Chapter 11 SURFACE AREAS AND VOLUMES Solutions

Table of contents

Exercise 11.1

Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π = 22 / 7)
Ans: Radius of cone r = 10.5/2 = 5.25 cm and Slant height (l) = 10 cm
Curved surface area of cone = πrl
= 22/7 × 5.25 × 10
= 22 × 0.75 × 10 = 165 cm²
Hence, the curved surface area of cone is 165 cm².

Q2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22 / 7)
Ans: 

Radius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m² 
= 22/7 × 12 × (21 + 12) m² 
= (22/7 × 12 × 33) m² 
= 1244.57 m²

Q3. 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. (Assume π = 22 / 7)
Ans: (i) Curved surface area of cone = 308 cm² and slant height / = 14 cm
Let, the radius of base of cone = r cm
Curved surface area of cone = πrl
⇒ 308 = 22/7 × r × 14
⇒ 308 = 44r
⇒ r = 308/44 = 7 cm
Hence, the radius of base of cone is 7 cm.
(ii) Total surface area of cone = πr(r + l)
= 22/7 × 7 × (7 + 14)
= 22 × 21 = 462 cm² 
Hence, the total surface area of cone is 462 cm².

Q4. 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 Rs 70. (Assume π = 22 / 7)
Ans: (i) Radius of cone r = 24 m and height h = 10 m
Let, the slant height = l m
We know that, l² = r² + h²
⇒ l² =24² +10² = 576 + 100 = 676
l = √676 = 26m
(ii) Area of canvas to make the tent = πrl
22/7 × 24 × 26 m²
Cost of 1 m² canvas = ₹ 70
Therefore, the cost of 22/7 × 24 × 26 m² canvas = ₹ 70 × 22/7 × 24 × 26 = ₹ 137280
Hence, the cost of canvas to make the tent is ₹ 137280.

Q5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? 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]
Ans: Radius of cone r = 6 m and height h = 8 m
Let, the slant height = l m
We know that, l² = r² + h² 
⇒ l² = 6² + 8² = 36 + 64 = 100 ⇒ l = √100 = 10 m
Area of tarpaulin to make the tent = πrl
= 3.14 × 6 × 10 = 188.40m² 
Let, the length of 3 m wide tarpaulin = L, therefore, the area of tarpaulin required = 3 × L
According to question,
3 × L = 188.40 ⇒ L = 188.40/3 = 62.80 m
Extra tarpaulin for stitching margins and wastage = 20 cm = 0.20 m
Therefore, the total length of tarpaulin = 62.80 + 0.20 = 63 m
Hence, the length of 3 m wide tarpaulin is 63 m to make the tent.

Q6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m². (Assume π = 22 / 7).
Ans: Radius of conical tomb r = 14/2 = 7 m and slant height / = 25 m/
Curved surface area of conical tomb = πrl
= 22/7 × 7 × 25 = 550 m² 
Cost of white washing at the rate of ₹ 210 per 100 m² = ₹550 × 210/100 = ₹1155
Hence, the cost of white washing curved surface area is ₹ 1155.

Q.7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π = 22 / 7)
Ans: Radius of conical cap, r = 7 cm
Height of conical cap, h = 24cm We know that, l² = (r² + h²)
⇒ l² = 7² + 24² = 49 + 576 = 625 ⇒ l = √625 = 25 cm
Area of sheet required to make 1 cap = πrl
= 22/7 × 7 × 25 = 550 cm²
Therefore, area of sheet required to make 10 such caps = 10 × 550 = 5500 cm²
Hence, area of sheet to make 10 such caps is 5500 cm².

Q8. 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).
Ans: Radius of cone r = 40/2 = 20 cm = 0.2 m and height h = 1 m
Let the slant height = l m
 We know that l² = r² + h² 
⇒ l²= (0.2)² + 1² = 0.04 + 1 = 1.04
⇒ l= √1.04 = 1.02 m
Curved surface area of cone = πrl
= 3.14 × 0.2 × 1.02 = 6.4056m² 
Curved surface area of 50 cones
= 50 × 6.4056
= 32.028 m² 
Cost of painting at the rate of ₹12 per m² 
= ₹12 × 32.028
= ₹384.34 (approx.)
Hence, the cost of painting the curved surface of 50 cones is ₹384.34

Exercise 11.2

Q1. Find the surface area of a sphere of radius:
(i) 10.5cm
(ii) 5.6cm
(iii) 14cm
(Assume π = 22 / 7)
Ans: (i) Radius of sphere r = 10.5 cm
Surface area of sphere = 4πr² 
= 4 × 22/7 × 10.5 × 10.5 = 4 × 22 × 1.5 × 10.5 = 1386.00 cm²
Hence, the surface area of sphere is 1386 cm².
(ii) Radius of sphere r = 5.6 cm
Surface area of sphere = 4πr² 
= 4 × 22/7 × 5.6 × 5.6 = 4 × 22 × 0.8 × 5.6 = 394.24 cm² 
Hence, the surface area of sphere is 394.24 cm².
(iii) Radius of sphere r = 14 cm
Surface area of sphere = 4πr² 
= 4 × 22/7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm² 
Hence, the surface area of sphere is 2464 cm².

Q2. Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
(Assume π = 22 / 7)
Ans: (i) Radius of sphere r = 14/2 = 7 cm
Surface area of sphere = 4πr² 
= 4 × 22/7 × 7 × 7 = 4 × 22 × 7 = 616 cm² 
Hence, the surface area of sphere is 616 cm².
(ii) Radius of sphere r = 21/2 = 10.5 cm
Surface area of sphere = 4πr² 
= 4 × 22/7 × 10.5 × 10.5 = 4 × 22 × 1.5 × 10.5 = 1386 cm² 
Hence, the surface area of sphere is 1386 cm².
(iii) Radius of sphere r = 3.5/2 = 1.75 cm
Surface area of sphere = 4π² 
= 4 × 22/7 × 1.75 × 1.75 = 4 × 22 × 0.25 × 1.75
= 38.50 cm² 
Hence, the surface area of sphere is 38.5 cm².

Q3. Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Ans: Radius of hemisphere r = 10 cm
Surface area of hemisphere = 3πr²
= 3 × 3.14 × 10 × 10 = 942 cm²
Hence, the total surface area of hemisphere is 942 cm².

Q4. 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.
Ans: 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

Q5. 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 Rs 16 per 100 cm². (Assume π = 22 / 7)
Ans: 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 = ₹16 per 100 cm²
Therefor, cost of 1 cm² = ₹16/100 
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72

Q6. Find the radius of a sphere whose surface area is 154 cm². (Assume π = 22/7)
Ans: Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = 154/(4 × 22/7)
⇒ r² = 49/4
⇒ r = 7/2 = 3.5 cm

Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Ans: Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
= (1/64)/(1/4)
= 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

Q8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π = 22 / 7)
Ans: 
Internal radius of hemispherical bowl r = 5 cm and thickness = 0.25 cm
Therefore,
The outer radius of hemispherical bowl
= R = 5 + 0.25 = 5.25 cm
Outer curved surface area of hemispherical bowl = 2πR² 
= 2 × 22/7 × 5.25 × 5.25
= 2 × 22 × 0.75 × 5.25
= 173.25 cm² 
Hence, the outer curved surface area of hemispherical bowl is 173.25 cm².

Q9. A right circular cylinder just encloses a sphere of radius r (see Fig). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in(i) and (ii).
Ans: (i) Radius of sphere = radius of cylinder = r
Hence, the surface area of sphere = 4πr² 
(ii) Radius of cylinder = r and height h = diameter of sphere = 2r
Hence, the curved surface area of cylinder = 2πrh = 2πr(2r) = 4πr² 
(iii) Now, Surface area of sphere/Curved surface area of cylinder = 4πr²/4πr² = 1/1
Hence, the ratio of surface area of sphere to curved surface area of cylinder is 1 : 1.

Exercise 11.3

Q1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm (Assume π = 22 / 7)
Ans: (i) Radius (r) = 6 cm
Height (h) = 7 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 6 × 6 × 7) cm³ 
= 264 cm³ 
(ii) Radius (r) = 3.5 cm
Height (h) = 12 cm
Volume of the cone = 1/3 πr²h
Volume of the cone = 1/3 πr²h
= 154 cm³

Q2. 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 (Assume π = 22 / 7)
Ans: (i) Radius (r) = 7 cm
Slant height (l) = 25 cm
Let h be the height of the conical vessel.
∴ h = √l² – r²
⇒ h = √25² – 7² 
⇒ h = √576 
⇒ h = 24 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 7 × 7 × 24) cm³
= 1232 cm³  
Capacity of the vessel = (1232/1000) = 1.232 litres
(ii) Height (h) = 12 cm
Slant height (l) = 13 cm
Let r be the radius of the conical vessel.
∴ r = √l² – h² 
⇒ r = √13² – 12²
⇒ r = √25 
⇒ r = 5 cm
Volume of the cone = 1/3 πr²h
= (1/3 × 22/7 × 5 × 5 × 12) cm³ 
= (2200/7) cm³ 
Capacity of the vessel = (2200/7000) l = 11/35 litres

Q3. The height of a cone is 15cm. If its volume is 1570cm³, find the diameter of its base. (Use π = 3.14)
Ans:  Height (h) = 15 cm
Volume = 1570 cm³
Let the radius of the base of cone be r cm
∴ Volume = 1570 cm³ 
⇒ 1/3 πr²h = 1570
⇒ 13 × 3.14 × r² × 15 = 1570
⇒ r² = 1570/(3.14×5) = 100
⇒ r = 10 cm 
Diameter of base = 2r = 2 x 10 = 20 cm 

Q4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Ans:  Given height of a cone, h = 9 cm
Volume of the cone = 48
(1/3)r²h = 48
(1/3)r² × 9 = 48
3r² = 48
r² = 48/3 = 16
r = 4
So diameter = 2 × radius
= 2 × 4
= 8 cm
Hence the diameter of the cone is 8 cm.

Q5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? (Assume π = 22 / 7)
Ans:  Radius of pit r = 3.5/2 = 1.75 m and height h = 12 m.
Volume of pit = 1/3 πr²h
= 1/3 × 22/7 × 1.75 × 1.75 × 12 = 38.5 m³
= 38.5 Kilolitres [∴ 1 m³ = 1 kilolitres]
Hence, the capacity of the pit is 38.5 kilolitres.

Q6. 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 (Assume π = 22 / 7)
Ans:  (i) Diameter of the base of the cone = 28 cm
Radius (r) = 28/2 cm = 14 cm
Let the height of the cone be h cm
Volume of the cone = 13πr²h = 9856 cm³ 
⇒ 1/3 πr²h = 9856
⇒ 1/3 × 22/7 × 14 × 14 × h = 9856
⇒ h = (9856 × 3)/(22/7 × 14 × 14)
⇒ h = 48 cm
(ii) Radius (r) = 14 m
Height (h) = 48 cm
Let l be the slant height of the cone
l² = h² + r²
⇒ l² = 48² + 14²
⇒ l² = 2304+196
⇒ l² = 2500
⇒ ℓ = √2500 = 50 cm
(iii) Radius (r) = 14 m
Slant height (l) = 50 cm
Curved surface area = πrl
= (22/7 × 14 × 50) cm²
= 2200 cm²

Q7. 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.
Ans: If the triangle is revolved about 12 cm side, a cone will be formed.

Therefore, the radius of cone r = 5 cm, height h = 12 cm and slant height l =13 cm.
Volume of solid (cone) = 1/3 πr²h 
= 1/3 × π × 5 × 5 × 12 = 100 π cm³ 
Hence, the volume of solid is 100π cm³.

Q8. If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Ans: 

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.

Volume of cone = (1/3) πr²h; where r is the radius and h be the height of cone

= (1/3) × π × 12 × 12 × 5

= 240 π

The volume of the cones of formed is 240π cm³.

So, required ratio = (result of question 7) / (result of question 8) = (100π) / (240π) = 5 / 12 = 5 : 12.

Q9. 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. (Assume π = 22 / 7)
Ans:  Radius (r) of heap = (10.5 / 2) m = 5.25

Height (h) of heap = 3m

Volume of heap = (1 / 3)πr²h

= (1 / 3) × (22 / 7) × 5.25 × 5.25 × 3

= 86.625 m³

The volume of the heap of wheat is 86.625 m³. Again,

Area of canvas required = CSA of cone = πrl, where l = 

After substituting the values, we have

= (22 / 7) × 5.25 × 6.05

= 99.825

Therefore, the area of the canvas is 99.825 m².

Exercise 11.4

Q1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m (Assume π = 22 / 7)
Ans: (i) Radius of sphere, r = 7 cm
Using, Volume of sphere = (4 / 3) πr³
= (4 / 3) × (22 / 7) × 7³ = 4312 / 3
Hence, volume of the sphere is 
(ii) Radius of sphere, r = 0.63 m
Using, volume of sphere = (4 / 3) πr³
= (4 / 3) × (22 / 7) × 0.63³ = 1.0478
Hence, volume of the sphere is 1.05 m³ (approx).

Q2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m (Assume π = 22 / 7)
Ans: (i) Radius of spherical ball r = 28/2 = 14 cm
Volume of water displaced by spherical ball = 4/3 πr³
= 4/3 × 22/7 × 14 × 14 × 14
= 4/3 × 22 × 2 × 14 × 14 = 

or =11498.67 cm³

(ii) Radius of spherical ball r = 0.21/2 = 0.105 m

Volume of water displaced by spherical ball = 4/3 πr³ 

= 4/3 × 22/7 × 0.105 × 0.105 × 0.105 = 4 × 22 × 0.005 × 0.63 × 0.63 = 0.004861 m³ 

Hence, the volume of water displaced by spherical ball is 0.004861 m³.

Q3. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³? (Assume π = 22 / 7)
Ans:  Radius of metallic ball r = 4.2/2 = 2.1 cm
Therefore, the volume of metallic ball = 4/3 πr³ 
= 4/3 × 22/7 × 2.1 × 2.1 × 2.1 
= 4 × 22 × 0.1 × 2.1 × 2.1 = 38.808 cm³
Here, the mass of 1 cm³ = 8.9 g 
So, the mass of 38.808 cm³ = 8.9 × 38.808 = 345.39 g (approx.)
Hence, the mass of the ball is 345.39 gram.

Q4. 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?
Ans:  Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2
Diameter of moon will be d / 4 and the radius of moon will be d/8
Find the volume of the moon:
Volume of the moon = (4 / 3) πr³ = (4 / 3) π (d / 8)³ = 4 / 3π(d³ / 512)
Find the volume of the earth:
Volume of the earth = (4 / 3) πr³ = (4 / 3) π (d / 2)³ = 4 / 3π(d³ / 8)
Fraction of the volume of the earth is the volume of the moon

Volume of moon is of the 1 / 64 volume of earth.

Q5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22 / 7)
Ans:  Diameter of hemispherical bowl = 10.5 cm
Radius of hemispherical bowl, r = 10.5 / 2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = (2 / 3) πr³
Volume of the hemispherical bowl = (2 / 3) × (22 / 7) × 5.25³ = 303.1875
Volume of the hemispherical bowl is 303.1875 cm³
Capacity of the bowl = (303.1875) / 1000 L = 0.303 litres(approx.)
Therefore, hemispherical bowl can hold 0.303 litres of milk.

Q6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)
Ans:  Inner Radius of the tank, (r) = 1m
Outer Radius (R) = 1.01m
Volume of the iron used in the tank = (2 / 3) π (R³ – r³)
Put values,
Volume of the iron used in the hemispherical tank = (2 / 3) × (22 / 7) × (1.01³ – 1³) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348 m³.

Q7. Find the volume of a sphere whose surface area is 154 cm². (Assume π = 22 / 7)
Ans:  Surface area of sphere A = 154 cm²
Let, the radius of sphere = r cm
We know that the surface area of sphere = 4πr² 

Volume of surface 4/3 πr³

Hence, the volume of sphere is .

Q8. 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 meter, find the
(i) Inside surface area of the dome
(ii) volume of the air inside the dome (Assume π = 22 / 7)
Ans: (i) Let the internal radius of dome = r m
Internal surface area of dome = 2πr² 
Cost of white washing at the rate of ₹2 = 2πr² × ₹2 = ₹4πr²
⇒ ₹4πr² = ₹498.96
⇒ 4 × 22/7 × r² = 498.96

Therefore, the internal surface of dome = 2πr² 
= 2 × 22/7 × (6.3)²
= 2 × 22/7 × 6.3 × 6.3 
= 2 × 22 × 0.9 × 6.3
= 249.48 m² 
Hence, the inside surface area of the dome is 249.48 m².
(ii) Volume of the air inside the dome = 2/3 πr³ 
= 2/3 × 22/7 × (6.3)³ 
= 2/3 × 22/7 × 6.3 × 6.3 × 6.3
= 2 × 22 × 0.3 × 6.3 × 6.3
= 523.9 cm³ 
= Hence, the volume of the air inside the dome is 523.9 cm³.

Q9. 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 Sand S’.
Ans:  Volume of the solid sphere = (4 / 3)πr³
Volume of twenty seven solid sphere = 27 × (4 / 3)πr³ = 36 πr³
(i) New solid iron sphere radius = r’
Volume of this new sphere = (4/3)π(r’)³
(4 / 3)π(r’)³ = 36 πr³
(r’)³ = 27r³
r’= 3r
Radius of new sphere will be 3r (thrice the radius of original sphere)
(ii) Surface area of iron sphere of radius r, S = 4πr²
Surface area of iron sphere of radius r’= 4π (r’)²
Now
S / S’ = (4πr²) / ( 4π (r’)²)
S / S’ = r² / (3r’)² = 1 / 9
The ratio of S and S’ is 1: 9.

Q10. 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? (Assume π = 22 / 7)
Ans:  Radius of capsule r = 3.5/2 = 1.75 mm
Volume of medicine to fill the capsule = 4/3πr³  
= 4/3 × 22/7 × 1.75 × 1.75 × 1.75
= 4/3 × 22 × 0.25 × 1.75 × 1.75
= 22.46 mm³ (approx.)
Hence, 22.46 mm³ medicine is required to fill this capsule.