NCERT Solutions for Class 9 Maths - Surface Areas and Volumes (Exercise 11.3)

Q1. Find the volume of the right circular cone with
(i) radius 6cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm (Assume π = 22 / 7)
Ans: Volume of cone = (1 / 3) πr²h cube units
Where r be radius and h be the height of the cone
(i) Radius of cone, r = 6 cm
Height of cone, h = 7cm
Say, V be the volume of the cone, we have
V = (1 / 3) × (22 / 7) × 36 × 7
= (12 × 22)
= 264
The volume of the cone is 264 cm³.
(ii) Radius of cone, r = 3.5cm
Height of cone, h = 12cm
Volume of cone = (1 / 3) × (22 / 7) × 3.52 × 7 = 154
Hence, The volume of the cone is 154 cm³.

Q2. Find the capacity in litres of a conical vessel with
(i) radius 7cm, slant height 25 cm
(ii) height 12 cm, slant height 12 cm (Assume π = 22 / 7)
Ans:  (i) Radius of cone, r = 7 cm
Slant height of cone, l = 25 cm

or h = 24
Height of the cone is 24 cm
Now,
Volume of cone, V = (1/3) πr²h (formula)
V = (1 / 3) × (22 / 7) × 72 × 24
= (154 × 8)
= 1232
So, the volume of the vessel is 1232 cm3
Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm³)
= 1.232 Liters.
(ii) Height of cone, h = 12 cm
Slant height of cone, l = 13 cm

r = 5
Hence, the radius of cone is 5 cm.
Now, Volume of cone,
V = (1 / 3)πr²h V = (1 / 3) × (22 / 7) × 52 × 12 cm³
= 2200 / 7
Volume of cone is 2200 / 7 cm³
Now, Capacity of the conical vessel= 2200 / 7000 litres (1L = 1000 cm³)
= 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 of the cone, h = 15 cm
Volume of cone = 1570 cm3
Let r be the radius of the cone
As we know: Volume of cone, V = (1 / 3) πr²h
So, (1 / 3) πr²h = 1570
(1 / 3) × 3.14 × r² × 15 = 1570
r² = 100
r = 10
Radius of the base of cone 10 cm.

Q4. If the volume of a right circular cone of height 9cm is 48πcm3, find the diameter of its base.
Ans: Height of cone, h = 9cm
Volume of cone = 48π cm³
Let r be the radius of the cone.
As we know: Volume of cone, V = (1 / 3) πr²h
So, 1/3 π r²(9) = 48 π
r² = 16
r = 4
Radius of cone is 4 cm.
So, diameter = 2 × Radius = 8
Thus, diameter of base is 8cm.

Q5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters? (Assume π = 22 / 7)
Ans: Diameter of conical pit = 3.5 m
Radius of conical pit, r = diameter / 2 = (3.5 / 2)m = 1.75m
Height of pit, h = Depth of pit = 12m
Volume of cone, V = (1 / 3) πr2h
V = (1 / 3) × (22 / 7) × (1.75)² × 12 = 38.5
Volume of cone is 38.5 m3
Hence, capacity of the pit = (38.5 × 1) kiloliters = 38.5 kiloliters.

Q6. The volume of a right circular cone is 9856cm³. If the diameter of the base is 28cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone (Assume π = 22 / 7)
Ans:  Volume of a right circular cone = 9856 cm³
Diameter of the base = 28 cm
(i) Radius of cone, r = (28  /2) cm = 14 cm
Let the height of the cone be h
Volume of cone, V = (1 / 3) πr²h
(1 / 3) πr²h = 9856
(1 / 3) × (22 / 7) × 14 × 14 × h = 9856
h = 48
The height of the cone is 48 cm.

Slant height of the cone is 50 cm.
(iii) curved surface area of cone = πrl
= (22 / 7) × 14 × 50 = 2200
curved surface area of the cone is 2200 cm².

Q7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Ans: Height (h)= 12 cm
Radius (r) = 5 cm, and
Slant height (l) = 13 cm
Volume of cone, V = (1 / 3) πr²h
V = (1 / 3) × π × 5² × 12 = 100π
Volume of the cone so formed 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
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.