Surface Areas and Volumes - NCERT Solutions

Exercise 11.1
Exercise 11.2
Exercise 11.3
Exercise 11.4
Examples

Exercise 11.1

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Given: Diameter = 10.5 cm, so radius (r) = 10.5/2 = 5.25 cm

Slant height (l) = 10 cm

Curved Surface Area = πrl

= (22/7) × 5.25 × 10

= (22/7) × 52.5

= 22 × 7.5 = 165 cm²

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Given: Slant height (l) = 21 m

Diameter = 24 m, so radius (r) = 12 m

Total Surface Area = πr(l + r)

= (22/7) × 12 × (21 + 12)

= (22/7) × 12 × 33

= (22/7) × 396

= 22 × 56.57 = 1244.54 m² (approx.)

3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.

(i) Curved Surface Area = πrl = 308 cm²

(22/7) × r × 14 = 308

44r = 308

r = 308/44 = 7 cm

(ii) Total Surface Area = πr(l + r)

= (22/7) × 7 × (14 + 7)

= 22 × 21 = 462 cm²

4. A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70.

(i) Height (h) = 10 m, Radius (r) = 24 m

Slant height (l) = √(r² + h²) = √(24² + 10²) = √(576 + 100) = √676 = 26 m

(ii) Curved Surface Area = πrl = (22/7) × 24 × 26

= (22/7) × 624 = 1961.14 m² (approx.)

Cost = 1961.14 × 70 = ₹ 137,279.80

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

Height (h) = 8 m, Radius (r) = 6 m

Slant height (l) = √(r² + h²) = √(36 + 64) = √100 = 10 m

Curved Surface Area = πrl = 3.14 × 6 × 10 = 188.4 m²

Width of tarpaulin = 3 m

Length required = Area/Width = 188.4/3 = 62.8 m

Extra length for margins = 20 cm = 0.2 m

Total length required = 62.8 + 0.2 = 63 m

6. 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 ₹ 210 per 100 m².

Slant height (l) = 25 m

Diameter = 14 m, so radius (r) = 7 m

Curved Surface Area = πrl = (22/7) × 7 × 25 = 550 m²

Cost = (550/100) × 210 = 5.5 × 210 = ₹ 1155

7. 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 (r) = 7 cm, Height (h) = 24 cm

Slant height (l) = √(r² + h²) = √(49 + 576) = √625 = 25 cm

Curved Surface Area for one cap = πrl = (22/7) × 7 × 25 = 550 cm²

Area for 10 caps = 10 × 550 = 5500 cm²

8. 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 ₹ 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)

Diameter = 40 cm = 0.4 m, so radius (r) = 0.2 m

Height (h) = 1 m

Slant height (l) = √(r² + h²) = √(0.04 + 1) = √1.04 = 1.02 m

Curved Surface Area of one cone = πrl = 3.14 × 0.2 × 1.02 = 0.64056 m²

Total area for 50 cones = 50 × 0.64056 = 32.028 m²

Cost = 32.028 × 12 = ₹ 384.34 (approx.)

Exercise 11.2

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

(i) Surface Area = 4πr² = 4 × (22/7) × 10.5 × 10.5 = 1386 cm²

(ii) Surface Area = 4πr² = 4 × (22/7) × 5.6 × 5.6 = 394.24 cm²

(iii) Surface Area = 4πr² = 4 × (22/7) × 14 × 14 = 2464 cm²

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

(i) Diameter = 14 cm, so radius = 7 cm

Surface Area = 4πr² = 4 × (22/7) × 7 × 7 = 616 cm²

(ii) Diameter = 21 cm, so radius = 10.5 cm

Surface Area = 4πr² = 4 × (22/7) × 10.5 × 10.5 = 1386 cm²

(iii) Diameter = 3.5 m, so radius = 1.75 m

Surface Area = 4πr² = 4 × (22/7) × 1.75 × 1.75 = 38.5 m²

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

Total Surface Area of hemisphere = 3πr²

= 3 × 3.14 × 10 × 10

= 3 × 3.14 × 100 = 942 cm²

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

Surface Area when radius = 7 cm: 4π(7)² = 4π × 49

Surface Area when radius = 14 cm: 4π(14)² = 4π × 196

Ratio = (4π × 49) : (4π × 196) = 49 : 196 = 1 : 4

5. 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 ₹ 16 per 100 cm².

Inner diameter = 10.5 cm, so inner radius = 5.25 cm

Inner surface area (hemisphere) = 2πr² = 2 × (22/7) × 5.25 × 5.25

= 2 × (22/7) × 27.5625 = 173.25 cm²

Cost = (173.25/100) × 16 = ₹ 27.72

6. Find the radius of a sphere whose surface area is 154 cm².

Surface Area = 4πr² = 154

4 × (22/7) × r² = 154

(88/7) × r² = 154

r² = (154 × 7)/88 = 12.25

r = √12.25 = 3.5 cm

7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Let diameter of earth = D, then diameter of moon = D/4

Radius of earth = D/2, Radius of moon = D/8

Surface Area of earth = 4π(D/2)² = πD²

Surface Area of moon = 4π(D/8)² = 4πD²/64 = πD²/16

Ratio = (πD²/16) : (πD²) = 1 : 16

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Inner radius = 5 cm

Thickness = 0.25 cm

Outer radius = 5 + 0.25 = 5.25 cm

Outer curved surface area = 2πr² = 2 × (22/7) × 5.25 × 5.25

= 2 × (22/7) × 27.5625 = 173.25 cm²

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 11.10). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).

(i) Surface Area of sphere = 4πr²

(ii) Height of cylinder = 2r, Radius of cylinder = r

Curved Surface Area of cylinder = 2πrh = 2πr × 2r = 4πr²

(iii) Ratio = (4πr²) : (4πr²) = 1 : 1

Exercise 11.3

1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm

(i) Volume = (1/3)πr²h = (1/3) × (22/7) × 6 × 6 × 7 = 264 cm³

(ii) Volume = (1/3)πr²h = (1/3) × (22/7) × 3.5 × 3.5 × 12 = 154 cm³

2. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

(i) Radius (r) = 7 cm, Slant height (l) = 25 cm

Height (h) = √(l² - r²) = √(625 - 49) = √576 = 24 cm

Volume = (1/3)πr²h = (1/3) × (22/7) × 7 × 7 × 24 = 1232 cm³

Capacity = 1232/1000 = 1.232 litres

(ii) Height (h) = 12 cm, Slant height (l) = 13 cm

Radius (r) = √(l² - h²) = √(169 - 144) = √25 = 5 cm

Volume = (1/3)πr²h = (1/3) × (22/7) × 5 × 5 × 12 = 314.29 cm³

Capacity = 314.29/1000 = 0.314 litres

3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Volume = (1/3)πr²h = 1570

(1/3) × 3.14 × r² × 15 = 1570

15.7 × r² = 1570

r² = 1570/15.7 = 100

r = 10 cm

4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Volume = (1/3)πr²h = 48π

(1/3) × π × r² × 9 = 48π

3r² = 48

r² = 16

r = 4 cm

Diameter = 8 cm

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Diameter = 3.5 m, so radius = 1.75 m

Height = 12 m

Volume = (1/3)πr²h = (1/3) × (22/7) × 1.75 × 1.75 × 12 = 38.5 m³

1 m³ = 1 kilolitre

Capacity = 38.5 kilolitres

6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone

Diameter = 28 cm, so radius = 14 cm

Volume = (1/3)πr²h = 9856

(1/3) × (22/7) × 14 × 14 × h = 9856

(1/3) × 22 × 28 × h = 9856

205.33h = 9856

h = 9856/205.33 = 48 cm

(i) Height = 48 cm

(ii) Slant height (l) = √(r² + h²) = √(196 + 2304) = √2500 = 50 cm

(iii) Curved Surface Area = πrl = (22/7) × 14 × 50 = 2200 cm²

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

When revolved about side 12 cm, we get a cone with:

Height (h) = 12 cm, Radius (r) = 5 cm

Volume = (1/3)πr²h = (1/3) × π × 25 × 12 = 100π cm³

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

When revolved about side 5 cm, we get a cone with:

Height (h) = 5 cm, Radius (r) = 12 cm

Volume = (1/3)πr²h = (1/3) × π × 144 × 5 = 240π cm³

Ratio = 100π : 240π = 5 : 12

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Diameter = 10.5 m, so radius = 5.25 m

Height = 3 m

Volume = (1/3)πr²h = (1/3) × (22/7) × 5.25 × 5.25 × 3 = 86.625 m³

Slant height (l) = √(r² + h²) = √(27.5625 + 9) = √36.5625 = 6.046 m

Canvas area required = πrl = (22/7) × 5.25 × 6.046 = 99.825 m²

Exercise 11.4

1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m

(i) Volume = (4/3)πr³ = (4/3) × (22/7) × 7 × 7 × 7 = 1437.33 cm³

(ii) Volume = (4/3)πr³ = (4/3) × (22/7) × 0.63 × 0.63 × 0.63 = 1.0478 m³

2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m

(i) Diameter = 28 cm, so radius = 14 cm

Volume = (4/3)πr³ = (4/3) × (22/7) × 14 × 14 × 14 = 11498.67 cm³

Water displaced = 11498.67 cm³

(ii) Diameter = 0.21 m, so radius = 0.105 m

Volume = (4/3)πr³ = (4/3) × (22/7) × 0.105 × 0.105 × 0.105 = 0.004851 m³

Water displaced = 0.004851 m³ = 4.851 litres

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Diameter = 4.2 cm, so radius = 2.1 cm

Volume = (4/3)πr³ = (4/3) × (22/7) × 2.1 × 2.1 × 2.1 = 38.808 cm³

Mass = Volume × Density = 38.808 × 8.9 = 345.39 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Let diameter of earth = D, then diameter of moon = D/4

Radius of earth = D/2, Radius of moon = D/8

Volume of earth = (4/3)π(D/2)³ = (4/3)π(D³/8) = (1/6)πD³

Volume of moon = (4/3)π(D/8)³ = (4/3)π(D³/512) = (1/384)πD³

Fraction = [(1/384)πD³] / [(1/6)πD³] = (1/384) × 6 = 6/384 = 1/64

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Diameter = 10.5 cm, so radius = 5.25 cm

Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × 5.25 × 5.25 × 5.25

= (2/3) × (22/7) × 144.703 = 303.1875 cm³

Capacity = 303.1875/1000 = 0.303 litres

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Inner radius = 1 m = 100 cm

Thickness = 1 cm

Outer radius = 100 + 1 = 101 cm

Volume of iron = Volume of outer hemisphere - Volume of inner hemisphere

= (2/3)π(101)³ - (2/3)π(100)³

= (2/3)π(1030301 - 1000000)

= (2/3)π(30301) = 63487.81 cm³

7. Find the volume of a sphere whose surface area is 154 cm².

Surface Area = 4πr² = 154

4 × (22/7) × r² = 154

(88/7) × r² = 154

r² = (154 × 7)/88 = 12.25

r = 3.5 cm

Volume = (4/3)πr³ = (4/3) × (22/7) × 3.5 × 3.5 × 3.5 = 179.67 cm³

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.

(i) Cost = ₹ 4989.60, Rate = ₹ 20 per m²

Inside surface area = 4989.60/20 = 249.48 m²

Surface area of hemisphere = 2πr² = 249.48

2 × (22/7) × r² = 249.48

(44/7) × r² = 249.48

r² = (249.48 × 7)/44 = 39.69

r = √39.69 = 6.3 m

(ii) Volume of air = (2/3)πr³ = (2/3) × (22/7) × 6.3 × 6.3 × 6.3 = 523.908 m³

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r' of the new sphere, (ii) ratio of S and S'.

(i) Volume of 27 spheres = 27 × (4/3)πr³ = 36πr³

Volume of new sphere = (4/3)π(r')³

So, (4/3)π(r')³ = 36πr³

(r')³ = 27r³

r' = 3r

(ii) S = 4πr², S' = 4π(r')² = 4π(3r)² = 36πr²

Ratio S : S' = 4πr² : 36πr² = 1 : 9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Diameter = 3.5 mm, so radius = 1.75 mm

Volume = (4/3)πr³ = (4/3) × (22/7) × 1.75 × 1.75 × 1.75

= (4/3) × (22/7) × 5.359 = 22.458 mm³

Important Examples

Example 1: A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Let radius of both = r

Height of hemisphere = r

Volume of hemisphere = (2/3)πr³

Volume of cone = (1/3)πr²h

Given: (1/3)πr²h = (2/3)πr³

h = 2r

Ratio of heights (cone : hemisphere) = 2r : r = 2 : 1

Example 2: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Radius of cone = 1 cm, Height of cone = 1 cm

Radius of hemisphere = 1 cm

Volume of cone = (1/3)πr²h = (1/3)π(1)²(1) = π/3 cm³

Volume of hemisphere = (2/3)πr³ = (2/3)π(1)³ = 2π/3 cm³

Total volume = π/3 + 2π/3 = π cm³

Example 3: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Volume of spherical part = (4/3)πr³ = (4/3) × 3.14 × (4.25)³

= (4/3) × 3.14 × 76.7656 = 321.392 cm³

Volume of cylindrical neck = πr²h = 3.14 × (1)² × 8 = 25.12 cm³

Total volume = 321.392 + 25.12 = 346.512 cm³

The child's measurement of 345 cm³ is approximately correct.

Example 4: A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-putt.

Radius = 4.9 cm

Volume = (4/3)πr³ = (4/3) × (22/7) × 4.9 × 4.9 × 4.9

= (4/3) × (22/7) × 117.649 = 493 cm³ (approx.)

Mass = Volume × Density = 493 × 7.8 = 3845.4 g = 3.845 kg

Example 5: A hemispherical tank full of water is emptied by a pipe at the rate of 3 4/7 litres per second. How much time will it take to empty half the tank, if it is 3 m in diameter?

Diameter = 3 m, so radius = 1.5 m

Volume of tank = (2/3)πr³ = (2/3) × (22/7) × 1.5 × 1.5 × 1.5

= (2/3) × (22/7) × 3.375 = 7.071 m³

Half volume = 7.071/2 = 3.5355 m³ = 3535.5 litres

Rate of emptying = 3 4/7 = 25/7 litres per second

Time = 3535.5 ÷ (25/7) = 3535.5 × 7/25 = 989.94 seconds

= 16 minutes 30 seconds (approx.)

Example 6: A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Volume of rod = πr²h = π × (0.5)² × 8 = 2π cm³

Length of wire = 18 m = 1800 cm

Let radius of wire = R cm

Volume of wire = πR² × 1800

So, πR² × 1800 = 2π

R² = 2/1800 = 1/900

R = 1/30 cm

Diameter = 2/30 = 1/15 cm

Thickness = 1/15 cm = 0.067 cm

Example 7: 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.

Curved Surface Area = 2πrh = 88

2 × (22/7) × r × 14 = 88

88r = 88

r = 1 cm

Diameter = 2 cm

Example 8: It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find (i) inner curved surface area of the vessel, (ii) radius of the base, (iii) capacity of the vessel.

(i) Cost = ₹ 2200, Rate = ₹ 20 per m²

Curved Surface Area = 2200/20 = 110 m²

(ii) Curved Surface Area = 2πrh = 110

2 × (22/7) × r × 10 = 110

(440/7)r = 110

r = (110 × 7)/440 = 1.75 m

(iii) Capacity = πr²h = (22/7) × 1.75 × 1.75 × 10 = 96.25 m³

Example 9: The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Capacity = 15.4 litres = 0.0154 m³

Volume = πr²h = 0.0154

(22/7) × r² × 1 = 0.0154

r² = (0.0154 × 7)/22 = 0.0049

r = 0.07 m

Total Surface Area = 2πr(h + r) = 2 × (22/7) × 0.07 × (1 + 0.07)

= 0.44 × 1.07 = 0.4708 m²

Example 10: A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Length = 14 cm = 140 mm

Radius of pencil = 3.5 mm, Radius of graphite = 0.5 mm

Volume of graphite = πr²h = (22/7) × 0.5 × 0.5 × 140 = 110 mm³

Volume of pencil = πR²h = (22/7) × 3.5 × 3.5 × 140 = 5390 mm³

Volume of wood = 5390 - 110 = 5280 mm³