25cm]

= 1.3750 cm³

\(\therefore\) 1.3750 cm³ of ink is used in writing 310 words

\(\Rightarrow\) 1 cm³ of ink is used in writing \(\frac{310}{1.3750}\) words

and 200 cm³ of ink is used in writing \(\frac{310}{1.3750} \times 200\) words

= 45091 words

= 45100 words [To the nearest 100 words] Ans.

Teacher's Note

This problem illustrates how mathematical proportions help determine product usage rates and estimate supply needs in everyday situations.

Example 3

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal everywhere is 0.4 cm in thickness. Calculate the volume of the metal correct to one place of decimal.

Solution

Since, internal radius (r) = \(\frac{11.2}{2}\) cm = 5.6 cm

\(\Rightarrow\) External radius (R) = 5.6 cm + 0.4 cm = 6.0 cm

\(\therefore\) Volume of metal = \(\pi(R^2 - r^2)h\)

= \(\frac{22}{7}\) [(6)² - (5.6)²] \(\times\) 21 cm³ [Given, h = 21 cm]

= 306.24 cm³ = 306.2 cm³ Ans.

Example 4

Find the volume of the largest cylinder formed when a rectangular piece of paper 22 cm by 15 cm is rolled along its longer side.

Solution

It is clear from the figure, drawn alongside, that the circumference of the cross-section of the cylinder formed is 22 cm and its height is 15 cm.

Let the radius of the cylinder = r cm

\(\therefore 2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{7}{2}\) cm.

Volume of the cylinder = \(\pi r^2h\)

= \(\frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 15\) cm³ = 577.5 cm³ Ans.

Teacher's Note

Rolling paper into cylinders demonstrates how changing the shape of a material affects its geometric properties while conserving volume.

Example 5

When a metal cube is completely submerged in water contained in a cylindrical vessel with diameter 30 cm, the level of water rises by \(1\frac{41}{99}\) cm.

Find: (i) the length of edge of the cube. (ii) the total surface area of the cube.

Solution

(i) \(\therefore\) Radius (r) of the vessel = \(\frac{30}{2}\) cm = 15 cm

and, rise in level (h) of the water = \(1\frac{41}{99}\) cm = \(\frac{140}{99}\) cm

\(\therefore\) Volume of water that rises = \(\pi r^2h\)

= \(\frac{22}{7} \times (15)^2 \times \frac{140}{99}\) cm³ = 1000 cm³

Let the length of edge of the cube = a cm

\(\therefore\) Its volume = a³ cm³

Clearly, volume of the cube submerged = vol. of water that rises by \(1\frac{41}{99}\) cm

\(\Rightarrow\) a³ = 1000 = 10³

\(\therefore\) a = 10

\(\Rightarrow\) The length of the edge of the cube = 10 cm Ans.

(ii) Total surface area of the cube = 6a² = 6 \(\times\) 10² cm² = 600 cm² Ans.

Teacher's Note

Displacement of water by submerged objects is a practical method used in science and engineering to measure volumes of irregular shapes.

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