Surface Areas and Volumes – Chapter 11 NCERT Solutions for Class 9

NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes is an extremely important chapter from exam point of view for the students. They need to get ready for battle with Surface Areas and Volumes to ensure maximum marks in their scorecard. Students struggling with chapter can get access to the answers below. These solutions are created by keeping in mind the exact need which will help them to get a better understanding of the chapter.

Overview of the Exercises of NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes

• Exercise 11.1: The first exercise of  Surface Areas and Volumes focuses on introducing a three-dimensional image of the right circular cone.
• Exercise 11.2: The second exercise of the chapter is built on the concept of a sphere’s surface area.
• Exercise 11.3: The third exercise of Surface Areas and Volumes focuses on how the volume of a right circular cone is equal to one-third the volume of the cylinder.
• Exercise 11.4: The final exercise of the chapter deals with the volume of a sphere.

Surface Areas and Volumes  Class 9 : NCERT Solutions for Chapter 11 Exercise 11.1

Surface Areas and Volumes

1. 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)

Solution:

Radius of the base of cone = diameter/2 = (10.5/2)cm = 5.25cm

The slant height of the cone, say l = 10 cm

CSA of the cone is = πrl

= (22/7)×5.25×10 = 165 cm²

Therefore, the curved surface area of the cone is 165 cm².

2. Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m. (Assume π = 22/7)

Solution:

Radius of cone, r = 24/2 m = 12m

Slant height, l = 21 m

Formula: Total Surface area of the cone = πr(l+r)

Total Surface area of the cone = (22/7)×12×(21+12) m²

= 1244.57m²

3. Curved surface area of a cone is 308 cm^2, and its slant height is 14 cm. Find

(i) the radius of the base and (ii) the total surface area of the cone.

(Assume π = 22/7)

Solution:

The slant height of the cone, l = 14 cm

Let the radius of the cone be r.

(i) We know that the CSA of cone = πrl

Given: Curved surface area of a cone is 308 cm²

(308 ) = (22/7)×r×14

308 = 44 r

r = 308/44 = 7 cm

Therefore, the radius of the cone base is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base (πr²)

Total surface area of cone = 308+(22/7)×7² = 308+154 = 462 cm²

Therefore, the total surface area of the cone is 462 cm².

4. 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)

Solution:

Let ABC be a conical tent

Height of conical tent, h = 10 m

The radius of the conical tent, r = 24m

Let the slant height of the tent be l.

(i) In the right triangle ABO, we have

AB² = AO²+BO²(using Pythagoras theorem)

l² = h²+r²

= (10)²+(24)²

= 676

l = 26 m

Therefore, the slant height of the tent is 26 m.

(ii) CSA of tent = πrl

= (22/7)×24×26 m²

Cost of 1 m² canvas = Rs 70

Cost of (13728/7)m² canvas is equal to Rs (13728/7)×70 = Rs 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

5. What length of tarpaulin 3 m wide will be required to make a 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]

Solution:

Height of conical tent, h = 8m

The radius of the base of the tent, r = 6m

The slant height of the tent, l² = (r²+h²)

l² = (6²+8²) = (36+64) = (100)

or l = 10 m

Again, CSA of conical tent = πrl

= (3.14×6×10) m²

= 188.4m²

Let the length of the tarpaulin sheet required be L

As 20 cm will be wasted,

The effective length will be (L-0.2m).

The breadth of tarpaulin = 3m (given)

Area of sheet = CSA of the tent

[(L–0.2)×3] = 188.4

 

L-0.2 = 62.8

L = 63 m

Therefore, the length of the required tarpaulin sheet will be 63 m.

6. The slant height and base diameter of a 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)

Solution:

The slant height of the conical tomb, l = 25m

Base radius, r = diameter/2 = 14/2 m = 7m

CSA of conical tomb = πrl

= (22/7)×7×25 = 550

CSA of conical tomb= 550m²

Cost of white-washing 550 m² area, which is Rs (210×550)/100

= Rs. 1155

Therefore, the cost will be Rs. 1155 while white-washing the tomb.

7. A joker’s cap is in the form of a right circular cone with a base radius of 7 cm and a height of 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)

Solution:

The radius of the conical cap, r = 7 cm

Height of conical cap, h = 24cm

Slant height, l² = (r²+h²)

= (7²+24²)

= (49+576)

= (625)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22/7)×7×25

= 550 cm²

CSA of 10 caps = (10×550) cm² = 5500 cm²

Therefore, the area of the sheet required to make 10 such caps is 5500 cm².

8. 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 a height of 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)

Solution:

Given:

Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m

Height of cone, h = 1m

The slant height of the cone is l, and l² = (r²+h²)

Using given values, l² = (0.2²+1²)

= (1.04)

Or l = 1.02 m

The slant height of the cone is 1.02 m

Now,

CSA of each cone = πrl

= (3.14×0.2×1.02)

= 0.64056 m

CSA of 50 such cones = (50×0.64056) = 32.028

CSA of 50 such cones = 32.028 m²

Again,

Cost of painting 1 m² area = Rs 12 (given)

Cost of painting 32.028 m² area = Rs (32.028×12)

= Rs.384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all the cones is Rs. 384.34.

Surface Areas and Volumes  Class 9 : NCERT Solutions for Chapter 11 Exercise 11.2

Surface Areas and Volumes

1. Find the surface area of a sphere of radius:

(i) 10.5cm (ii) 5.6cm (iii) 14cm

(Assume π=22/7)

Solution:

Formula: Surface area of a sphere (SA) = 4πr²

(i) Radius of the sphere, r = 10.5 cm

SA = 4×(22/7)×10.5² = 1386

Surface area of the sphere is 1386 cm²

(ii) Radius of the sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.6² = 394.24

Surface area of a sphere is 394.24 cm²

(iii) Radius of the sphere, r = 14cm

SA = 4πr²

= 4×(22/7)×(14)²

= 2464

Surface area of the sphere is 2464 cm²

2. Find the surface area of a sphere of diameter:

(i) 14cm (ii) 21cm (iii) 3.5cm

(Assume π = 22/7)

Solution:

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for Surface area of sphere = 4πr²

= 4×(22/7)×7² = 616

Surface area of the sphere is 616 cm²

(ii) Radius (r) of the sphere = 21/2 = 10.5 cm

Surface area of a sphere = 4πr²

= 4×(22/7)×10.5² = 1386

Surface area of the sphere is 1386 cm²

Therefore, the surface area of a sphere having diameter 21cm is 1386 cm²

(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm

Surface area of a sphere = 4πr²

= 4×(22/7)×1.75² = 38.5

Surface area of the sphere is 38.5 cm²

3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]

Solution:

Radius of the hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr²

= 3×3.14×10² = 942

The total surface area of the given hemisphere is 942 cm².

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution:

Let r₁ and r₂ be the radii of the spherical balloon and spherical balloon when air is pumped into it, respectively. So,

r₁ = 7cm

r₂ = 14 cm

Now, required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4πr₁²/4πr₂²

= (r₁/r₂)²

= (7/14)² = (1/2)² = ¼

Therefore, the ratio between the surface areas is 1:4.

5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm². (Assume π = 22/7)

Solution:

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr²

= 2×(22/7)×(5.25)² = 173.25

Surface area of hemispherical bowl is 173.25 cm²

Cost of tin-plating 100 cm² area = Rs 16

Cost of tin-plating 1 cm² area = Rs 16 /100

Cost of tin-plating 173.25 cm² area = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm² is Rs 27.72.

6. Find the radius of a sphere whose surface area is 154 cm². (Assume π = 22/7)

Solution:

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr² = 154

r² = (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

7. The diameter of the moon is approximately one fourth of the diameter of the earth.

Find the ratio of their surface areas.

Step 1: Express the radii in terms of the Earth’s diameter

Given:

• The diameter of the Earth =
  $d$ 

  $d$

• The diameter of the Moon =
  $\frac{d}{4}$ 

  4d​

Now, the radii:

• Radius of the Earth,
  $R_{\text{Earth}} = \frac{d}{2}$ 

  REarth​=2d​

• Radius of the Moon,
  $R_{\text{Moon}} = \frac{d}{8}$ 

  RMoon​=8d​

Step 2: Calculate the surface areas

The surface area

$A$

A of a sphere is given by the formula:

 

$$A = 4\pi R^2$$

A=4πR2

Surface area of the Earth:

 

$$A_{\text{Earth}} = 4\pi \left(\frac{d}{2}\right)^2 = 4\pi \times \frac{d^2}{4} = \pi d^2$$

AEarth​=4π(2d​)2=4π×4d2​=πd2

Surface area of the Moon:

 

$$A_{\text{Moon}} = 4\pi \left(\frac{d}{8}\right)^2 = 4\pi \times \frac{d^2}{64} = \frac{\pi d^2}{16}$$

AMoon​=4π(8d​)2=4π×64d2​=16πd2​

Step 3: Find the ratio of their surface areas

The ratio of the surface area of the Moon to the surface area of the Earth is:

 

$$\text{Ratio} = \frac{A_{\text{Moon}}}{A_{\text{Earth}}} = \frac{\frac{\pi d^2}{16}}{\pi d^2} = \frac{1}{16}$$

Ratio=AEarth​AMoon​​=πd216πd2​​=161​

Therefore, the ratio of the surface areas of the Moon to the Earth is

$\frac{1}{16}$

161​ or

$1:16$

1:16.

8. 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)

Solution:

Given:

Inner radius of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of hemispherical bowl = 2πr², where r is the radius of the hemisphere

= 2×(22/7)×(5.25)² = 173.25 cm²

Therefore, the outer curved surface area of the bowl is 173.25 cm².

9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in(i) and (ii).

Solution:

(i) Surface area of sphere = 4πr², where r is the radius of sphere

(ii) Height of cylinder, h = r+r =2r

Radius of cylinder = r

CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr²

(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)

= 4πr²/4πr² = 1/1

Ratio of the areas obtained in (i) and (ii) is 1:1.

Surface Areas and Volumes  Class 9 : NCERT Solutions for Chapter 11 Exercise 11.3

Surface Areas and Volumes

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 (Assume π = 22/7)

Solution:

Volume of cone = (1/3) πr²h cube units

Where r be radius and h be the height of the cone

(i) Radius of the cone, r = 6 cm

Height of cone, h = 7cm

Let 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 the cone, r = 3.5cm

Height of the cone, h = 12cm

Volume of the cone = (1/3)×(22/7)×3.5²×7 = 154

Hence,

The volume of the cone is 154 cm³.

2. Find the capacity in litres of a conical vessel with

(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

(Assume π = 22/7)

Solution:

(i) Radius of the cone, r =7 cm

Slant height of the 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 the cone, h = 12 cm

Slant height of the cone, l = 13 cm

r = 5

Hence, the radius of the cone is 5 cm.

Now, Volume of cone, V = (1/3)πr²h

V = (1/3)×(22/7)×52×12 cm³

= 2200/7

Volume of the cone is 2200/7 cm³

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm³)

= 11/35 litres

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

Solution:

Height of the cone, h = 15 cm

Volume of the cone =1570 cm³

Let r be the radius of the cone

As we know: Volume of the 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 the cone is 10 cm.

4. If the volume of a right circular cone of height 9cm is 48πcm³, find the diameter of its base.

Solution:

Height of cone, h = 9cm

Volume of cone =48π cm³

Let r be the radius of the cone.

As we know: Volume of the cone, V = (1/3) πr²h

So, 1/3 π r²(9) = 48 π

r² = 16

r = 4

Radius of the cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of the base is 8 cm.

5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters?

(Assume π = 22/7)

Solution:

Diameter of the conical pit = 3.5 m

Radius of the conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m

Height of the pit, h = Depth of the pit = 12m

Volume of the cone, V = (1/3) πr²h

V = (1/3)×(22/7) ×(1.75)²×12 = 38.5

Volume of the cone is 38.5 m³

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

6. 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)

Solution:

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 the 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 the cone = πrl

= (22/7)×14×50

= 2200

The curved surface area of the cone is 2200 cm².

7. 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.

Solution:

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

The volume of the cone so formed is 100π cm³.

8. 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.

Solution:

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius 12 cm, height 5 cm, and slant height 13 cm.

Volume of cone = (1/3) πr²h; where r is the radius and h is the height of cone

= (1/3)×π×12×12×5

= 240 π

The volume of the cones formed is 240π cm³.

So, the required ratio = (result of question 7) / (result of question 8) = (100π)/(240π) = 5/12 = 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.

(Assume π = 22/7)

Solution:

Given:

• Radius of the heap
  $r = 5.25$ 

  $r=5.25$ m

• Height of the heap
  $h = 3$ 

  $h=3$ m

Step 1: Calculate the slant height

$l$

l of the cone

The slant height

$l$

l of a cone is calculated using the Pythagorean theorem:

 

$$l = \sqrt{r^2 + h^2}$$

l=r2+h2​

Substitute the values:

 

$$l = \sqrt{(5.25)^2 + (3)^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \approx 6.05 \text{ m}$$

l=(5.25)2+(3)2​=27.5625+9​=36.5625​≈6.05 m

Step 2: Calculate the surface area of the cone (area of the canvas)

The surface area

$A$

A of a cone (which is the area of the canvas) is given by:

 

$$A = \pi r l$$

A=πrl

Substitute the values:

 

$$A = \frac{22}{7} \times 5.25 \times 6.05 \approx \frac{22 \times 5.25 \times 6.05}{7}$$

A=722​×5.25×6.05≈722×5.25×6.05​

Let’s calculate this:

 

$$A \approx \frac{22 \times 31.7625}{7} \approx \frac{698.775}{7} \approx 99.82 \text{ m}^2$$

A≈722×31.7625​≈7698.775​≈99.82 m2

Final Answer:

The volume of the heap of wheat is

$86.625 \, \text{m}^3$

86.625m3, and the area of the canvas required to cover the heap is approximately

$99.82 \, \text{m}^2$

99.82m2.

Surface Areas and Volumes  Class 9 : NCERT Solutions for Chapter 11 Exercise 11.4

Surface Areas and Volumes

1. Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

(Assume π =22/7)

Solution:

(i) Radius of the sphere, r = 7 cm

Using, Volume of sphere = (4/3) πr³

= (4/3)×(22/7)×7³

= 4312/3

Hence, the volume of the sphere is 4312/3 cm³

(ii) Radius of the sphere, r = 0.63 m

Using, volume of sphere = (4/3) πr³

= (4/3)×(22/7)×0.63³

= 1.0478

Hence, the volume of the sphere is 1.05 m³ (approx).

2. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

(Assume π =22/7)

Solution:

(i) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr³

Volume of the ball = (4/3)×(22/7)×14³ = 34496/3

Hence, the volume of the ball is 34496/3 cm³

(ii) Diameter = 0.21 m

Radius of the ball =0.21/2 m= 0.105 m

Volume of the ball = (4/3 )πr³

Volume of the ball = (4/3)× (22/7)×0.105³ m³

Hence, the volume of the ball = 0.004851 m³

3.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)

Solution:

Given,

Diameter of a metallic ball = 4.2 cm

Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm

Volume formula = 4/3 πr³

Volume of the metallic ball = (4/3)×(22/7)×2.1 cm³

Volume of the metallic ball = 38.808 cm³

Now, using the relationship between density, mass and volume,

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Mass of the ball is 345.39 g (approx).

4. 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?

Solution:

Let the diameter of the earth be “d”. Therefore, the radius of the earth will be d/2

The diameter of the moon will be d/4, and the radius of the 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

Answer: The volume of the moon is 1/64 volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)

Solution:

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, the hemispherical bowl can hold 0.303 litres of milk.

6. 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)

Solution:

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, the volume of the iron used in the hemispherical tank is 0.06348 m³.

7. Find the volume of a sphere whose surface area is 154 cm². (Assume π = 22/7)

Solution:

Let r be the radius of a sphere.

The surface area of the sphere = 4πr²

4πr² = 154 cm² (given)

r² = (154×7)/(4 ×22)

r = 7/2

Radius is 7/2 cm

Now,

Volume of the sphere = (4/3) πr³

8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the

(i) inside surface area of the dome (ii) volume of the air inside the dome

(Assume π = 22/7)

Solution:

(i) Cost of white-washing the dome from inside = Rs 4989.60

Cost of white-washing 1m² area = Rs 20

CSA of the inner side of dome = 498.96/2 m²  = 249.48 m²

(ii) Let the inner radius of the hemispherical dome be r.

CSA of inner side of dome = 249.48 m² (from (i))

Formula to find CSA of a hemisphere = 2πr²

2πr² = 249.48

2×(22/7)×r² = 249.48

r² = (249.48×7)/(2×22)

r² = 39.69

r = 6.3

So, the radius is 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

Using the formula, the volume of the hemisphere = 2/3 πr³

= (2/3)×(22/7)×6.3×6.3×6.3

= 523.908

= 523.9(approx.)

Answer: The volume of air inside the dome is 523.9 m³.

9. 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’.

Solution:

Volume of the solid sphere = (4/3)πr³

Volume of twenty seven solid sphere = 27×(4/3)πr³ = 36 π r³

(i) New solid iron sphere radius = r’

Volume of this new sphere = (4/3)π(r’)³

(4/3)π(r’)³ = 36 π r³

(r’)³ = 27r³

r’= 3r

Radius of the new sphere will be 3r (thrice the radius of the original sphere)

(ii) Surface area of the iron sphere of radius r, S =4πr²

Surface area of the 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.

10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm³) is needed to fill this capsule? (Assume π = 22/7)

Solution:

Diameter of capsule = 3.5 mm

Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm

Volume of spherical capsule = 4/3 πr³

Volume of spherical capsule = (4/3)×(22/7)×(1.75)³ = 22.458

Answer: The volume of the spherical capsule is 22.46 mm³.

 

From this exercise of  Surface Areas and Volumes of NCERT Solutions for Class 9 Maths, students will learn how to find the surface area and volume of various geometrical objects in a simplified way.