NCERT Solutions Class 9 Chapter 11 Surface Areas And Volumes Exercise 11.3 (original) (raw)
Last Updated : 26 Apr, 2024
**Question 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**
**Solution:
**Volume of cone (V) = (1/3) × πr 2 h
****(i)** Given values,
Radius of cone (r) = 6 cm
Height of cone (h) = 7 cm
V = (1/3) × (22/7) × 6 × 6 × 7 (using π=22/7)
V = 22 × 6 × 2
**V = 264 cm 3
****(ii)** Given values,
Radius of cone (r) = 3.5 cm
Height of cone (h) = 12 cm
V = (1/3) × (22/7) × 3.5 × 3.5 × 12 (using π=22/7)
**V = 154 cm 3
**Question 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**
**Solution:
**Volume of cone (V) = (1/3) × πr 2 h
****(i)** Given values,
Radius of cone (r) = 7 cm
Slant height of cone (l) = 25 cm
**h = √(l 2 - r 2 )
h = √(252 - 72)
h = √576
**h = 24 cm
V = (1/3) × (22/7) × 7 × 7 × 24 (using π=22/7)
V = 22 × 7 × 8
**V = 1232 cm 3
****(ii)** Given values,
Height of cone (h) = 12 cm
Slant height of cone (l) = 13 cm
**r = √(l 2 - h 2 )
r = √(132 - 122)
r = √25
**r = 5 cm
V = (1/3) × (22/7) × 5 × 5 × 12 (using π=22/7)
V = 2200/7 cm3
**V = 314.28 cm 3
**Question 3. The height of a cone is 15 cm. If its volume is 1570 cm 3 , find the radius of the base. (Use π = 3.14)
**Solution:
Given values,
Height of cone (h) = 15 cm
Volume of cone (V) = 1570 cm3
**V = (1/3) × πr 2 h
1570 = (1/3) × 3.14 × r2 × 15 (using π=3.14)
r2 = 1570 × 3 / (3.14 × 15)
r2 = 100
r = √100
**r = 10 cm
**Question 4. If the volume of a right circular cone of height 9 cm is 48 π **cm 3 , find the diameter of its base.
**Solution:
Given values,
Height of cone (h) = 9 cm
Volume of cone (V) = 48π cm3
V = (1/3) × πr2h
48 × π = (1/3) × π × r2 × 9
r2 = 48 × 3 / 9 (canceling π from both sides)
r2 = 16
r = √16
**r = 4 cm
**Diameter = 2 times radius = 2 × r
= 2 × 4
**= 8 cm
**Question 5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters?
**Solution:
Given values,
Radius of cone (r) = 3.5/2 m
Height of cone (h) = 12 m
**Volume of cone = (1/3) × πr 2 h
= (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 (taking π=22/7)
= (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2)
= 38.5 m3
**Capacity of conical pit in kilo liters:
**1000 m 3 = 1 liter
38.5 m3 = 1000 × 38.5 liters
**= 38,500 liters
**= 38.5 kilo liters
**Question 6. The volume of a right circular cone is 9856 cm 3 . 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**
**Solution:
Given values,
Radius of cone (r) = 28/2 = 14 cm
Volume of cone (V) = 9856 cm3
****(i)** **Volume of cone = (1/3) × πr 2 h
9856 = (1/3) × (22/7) × 14 × 14 × h (taking π=22/7)
h = (9856 × 3 × 7) / (22 × 14 × 14)
**h = 48 cm
****(ii)** Let slant height = l
**l 2 **= h 2 + r 2
l = √(h2 + r2)
l = √(482 + 142)
l = √(2304 + 196)
l = √2500
**l = 50 cm
****(iii) curved surface area of the cone = πrl**
= π × 14 × 50 cm2
= 22/7 × 700 (taking π=22/7)
**= 2,200 cm 2
**Question 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.
**Solution:
Here, after revolving the triangle about the side 12 cm, we get
Radius of cone (r) =5 cm
Height of cone (h) = 12 cm
**Volume of cone = (1/3) × πr 2 h
= (1/3) × π × 5 × 5 × 12
= (12 × π × 5 × 5) / 3
**V = 100π cm 3
**Question 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.
**Solution:
Here, after revolving the triangle about the side 5 cm, we get
Radius of cone (r) =12 cm
Height of cone (h) = 5 cm
Volume of cone = (1/3) × πr2h
= (1/3) × π × 12 × 12 × 5
= (12 × π × 12 × 5) / 3
**V = 240π cm 3
Volume in Question 7 = 100π cm3
Ratio = (Volume in Question 8) / (Volume in Question 7)
= 240π/100π
= 12/5
**Hence, the ratio obtained = 12 : 5
**Question 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.
**Solution:
Given values,
Radius of cone (r) = 10.5/2 = 105/20 m
Height of cone (h) = 3 m
Volume of cone = (1/3) × πr2h
= (1/3) × (22/7) × (105/2) × (105/2) × 3 (taking π=22/7)
= (22 × 105 × 105 × 3) / (3 × 20× 20× 7)
**V = 86.625 m 3
**Area of the canvas = surface area of cone = πrl
**Slant height (l) = **√(h 2 **+ r 2 )
**l = **√(3 2 **+ (10.5/2) 2)
l = √(9+ (110.25/4))
l = √(146.25/4)
l = √36.56
**l = 6.05 m (approx.)
**Surface area of cone = π × (105/20) × 6.05 m2
= (22/7) × (635.25/20) (taking π=22/7)
**= 99.82 m 2
