Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  NCERT Solutions: Surface Areas & Volumes

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

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 / = 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:
 

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and VolumesRadius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m
= 22/7 × 12 × (21 + 12) m
= (22/7 × 12 × 33) m
= 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.
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes(ii) Total surface area of cone = πr(r + l)
= 22/7 × 7 × (7 + 14)
= 22 × 21 = 462 cm
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 =24+102 = 576 + 100 = 676
l = √676 = 26m
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes(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 + h
⇒ l2 = 62 + 82 = 36 + 64 = 100 ⇒ l = √100 = 10 m
Area of tarpaulin to make the tent = πrl
= 3.14 × 6 × 10 = 188.40m
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 m
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 NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and VolumesWe 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
 NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and VolumesWe know that l2 = r2 + h
⇒ 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 = 6.4056m
Curved surface area of 50 cones
= 50 × 6.4056
= 32.028 m
Cost of painting at the rate of ₹12 per m
= ₹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 = 4πr
= 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 = 4πr
= 4 × 22/7 × 5.6 × 5.6 = 4 × 22 × 0.8 × 5.6 = 394.24 cm
Hence, the surface area of sphere is 394.24 cm2.
(iii) Radius of sphere r = 14 cm
Surface area of sphere = 4πr
= 4 × 22/7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm
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 = 4πr
= 4 × 22/7 × 7 × 7 = 4 × 22 × 7 = 616 cm
Hence, the surface area of sphere is 616 cm2.
(ii) Radius of sphere r = 21/2 = 10.5 cm
Surface area of sphere = 4πr
= 4 × 22/7 × 10.5 × 10.5 = 4 × 22 × 4.5 × 10.5 = 1386 cm
Hence, the surface area of sphere is 1386 cm2.
(iii) Radius of sphere r = 3.5/2 = 1.75 cm
Surface area of sphere = 4π
= 4 × 22/7 × 1.75 × 1.75 = 4 × 22 × 0.25 × 1.75
= 38.50 cm
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 cm= ₹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
⇒ 4πr= 154
⇒ 4 × 22/7 × r= 154
⇒ r= 154/(4 × 22/7)
⇒ r= 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: 

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and VolumesInternal 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 = 2πR
= 2 × 22/7 × 5.25 × 5.25
= 2 × 22 × 0.75 × 5.25
= 173.25 cm
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).

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and VolumesAns: (i) Radius of sphere = radius of cylinder = r
Hence, the surface area of sphere = 4πr
(ii) Radius of cylinder = r and height h = diameter of sphere = 2r
Hence, the curved surface area of cylinder = 2πrh = 2πr(2r) = 4πr
(iii) Now, Surface area of sphere/Curved surface area of cylinder = 4πr2/4πr= 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) cm
= 264 cm
(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 - 7
⇒ 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 - h
⇒ r = √132 - 122
⇒ r = √25 
⇒ r = 5 cm
Volume of the cone = 1/3 πr2h
= (1/3 × 22/7 × 5 × 5 × 12) cm
= (2200/7) cm
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 cm
⇒ 1/3 πr2h = 1570
⇒ 13 × 3.14 × r2 × 15 = 1570
⇒ r2 = 1570/(3.14×5) = 100
⇒ r = 10

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)r× 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 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 = h+ r2
⇒ l2 = 48+ 142
⇒ l2 = 2304+196
⇒ l= 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.
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

Therefore, the radius of cone r = 5 cm, height h = 12 cm and slant height l =13 cm.
Volume of solid (cone) = 1/3 πr2
= 1/3 × π × 5 × 5 × 12 = 100 π cm
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: 

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and VolumesA 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 = NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

After substituting the values, we have

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

= (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 
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes(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 = NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes
(ii) Radius of spherical ball r = 0.21/2 = 0.105 m
Volume of water displaced by spherical ball = 4/3 πr
= 4/3 × 22/7 × 0.105 × 0.105 = 4 × 22 × 0.005 × 0.63 × 0.63 = 0.004861 m
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 πr
= 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
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes
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 = 4πr
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

Volume of surface 4/3 πr3

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

Hence, the volume of sphere is NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes.


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 = 2πr
Cost of white washing at the rate of ₹2 = 2πr2 × ₹2 = ₹4πr2
⇒ ₹4πr2 = ₹498.96
⇒ 4 × 22/7 × r2 = 498.96
NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes
Therefore, the internal surface of dome = 2πr
= 2 × 22/7 × (6.3)2
= 2 × 22/7 × 6.3 × 6.3 
= 2 × 22 × 0.9 × 6.3
= 249.48 m
Hence, the inside surface area of the dome is 249.48 m2.
(ii) Volume of the air inside the dome = 2/3 πr
= 2/3 × 22/7 × (6.3)
= 2/3 × 22/7 × 6.3 × 6.3 × 6.3
= 2 × 22 × 0.3 × 6.3 × 6.3
= 523.9 cm
= 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 mmmedicine is required to fill this capsule.

The document NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
44 videos|412 docs|55 tests

Top Courses for Class 9

FAQs on NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

1. What is the formula for calculating the surface area of a cylinder?
Ans.The surface area of a cylinder can be calculated using the formula: Surface Area = 2πr(h + r), where r is the radius of the base and h is the height of the cylinder.
2. How do you find the volume of a cone?
Ans.The volume of a cone can be found using the formula: Volume = (1/3)πr²h, where r is the radius of the base and h is the height of the cone.
3. What are the steps to calculate the surface area of a sphere?
Ans.To calculate the surface area of a sphere, use the formula: Surface Area = 4πr², where r is the radius of the sphere. First, measure the radius, square it, multiply by 4, and then multiply by π (approximately 3.14).
4. Can you explain how to find the volume of a hemisphere?
Ans.The volume of a hemisphere is calculated using the formula: Volume = (2/3)πr³, where r is the radius of the hemisphere. Measure the radius, cube it, multiply by 2/3, and then by π.
5. What is the importance of understanding surface areas and volumes in real life?
Ans.Understanding surface areas and volumes is important in real life for various applications such as construction, manufacturing, and packaging. It helps in determining the amount of materials needed, optimizing space usage, and ensuring proper fitting of objects.
44 videos|412 docs|55 tests
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

past year papers

,

Sample Paper

,

MCQs

,

mock tests for examination

,

Viva Questions

,

Objective type Questions

,

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

,

study material

,

Semester Notes

,

Summary

,

ppt

,

Previous Year Questions with Solutions

,

Extra Questions

,

Free

,

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

,

video lectures

,

NCERT Solutions for Class 9 Maths Chapter 11 - Surface Areas and Volumes

,

pdf

,

Exam

,

practice quizzes

,

Important questions

,

shortcuts and tricks

;