Exercise 11.3 Solutions (Class 9)
Surface Areas and Volumes
(Assume π = 22/7, unless stated otherwise)
1. Find the volume of the right circular cone with:
Formula: Volume of a cone = (1/3)πr²h
(i) radius 6 cm, height 7 cm
Volume = (1/3) × (22/7) × 6² × 7 = (1/3) × 22 × 36 = 264 cm³.
(ii) radius 3.5 cm, height 12 cm
Volume = (1/3) × (22/7) × (3.5)² × 12 = 4 × (22/7) × 12.25 = 4 × 22 × 1.75 = 154 cm³.
2. Find the capacity in litres of a conical vessel with:
(i) radius 7 cm, slant height 25 cm
First, find height h = √(l² - r²) = √(25² - 7²) = √(625 - 49) = √576 = 24 cm.
Volume = (1/3) × (22/7) × 7² × 24 = 1232 cm³.
Capacity in litres = 1232 / 1000 = 1.232 litres.
(ii) height 12 cm, slant height 13 cm
First, find radius r = √(l² - h²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
Volume = (1/3) × (22/7) × 5² × 12 = 2200/7 cm³.
Capacity in litres = (2200/7) / 1000 = 22/70 = 11/35 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)
Given: h = 15 cm, Volume = 1570 cm³.
Volume = (1/3)πr²h => 1570 = (1/3) × 3.14 × r² × 15.
1570 = 5 × 3.14 × r² => 1570 = 15.7 × r².
r² = 1570 / 15.7 = 100.
r = √100 = 10 cm.
Answer: The radius of the base is 10 cm.
4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Given: h = 9 cm, Volume = 48π cm³.
Volume = (1/3)πr²h => 48π = (1/3) × π × r² × 9.
48 = 3r² => r² = 16 => r = 4 cm.
Diameter = 2 × r = 2 × 4 = 8 cm.
Answer: The diameter of the base is 8 cm.
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Given: Diameter = 3.5 m, Depth (h) = 12 m.
Radius (r) = 3.5 / 2 = 1.75 m.
Volume = (1/3)πr²h = (1/3) × (22/7) × (1.75)² × 12
= 4 × (22/7) × 3.0625 = 38.5 m³.
Since 1 m³ = 1 kilolitre (kl), the capacity is 38.5 kl.
Answer: The capacity of the pit is 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.
Given: Volume = 9856 cm³, Diameter = 28 cm.
Radius (r) = 28 / 2 = 14 cm.
(i) height of the cone (h)
Volume = (1/3)πr²h => 9856 = (1/3) × (22/7) × 14² × h
9856 = (1/3) × 616 × h => h = (9856 × 3) / 616 = 48 cm.
(ii) slant height of the cone (l)
l = √(r² + h²) = √(14² + 48²) = √(196 + 2304) = √2500 = 50 cm.
(iii) curved surface area of the cone (CSA)
CSA = 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 the triangle is revolved about the side of 12 cm, a cone is formed.
The height of the cone (h) will be 12 cm.
The radius of the cone’s base (r) will be 5 cm.
The slant height (l) will be 13 cm.
Volume = (1/3)πr²h = (1/3) × π × 5² × 12 = 100π cm³.
Answer: The volume of the solid is 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 the side of 5 cm, a new cone is formed.
The height of this cone (h) will be 5 cm.
The radius of the cone’s base (r) will be 12 cm.
The slant height (l) will be 13 cm.
Volume (V₂) = (1/3)πr²h = (1/3) × π × 12² × 5 = 240π cm³.
Volume from Q7 (V₁) = 100π cm³.
Ratio = V₁ / V₂ = (100π) / (240π) = 10/24 = 5/12.
Answer: The volume is 240π cm³ and the ratio of volumes is 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.
Given: Diameter = 10.5 m, Height (h) = 3 m.
Radius (r) = 10.5 / 2 = 5.25 m.
Step 1: Find the Volume.
Volume = (1/3)πr²h = (1/3) × (22/7) × (5.25)² × 3
= (22/7) × 27.5625 = 86.625 m³.
Step 2: Find the slant height (l) for the canvas area.
l = √(r² + h²) = √((5.25)² + 3²) = √(27.5625 + 9) = √36.5625 ≈ 6.05 m.
Step 3: Find the area of canvas required (CSA).
CSA = πrl = (22/7) × 5.25 × 6.05 ≈ 99.825 m².
Answer: Volume is 86.625 m³ and the area of canvas required is approx. 99.825 m².
