Surface Areas and Volumes NCERT Solution, study material, CBSE Notes

NCERT Solution: Surface Areas and Volumes

The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm²
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × 22/7 × 0.7) (1 + 0.7) m²
= (2 × 22 × 0.1 × 1.7) m²
=7.48 m²

A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Let R be external radius and r be the internal radius h be the length of the pipe.

R = 4.4/2 cm = 2.2 cm

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm²
= 2 × 22/7 × 2 × 77cm²
= 968 cm²

(ii) Outer curved surface = 2πRh cm²

= 2 × 22/7 × 2.2 × 77 cm²
= 1064.8 cm²

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases

= 2πrh + 2πRh + 2π(R^{2 –} r²)
= [968 + 1064.8 + (2 × 22/7) (4.84 – 4)] cm²
= (2032.8 + 44/7 × 0.84) cm²
= (2032.8 + 5.28) cm²= 2038.08 cm²

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Answer

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × 22/7 × 0.42 × 1.2) m²
= 3.168 m²
Area of the playground = (500 × 3.168) m²= 1584 m²

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of rs12.50 per m²

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = rs12.50 per m²
Curved surface = 2πrh
= (2 × 22/7 × 0.25 × 3.5) m²
=5.5 m²
Total cost of painting = (5.5 × 12.5) = rs 68.75

Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m²
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of rs 40 / m²

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = rs 40 per m²
(i) Curved surface = 2πrh
= (2 × 22/7 × 1.75 × 10) m²
= 110 m²

(ii) Cost of plastering = ��(110 × 40) = rs4400

In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
= (2 × 22/7 × 0.025 × 28) m²
= 4.4 m²