**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:**

Diameter of the base of the cone is 10.5 cm. To find the radius, we need to divide the diameter by 2.

Radius of the base of the cone, r = diameter / 2 = 10.5 / 2 = 5.25 cm

The slant height of the cone is given as 10 cm. Let's denote it as l.

Slant height of the cone, l = 10 cm

Now, we can find the curved surface area (CSA) of the cone using the formula:

CSA = π r l

where π (pi) is approximately equal to 22 / 7.

CSA = (22 / 7) × 5.25 × 10

CSA = 165 cm²

Hence, the curved surface area of the 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)Solution:**

The radius of the cone, r = 24 / 2 m = 12m

Slant height, l = 21 m

To find the total surface area of a cone, we use the formula: Total Surface area of the cone = πr(l + r)

Total Surface area of the cone = (22 / 7) × 12 × (21 + 12) m^{2}

= (22 / 7) × 12 × 33 m^{2}

= 1244.57 m^{2}

Thus, the total surface area of the cone is 1244.57 m^{2}.

**Q3. Curved surface area of a cone is 308 cm ^{2} 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:**Slant height of cone, l = 14 cm

Let the radius of the cone be r.

**(i)**We know, CSA of cone = πrl

**Given**: Curved surface area of a cone is 308 cm

^{2}

(308) = (22 / 7) × r × 14

308 = 44r

r = 308 / 44 = 7cm

The radius of the cone base is 7 cm.

**(ii)** To find the total surface area of the cone, we need to add the curved surface area (CSA) and the area of the base (πr^{2}).

Total surface area of cone = CSA of cone + Area of base (πr^{2})

Total surface area of cone = 308 + (22 / 7) × 7^{2} = 308 + 154

Therefore, the total surface area of the cone is 462 cm^{2}.

**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 m ^{2} canvas is Rs 70. (Assume π = 22 / 7)Ans:** Let ABC be a conical tent with vertex A, base center O, and base circumference point B.

Height of conical tent (AO), h = 10 m

Radius of conical tent (BO), r = 24m

Let the slant height of the tent (AB) be l.

**(i)**In right triangle ABO, we have

AB

^{2}= AO

^{2}+ BO

^{2}(using Pythagoras theorem)

l

^{2}= h

^{2}+ r

^{2}

= (10)

^{2}+ (24)

^{2}

= 100 + 576

= 676

l = √676 = 26m

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

**(ii)**The curved surface area (CSA) of the conical tent can be calculated using the formula:

CSA = πrl

= (22/7) × 24 × 26 m

^{2}

=13728/7 m

^{2}

Now, let's calculate the cost of the canvas required to make the tent.

Cost of 1 m

^{2}canvas = Rs 70

Cost of (13728 / 7) m

^{2}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.

**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:**Height of conical tent, h = 8m

Radius of base of tent, r = 6m

Slant height of tent, l^{2} = (r^{2} + h^{2})

l^{2} = (62 + 82) = (36 + 64) = (100)

or l = 10

Again, CSA of conical tent = πrl

= (3.14 × 6 × 10) m^{2}

= 188.4m^{2}

Let the length of tarpaulin sheet required be L

As 20 cm will be wasted, therefore,

Effective length will be (L - 0.2m).

Breadth of tarpaulin = 3m (given)

Area of sheet = CSA of tent

[(L – 0.2) × 3] = 188.4

L - 0.2 = 62.8

L = 63

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

**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 m ^{2}. (Assume π = 22 / 7).**

**Ans:**Slant height of conical tomb, l = 25m

Base radius, r = diameter / 2 = 14 / 2 m = 7mCSA of conical tomb = πrl= (22 / 7) × 7 × 25 = 550

CSA of conical tomb = 550m

^{2}

Cost of white-washing 550 m

^{2}area, which is Rs (210 × 550) / 100

= Rs.1155

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

**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 = 24cmSlant height, l

^{2}= (r

^{2}+ h

^{2})

= (7

^{2}+ 24

^{2})

= (49 + 576)

= (625)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22 / 7) × 7 × 25

= 550

CSA of 10 caps = (10 × 550) cm

^{2}= 5500 cm

^{2}

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

^{2}.

**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 m ^{2}, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) = 1.02).Ans:**

**Given**:

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

Height of cone, h = 1m

Slant height of cone is l, and l

^{2}= (r

^{2}+ h

^{2})

Using given values, l

^{2}= (0.22 + 12)

= (1.04)

Or l = 1.02

Slant height of the cone is 1.02 m

Now,

CSA of each cone = πrl

= (3.14 × 0.2 × 1.02)

= 0.64056m

^{2}

CSA of 50 such cones = (50 × 0.64056)

CSA of 50 such cones = 32.028 m

^{2}

Again,

Cost of painting 1 m

^{2}area = Rs 12 (given)

Cost of painting 32.028 m

^{2}area

= Rs (32.028 × 12)

= Rs.384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all these cones is Rs. 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: Formula:**Surface area of sphere (SA) = 4πr

^{2}

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

SA = 4 × (22 / 7) × 10.52 = 1386

Surface area of sphere is 1386 cm

^{2}

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

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

Surface area of sphere is 394.24 cm

^{2}

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

SA = 4πr

^{2}

= 4 × (22 / 7) × (14)

^{2}

= 2464

Surface area of sphere is 2464 cm

^{2}.

**Q2. Find the surface area of a sphere of diameter:(i) 14cm(ii) 21cm(iii) 3.5cm(Assume π = 22 / 7)**

**Ans:**

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

Formula for Surface area of sphere = 4πr2

= 4 × (22 / 7) × 7

^{2}= 616

Surface area of a sphere is 616 cm

^{2}

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

Surface area of sphere = 4πr

^{2}

= 4 × (22 / 7) × 10.52 = 1386

Surface area of a sphere is 1386 cm

^{2}

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

^{2}

**(iii)**Radius(r) of sphere = 3.5 / 2 = 1.75 cm Surface area of sphere = 4πr2

= 4 × (22 / 7) × 1.752 = 38.5

Surface area of a sphere is 38.5 cm

^{2}.

**Q3. Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]Ans:** Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr

^{2}

= 3 × 3.14 × 10

^{2}= 942

The total surface area of given hemisphere is 942 cm

^{2}.

**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

_{1}and r

_{2}be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So

r

_{1}= 7cm

r

_{2}= 14 cm

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

= 4r

_{1}

^{2}/ 4r

_{2}

^{2}

= (r

_{1}/ r

_{2})

^{2}

= (7 / 14)

^{2}= (1 / 2)

^{2}= 1 / 4

Therefore, the ratio between the surface areas is 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 cm ^{2}. (Assume π = 22 / 7)Ans:** Inner radius of hemispherical bowl, say r = diameter / 2 = (10.5) / 2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr

^{2}

= 2 × (22 / 7) × (5.25)

^{2}= 173.25

Surface area of hemispherical bowl is 173.25 cm

^{2}

Cost of tin-plating 100 cm

^{2}area = Rs 16

Cost of tin-plating 1 cm

^{2}area = Rs 16 /100

Cost of tin-plating 173.25 cm

^{2}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 cm

^{2}is Rs 27.72.

**Q6. Find the radius of a sphere whose surface area is 154 cm ^{2}. (Assume π = 22/7)Ans:** Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr

^{2}= 154

r

^{2}= (154 × 7) / (4 × 22) = (49 / 4)

r = (7 / 2) = 3.5

The radius of the sphere is 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:** If diameter of earth is said d, then the diameter of moon will be d / 4 (as per given statement)

Radius of earth = d / 2

Radius of moon = 1 / 2 × d / 4 = d / 8

Surface area of moon = 4π(d / 8)

^{2}

Surface area of earth = 4π(d / 2)

^{2}

Ratio of their Surace areas = 4π(d / 8)

^{2}/ 4π (d / 2)

^{2}= 4 / 64 = 1 / 16

The ratio between 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:**

**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 = 2πr

^{2}, where r is radius of hemisphere

= 2 × (22 / 7) × (5.25)

^{2}= 173.25

Therefore, the outer curved surface area of the bowl is 173.25 cm

^{2}.

**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).**

**Ans:**

**(i)**Surface area of sphere = 4πr

^{2}, 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)

= 4πr

^{2}

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

= 4r

^{2}/ 4r

^{2}

= 1 / 1

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

**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) πr

^{2}h 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 cm

^{3}.

**(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 cm

^{3}.

**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) πr

^{2}h (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 cm

^{3})

= 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)πr

^{2}h V = (1 / 3) × (22 / 7) × 52 × 12 cm

^{3}

= 2200 / 7

Volume of cone is 2200 / 7 cm

^{3}

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

^{3})

= 11 / 35 litres.

**Q3. The height of a cone is 15cm. If its volume is 1570cm ^{3}, 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) πr

^{2}h

So, (1 / 3) πr

^{2}h = 1570

(1 / 3) × 3.14 × r

^{2}× 15 = 1570

r

^{2}= 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 cm

^{3}

Let r be the radius of the cone.

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

^{2}h

So, 1/3 π r

^{2}(9) = 48 π

r

^{2}= 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 9856cm ^{3}. 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 cm

^{3}

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) πr

^{2}h

(1 / 3) πr

^{2}h = 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 cm

^{2}.

**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) πr

^{2}h

V = (1 / 3) × π × 5

^{2}× 12 = 100π

Volume of the cone so formed is 100π cm

^{3}.

**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) πr

^{2}h; 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 cm

^{3}.

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)πr

^{2}h

= (1 / 3) × (22 / 7) × 5.25 × 5.25 × 3

= 86.625

The volume of the heap of wheat is 86.625 m

^{3}. 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 m

^{2}.

**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) πr

^{3}

= (4 / 3) × (22 / 7) × 7

^{3}= 4312 / 3

Hence, volume of the sphere is 4312 / 3 cm

^{3}

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

Using, volume of sphere = (4 / 3) πr

^{3}

= (4 / 3) × (22 / 7) × 0.63

^{3}= 1.0478

Hence, volume of the sphere is 1.05 m

^{3}(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)**Diameter = 28 cm

Radius, r = 28 / 2 cm = 14cm

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

^{3}

Volume of the ball = (4 / 3) × (22 / 7) × 14

^{3}= 34496 / 3

Hence, volume of the ball is 34496 / 3 cm

^{3}

**(ii)**Diameter = 0.21 m

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

Volume of the ball = (4 / 3)πr

^{3}

Volume of the ball = (4 / 3) × (22 / 7) × 0.105

^{3}

Hence, volume of the ball = 0.004851 m

^{3}

**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 cm ^{3}? (Assume π = 22 / 7)Ans: **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 πr

^{3}

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

^{3}

Volume of the metallic ball = 38.808 cm

^{3}

Now, using 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).

**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) πr

^{3}= (4 / 3) π (d / 8)

^{3}= 4 / 3π(d

^{3}/ 512)

Find the volume of the earth:

Volume of the earth = (4 / 3) πr

^{3}= (4 / 3) π (d / 2)

^{3}= 4 / 3π(d

^{3}/ 8)

Fraction of the volume of the earth is the volume of the moon

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) πr

^{3}

Volume of the hemispherical bowl = (2 / 3) × (22 / 7) × 5.25

^{3}= 303.1875

Volume of the hemispherical bowl is 303.1875 cm

^{3}

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 (R

^{3}– r

^{3})

Put values,

Volume of the iron used in the hemispherical tank = (2 / 3) × (22 / 7) × (1.01

^{3}– 1

^{3}) = 0.06348

So, volume of the iron used in the hemispherical tank is 0.06348 m

^{3}.

**Q7. Find the volume of a sphere whose surface area is 154 cm ^{2}. (Assume π = 22 / 7)Ans:** Let r be the radius of a sphere.

Surface area of sphere = 4πr

^{2}

4πr

^{2}= 154 cm

^{2}(given)

r

^{2}= (154 × 7) / (4 × 22) r = 7 / 2

Radius is 7 / 2 cm

Now, Volume of the sphere = (4 / 3) πr

^{3}

Volume of the sphere = (4 / 3) x (22 / 7) x (7 / 2)

^{3}= 179 x 2 / 3

Volume of the sphere is 179 x 2 / 3 cm

^{3}

**Q8. A dome of a building is in the form of a hemisphere. 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)Ans: (i)** Cost of white-washing the dome from inside = Rs 4989.60

Cost of white-washing 1m

^{2}area = Rs 20

CSA of the inner side of dome = 498.96 / 2 m

^{2}= 249.48 m

^{2}

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

CSA of inner side of dome = 249.48 m

^{2}(from (i))

Formula to find CSA of a hemi sphere = 2πr

^{2}

2πr

^{2}= 249.48

2 × (22 / 7) × r

^{2}= 249.48

r

^{2}= (249.48 × 7) / (2 × 22)

r

^{2}= 39.69

r = 6.3

So, radius is 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

Using formula, volume of the hemisphere

= 2 / 3 πr

^{3}= (2 / 3) × (22 / 7) × 6.3 × 6.3 × 6.3

= 523.908

= 523.9(approx.)

Volume of air inside the dome is 523.9 m

^{3}.

**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)πr

^{3}

Volume of twenty seven solid sphere = 27 × (4 / 3)πr

^{3}= 36 πr

^{3}

**(i)**New solid iron sphere radius = r’

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

^{3}

(4 / 3)π(r’)

^{3}= 36 πr

^{3}

(r’)

^{3}= 27r

^{3}

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 = 4πr

^{2}

Surface area of iron sphere of radius r’= 4π (r’)

^{2}

Now

S / S’ = (4πr

^{2}) / ( 4π (r’)

^{2})

S / S’ = r

^{2}/ (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 3.5mm. How much medicine (in mm ^{3}) is needed to fill this capsule? (Assume π = 22 / 7)Ans:** 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 πr

^{3}

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

^{3}= 22.458

The volume of the spherical capsule is 22.46 mm

^{3}.