OP Malhotra Class 10 Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(a)

Practicing Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(a)

Question 1.
Find the area of curved surface and total surface area of the cylinders whose heights and radii are given below:
(i) h = 12 cm, r = 7 cm
(ii) h = 10 cm, r = 7 cm
(iii) h = 5 cm, r = 21 cm
(iv) h = 20 cm, r = 14 cm
(v) h = 16 m, r = 10.5 m
Solution:
(i) Height of a cylinder (h) = 12 cm and radius (r) = 7 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 7 × 12 = 528 cm²
Total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 7(12 + 7) cm²
= 44 × 19 = 836 cm²

(ii) Height of the cylinder (h) = 10 cm and radius (r) = 1 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 7 × 10 = 440 cm²
and total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 7(10 + 7) cm²
= 44 × 17 = 748 cm²

(iii) Height of the cylinder (h) = 5 cm and radius (r) = 21 cm
Curved surface area = 2nrh
= 2 × \(\frac { 22 }{ 7 }\) × 21 × 5 = 660 cm²
and total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 21(5 + 21) cm²
= 132 × 26 = 3432 cm²

(iv) Height of cylinder (h) = 20 cm and radius (r)= 14 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 14 × 20 cm² = 1760 cm²
Total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 14 × (20 + 14) cm²
= 88 × 34 cm² = 2992 cm²

(v) Height of cylinder (l) = 16 m
and radius (r) = 10.5 m = \(\frac { 21 }{ 2 }\) m
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\) × 16 m² = 1056 m²
and total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\) \(\left(16+\frac{21}{2}\right)\) m²
= 66 × \(\frac { 53 }{ 2 }\) = 1749 m²

Question 2.
A cylindrical tank 7 m in diameter, contains water to a depth of 4 m. Find the total area of the wet surface. (π = 3.14)
Solution:
Diameter of the base of a tank = 7 m
∴ Radius (r) = \(\frac { 7 }{ 2 }\) m
Depth (h) = 4 m
∴ Total surface area of the wet surface
= 2πrh + πr² = πr(2h + r) m²
= 3.14 × \(\frac { 7 }{ 2 }\)(2 × 4 + 3.5) m²
= 3.14 × \(\frac { 7 }{ 2 }\) × 11.5 m²
= 126.385 m² = 126.39 m²

Question 3.
Find the whole surface of a hollow cylinder open at both ends, whose external diameter is 14 cm, thickness 2 cm and height 20 cm.
Solution:
External diameter of a hollow cylinder = 14 cm
Radius (R) = \(\frac { 14 }{ 2 }\) = 7 cm
Thickness of cylinder = 2 cm
∴ Inner radius (r) = 7 – 2 = 5 cm
Height (h) = 20 cm
Inner and external surface area
2πrh + 2πRh
= 2πh(r + R)
=2 × π × 20(7 + 5) cm²
=40π × 12 = 480π cm²
Area of two rings (upper and lower)
= 2 × π(R² – r²) = 2(7² – 5²)
= 2π × (49 – 25) = 2π × 24 = 48π cm²
∴ Total surface area – 480π × 48π
= 528π cm²

Question 4.
Find the radius of the cylinder if area of its curved surface is 110 cm2, and height 5 cm. \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
Solution:
Area of curved surface of a cylinder = 110 cm²
and height (h) = 5 cm
Curved surface area = 2πrh
Then radius (r) = \(\frac{\text { Curved surface area }}{\text { 2πh }}\)

Question 5.
Find the height of the cylinder if area of its curved surface is 13.2 cm², and radius 6 cm.
Solution:
Area of curved surface of a cylinder = 13.2 cm²
Radius (r) = 6 cm
Curved surface area = 2πrh
∴ Height (h) = \(\frac{\text { Surface area }}{\text { 2πr }}\)

Question 6.
A garden roller is 75 cm in diameter and 105 cm in width. What area does it cover in 14 revolutions?
Solution:
Diameter of a garden roller = 75 cm
∴ Radius (r) = \(\frac { 75 }{ 2 }\) cm
and width (h) = 105 cm
∴ Curved surface area = 27πrh
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 75 }{ 2 }\) × 105 cm² = 24750 cm²
Area covered in 14 revolutions
= 24750 × 14 cm² = 346500 cm²
= \(\frac{346500}{100 \times 100}\) = 34.65 m²

Question 7.
10 cylindrical pillars of a building have to be painted. If the diameter of each pillar is 50 cm and the height 4 m, what will be the cost of painting these at the
rate of 50 paise per m²? (Use π = 3.14)
Solution:
Diameter of each cylinder = 50 cm
∴ Radius (r) = \(\frac { 50 }{ 2 }\) = 25 cm = \(\frac { 25 }{ 100 }\) m
and height (h) = 4 m
∴ Curved surface area = 2πrh
= 2 × 3.14 × \(\frac { 25 }{ 100 }\) × 4 m² = 6.28 m²
Area of 10 cylinders = 6.28 × 1o m² = 62.8m²
Rate of painting = 50 paise per m²
Total cost 62.8 × \(\frac { 50 }{ 100 }\) = ₹ 31.40

Question 8.
The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paise per m². \(\left(\text { Use } \pi=\frac{22}{7}\right)\)
Solution:
Diameter of a roller = 84 cm
∴ Radius (r) = \(\frac { 84 }{ 2 }\) = 42 cm
and length (h) 120 cm
Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 42 × 120 cm² = 31680 cm²
Area of leveling playground by 500 revolutions = 31680 × 500 cm²
= \(\frac{31680 \times 500}{100 \times 100}\) = 1584 cm²
Rate of leveling the ground =30 paise per m²
∴ Total cost = ₹ \(\frac{1584 \times 30}{100}\) = ₹ 475.20

Question 9.
The curved surface of a cylinder is 1000 cm². A wire of diameter 5 mm is wound round it so as to cover it completely. Find the length of the wire.
Solution:
Curved surface area of a cylinder = 1000cm²
It is covered by a wire of 5 mm diameter
Let h be the length of the ‘wire, then
2πrh = 1000 cm²
Let h be the height of the cylinder, then
Number of coils = \(\frac{\text { h }}{\text { Diameter of wire }}\) = \(\frac{h}{0.5}\)
Now total length of wire = 2 × 2πRh
= 2 × 1000 cm = \(\frac{2 \times 1000}{100}\) = 20 m

Question 10.
An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, find the whole surface of the pipe.
Solution:
Length of iron pipe = 20 cm
Exterior diameter = 25 cm
∴ Radius (R) = \(\frac{25}{2}\) cm
Thickness of pipe = 1 cm
∴ Inner radius (r) = \(\frac{25}{2}\) – 1 = \(\frac{23}{2}\)cm
∴ whole surface area

Question 11.
50 circular plates, each of radius 7 cm and thickness 0.5 cm, are placed one above the other to form a right circular cylinder. Find Its total surface area.
Solution:
Radius of each plate (r) = 7 cm
and thickness = 0.5 cm
∴ Height (thickness) of 50 plates (h)
= 0.5 × 50 = 25 cm
∴ Curved surface area of so formed cylinder
= 2 × πrh = 2 × \(\frac{22}{7}\) × 7 × 25 cm² = 1100 cm²
Area of top and bottom = 2 × πr²
= 2 × \(\frac{22}{7}\) × 7² = 308 cm²
Total surface area 1100 cm² + 308 cm²
= 1408 cm²