CBSE Class 9- Maths - Surface Areas and Volumes -NCERT EXERCISE 13.7 (#cbsenotes)

Surface Areas and Volumes 
NCERT EXERCISE 13.7

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

Answer:
(i) Given, r = 6 cm, h = 7 cm

 Volume of the cone = ⅓πr²h
= (1/3) × (22/7) × 6² × 7
= 264 cm³ (Answer)

(ii) Given, r = 3.5 cm, h = 12cm
 Volume of the cone = ⅓πr²h
  = (1/3) × (22/7) × 3.5² × 12
  = 154 cm³ (Answer)





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

Answer:
(i) Given r = 7 cm, l = 25cm
∴ h = √(l² - r²) = √(25² - 7²) = √(625 - 49) = √576
= 24 cm
 Volume of the cone = ⅓πr²h
  = (1/3) × (22/7) × 7² × 24
  = 1232 cm³
  = 1232 / 1000 = 1.232 litres (Answer)

 (ii) Given, h = 12 cm, l = 13 cm
∴ r = √(l² - h²) = √(13² - 12²) = √(169 - 144) = √25
= 5 cm
 Volume of the cone = ⅓πr²h
  = (1/3) × (22/7) × 5² × 12 ×(1/1000)
  = 11/35 litres (Answer)


Q3: The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Answer: Given h = 15 cm, volume = 1570 cm³

Volume of the cone = ⅓πr²h

⇒ 1570 = (1/3) × 3.14 × r² × 15
⇒ r² = (1570 × 3) / (3.14 × 15)
= 100
⇒ r = √(100) = 10 cm
Thus, radius of the base = 10 cm

Q4:  If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.

Answer: Given, h = 9 cm, volume = 48π cm³

Volume of the cone =  ⅓πr²h

⇒ 48π = ⅓π × r²× 9

⇒ r² = (48 π × 3) / (π × 9) = 16

⇒ r = 4

Thus, base diameter of the cone = 2 × 4 cm = 8 cm


Q5: A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Answer: Given, r = 3.5/2 = 1.75 m, h = 12 m

Capacity of the conical pit = ⅓πr²h

= (1/3) × (22/7) × 1.75² × 12

= 38.5 m³ = 38.5 kl (Answer)

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.

Answer: Given r = 28/2 = 14cm, Volume = 9856 cm³
(i) Volume of the cone =  ⅓πr²h
⇒ 9856 = (1/3) × (22/7) × 14² × h
⇒ h = (9856 × 3 × 7) / (22 × 14 × 14)
⇒ h = 48 cm (Answer)

(ii) Slant height l = √(h² + r²)
⇒ h =  √(48² + 14²)
⇒ h = √(2304 + 196) = √(2500)
⇒ h = 50 cm (Answer)

(iii) Curved surface area of the cone = πrl
= (22/7) × 14 × 50  = 2200 cm² (Answer)

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.

Answer: The solid formed is a cone of height (h) = 12 cm

base radius r = 5 cm

∴ Volume of the cone =  ⅓πr²h

= (1/3) × π × 5 × 5 × 12  = 100 π cm³ (Answer)

Q8: 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.

Answer: Here radius r of the cone = 12 cm and height h of the cone = 5 cm.

∴ Volume of the cone =  ⅓πr²h
=  (1/3) × π × 12² × 5 × 12
= 240 π cm³ (Answer)

required ratio = 100π / 240π = 5/12
= 5:12 (Answer)

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 required.

Answer: Given r = 10.5/2 = 5.25 m, h = 3m

Volume of the heap = ⅓πr²h
= (1/3) × (22/7) × 5.25² × 3
= 86.625 m³ (Answer)

Slant height l =  √(h² + r²)
= √(3² + 5.25²)
= √(9 + 27.5625) = √(36.5625)
= 6.05 m (approx)

Curved surface area of the cone = πrl
= (22/7) × 5.25 × 6.05
= 99.825 m² (Answer)