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

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of rs 210 per 100 m².

Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = πrl m²
=(227×25×7) m²
=550 m²
Rate of white- washing = rs 210 per 100 m²
Total cost of white-washing the tomb = rs (550 × 210/100) = rs1155

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Radius of the cone (r) = 7 cm
Height of the cone (h) = 24 cm
Let l be the slant height
∴ l = √h²+ r²
⇒ l = √24²+ 7²
⇒ l = √625
⇒ l = 25 m
Sheet required for one cap = Curved surface of the cone
= πrl cm²
= (22/7 × 7 × 25) cm²
= 550 cm²
Sheet required for 10 caps = 550 × 10 cm²= 5500 cm²

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is rs 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m
Height of the cone (h) = 1 m
Let l be the slant height of a cone.
∴ l = √h²+ r²
⇒ l = √1²+ 0.2²
⇒ l = √1.04
⇒ l = 1.02 m
Rate of painting = rs12 per m²

Curved surface of 1 cone = πrl m²
= (3.14 × 0.2 × 1.02) m²
= 0.64056 m²
Curved surface of such 50 cones = (50 × 0.64056) m²
= 32.028 m²
Cost of painting all these cones = rs(32.028 × 12)
= rs 384.34

1. Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
                    = (4 × 22/7 × 10.5 × 10.5) cm²
                    = 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
                     = (4 × 22/7 × 5.6 × 5.6) cm²
                     = 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
                     = (4 × 22/7 × 14 × 14) cm²
                   = 2464 cm²

Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
                    = (4 × 22/7 × 10.5 × 10.5) cm²
                    = 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
                     = (4 × 22/7 × 5.6 × 5.6) cm²
                     = 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
                     = (4 × 22/7 × 14 × 14) cm²
                   = 2464 cm²

Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

r = 10 cm
Total surface area of hemisphere = 3πr²
                                                     = (3 × 3.14 × 10 ×10) cm²
                                                     = 942 cm²

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr²/4πR²
= r²/R²
= (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of rs16 per 100 cm².

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
= (2 × 22/7 × 5.25 × 5.25) cm²
= 173.25 cm²
Rate of tin - plating is = rs16 per 100 cm²
Therefor, cost of 1 cm²=rs16/100
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= rs 27.72

NCERT Solution: Surface Areas and Volumes

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