EXERCISE 13.1

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1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box. 

(ii) The cost of sheet for it, if a sheet measuring 1m2 costs Rs 20.

Solution:

L = 1.5m,      B = 1.25m,     H = 65cm  Þ 0.65m

Surface area of cuboid =  2(l + b) × h + lb

⇒ 2(1.5 × 1.25) × 0.65 + 1.5 × 1.25

⇒ 2 × 2.75 × 0.65 × 1.875

⇒ 3.575 × 1.875

⇒ 5.45 m²

Cost of 1m² sheet = 5.45 × 20

⇒ 109.00

⇒ Rs.109 

2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of 7.50 per m².

Solution:

L = 5m,       B = 4m,      h = 3m

Surface area of cuboid = 2(l + b) × h + lb

⇒ 2(5 + 4) × 3 + 5 ×4

⇒ 2 × 9 × 3 + 20

⇒ 54 + 20

⇒ 74 m²

Cost of painting = 74 × 7.50

⇒ Rs. 555.70

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of 10 per m² is 15000, find the height of the hall.

[Hint : Area of the four walls = Lateral surface area.]

Solution:

Perimeter = 250m

10m² painted = Rs. 15000

Area of four walls = 15000/250 = 1500m²

Area of four walls = lateral surface area

1500cm² = 2(l + b) × h

1500 = 250m × h

h = 1500/250

h = 6m

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Solution:

L = 22.5,       B = 10cm,        H = 7.5cm

Surface area of cuboid = 2(lb + bh + hl)

⇒  2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)

⇒  2(225 + 75 + 168.75)

⇒  2(468.75)

⇒  937.50cm²

No. of bricks =

⇒ 100 bricks

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

(i) Which box has the greater lateral surface area and by how much?

(ii) Which box has the smaller total surface area and by how much?

Solution:

Cube = a = 10cm,         cuboid = l = 12.5cm,   b = 10cm,   h = 8cm

Lateral surface area of cube = 4a²

⇒ 4 × 10²

⇒ 400cm³

Lateral surface area of cuboid = 2(l + b) × h

⇒ 2(12.5 + 10) × 8

⇒ 2(22.5) × 8

⇒ 45 × 8

⇒ 360cm²

Area of cube = 400 – 360

⇒ 40 cm²

Cube has greater lateral surface area by 40cm²

Total surface area of cube = 6a²

⇒ 6 × 10²

⇒ 600cm²

Total surface area of cuboid = 2(lb + bh + hl)

⇒ 2(12.5 × 10 +10 × 8 + 8 × 12.5)

⇒ 2(125 + 80 + 100)

⇒ 2(305)

⇒ 610

Area of cube = area of cuboid

600 = 610

⇒ 610 – 600

⇒ 10 cm²

Cuboid has greater total surface area by 10m²

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

Solution:

(i) area of glass = total surface area of cuboid

2(lb + bh + hl)

⇒ 2(30 × 25 + 25 × 25 + 25 × 30)

⇒ 2(750 + 625 + 750)

⇒ 2(2125)

⇒ 42250cm²

(ii) tape needed for all 12 edges

4(l + b + h)

⇒ 4(30 + 25 + 25)

⇒ 4 × 80

⇒ 320cm²

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

L = 25cm,     B = 20cm,      H = 5cm

Total surface area of cuboid = 2(lb + bh + hl)

⇒ 2(25 × 20 + 20 × 5 + 5 × 25)

⇒ 2(500 + 100 + 125)

⇒ 2(725)

⇒ 1450cm²

L = 15cm,      B = 12cm,     H = 5cm

Total surface area of cuboid = 2(lb + bh + hl)

⇒ 2(15 × 12 + 12 × 5 + 5 ×15)

⇒ 2(180 + 60 + 75)

⇒ 2(315)

⇒ 630cm²

Total cardboard for both boxes

⇒ 1450 + 630

⇒ 2080cm²

Cardboard for 250 boxes = 250 × 2080

⇒ 520000 cm²

Laps 5% = 5% of 520000

⇒  × 520000

⇒ 26000cm²

Cardboard for purchasing = 520000 +26000

⇒ 546000cm²

Cost of cardboard = 546000 ×

⇒ 546 × 4

⇒ Rs. 2184

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

Solution:

L = 4m,      B = 3m,      H = 2.5m

Tarpaulin required for shelter = lb + 2bh + 2hl

⇒ 4 × 3 + 2 × 3 ×  + 2 ×  × 4

⇒ 12 + 15 + 20

⇒ 47m²

 

EXERCISE 13.2

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Assume that π = ​  , unless stated otherwise.

Q1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Solution: 

H = 14cm,    curved surface area of cylinder = 88cm²

Curved surface area of cylinder = 88cm²

⇒ 2 ×  × r × h = 88cm²

⇒ 2 ×  × r × 14 = 88cm²

⇒ 88r = 88

⇒ r = 88/88

⇒ r = 1cm

D = 2r   ⇒  2 ×1 = 2cm

Q2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?   

Solution:

D = 140cm  ⇒  r = 70cm = 0.7m,        H = 1m

Total surface area of cylinder = 2πr(r + h)

⇒ 2 × × 0.7 (0.7 + 1)

⇒ 2 × 22 × 0.1 × 1.7

⇒ 7.48m²

Hence, 7.48m² metal sheet required to make closed cylindrical tank.

Q3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its

(i) Inner curved surface area,                                   

(ii) Outer curved surface area,                        

(iii) Total surface area.

Solution:

H = 77cm,       H = 77cm

D = 4cm,         D = 4.4cm

R = 2cm,         R = 2.2cm

(i) interior curved surface area = 2πrh

⇒ 2 ×   × 2 ×77

⇒ 88 × 11

⇒ 968 cm²

(ii) exterior curved surface area = 2πrh

⇒ 5.28 cm²

Total surface area = interior surface area + exterior surface area + ring surface area

⇒ 1064.8 + 9.68 + 5.28 cm²

⇒ 2038.08 cm²

Q4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Solution:

D = 84cm   ⇒  r = 42cm,        H = 120cm

Curved surface area of cylinder = 2πrh

⇒ 2 ×  × 42 × 120

⇒ 44 × 6 × 120

⇒ 31680 cm²

Total area of playground

Q5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m².

Solution:

D = 50 cm , ⇒   R = 0.25m,     H = 3.5m

Curved surface area of cylinder = 2πrh

Q6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Solution:

Curved surface area of cylinder = 4.4 m² ,    radius = 0.7 m

Let the height of the cylinder be = h

2πrh = 4.4

Q7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find

(i) its inner curved surface area,

(ii) the cost of plastering this curved surface at the rate of 40 per m².

Solution:

Inner diameter of circular well = 3.5 m ⇒  r = 1.75 m

Depth of the well = 10 m

(i) inner surface area of the well = 2πrh

Q8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

The length of the pipe = 28 m

Q9. Find

(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.

(ii) how much steel was actually used, if of the steel actually used was wasted in making the tank.

Solution:

(i) diameter of the cylindrical petrol tank = 4.2 m

Radius of the tank = 2.1m ,    height = 4.5 m

Curved surface area of cylinder = 2πrh

Q10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:

Height of the folding = 2.5 cm

Height of the frame = 30 cm

Diameter = 20 cm  ⇒  radius = 10 cm

Now cloth required for covering for lampshade

= C.S.A of top + C.S.A of middle + C.S.A of bottom

Q11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Solution:

Radius of pen holder = 3 cm

Height of pen holder = 10.5 cm

Cardboard required for pen holder = CSA of pen holder + area of circular base

⇒  2πrh + πr²

⇒  πr(2h + r)

EXERCISE 13.3

Assume that π = , unless stated otherwise.

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Solution:

Diameter = 10.5 cm  Þ radius = 5.25,   height = 10 cm

Curved surface area of cone = πrl

Þ    × 5.25 × 10

Þ    × 52.5

Þ 165 cm²

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

Slant height = 21 m ,   diameter = 24 m   Þ radius = 12 m

Total surface area of cone = πr(l + r)

3. 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.

Solution:

CSA of cone = 308 cm²,    height = 14 cm

(i) CSA of cone = 308 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 70.

Solution:

Height = 10 cm,  radius = 24 m

l² = h² + r²

l² = 10² + 24²

l =

l = 26 m

(ii)  curved surface area of cone = πrl

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? 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:

Wide = 3 m , height = 8 , radius = 6 m

l² = r² + h²

l² = 6² + 8²

l =

l = 10 m

Curved surface area of cone = πrl

Þ 3.14 × 6 × 10

Þ 31.4 × 6

Þ 188.4 m²

6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of 210 per 100 m².

Solution:

Length = 25 m, diameter = 14 m, radius = 7 m

Curved surface area of cone = πrl

7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Radius = 7 cm, height = 24 cm

l² = r² + h²

l² = 7² + 24²

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 height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 )

Solution:

Diameter = 40 cm,  radius = 20 cm,  height = 1 m

l² = r² + h²

l² = 0.2² + 1²

 

Exercise 13.4

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Assume π = 22/7 , unless stated otherwise.

Q1. Find the surface area of a sphere of radius:

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

Solution:

Q2. Find the surface area of a sphere of diameter:

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

Solution:

(i) Surface area of sphere = 4πr²

Q3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Solution:

Total surface area of hemisphere = 3πr²

⇒ 3 × 3.14 × 10 × 10

⇒ 3 × 314

⇒ 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.

Solution:

Surface area of sphere = 4πr²

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 16 per 100 cm².

Solution:

Curved surface area of hemisphere = 2πr²  

​

Q6. Find the radius of a sphere whose surface area is 154 cm².

Solution:

Area = 154 cm²

Surface area of sphere = 4πr²

Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution:

Let the diameter of earth = x

Q8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Solution:

Inner radius = 5 cm,    width = 0.25 cm,   radius = 5.25 cm

Curved surface area of hemisphere = 2πr²

⇒ 44 × 0.75 × 5.25

⇒ 173.25

Q9. 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:

Radius of sphere = r,   radius of cylinder = r + r = 2r

(i) surface area of sphere = 4πr²

(ii) curved surface area of cylinder = 2πrh

⇒ 2πr(2r)

⇒ 4πr²

EXERCISE 13.5

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Q1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?

Solution:

Length = 4 cm,  breadth = 2.5 cm,  height = 1.5 cm

Volume of cuboid = l × b × h 

⇒ 4 × 2.5 × 1.5 cm

⇒ 10 × 1.5

⇒ 15 cm × 3

⇒ 180 cm³

Q2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 l)

Solution:

Length = 6 m,  width = 5 m,  depth = 4.5 m

Volume of cuboid = l × b × h

⇒ 6 × 5 × 4.5 m

⇒ 30 × 4.5

⇒ 135 m³

Water can be hold = 135 × 1000

⇒ 135000 l

Q3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Solution:

Length = 10 m,  width = 8 m,  volume = 380 cm²

Volume of cuboid = l × b × h

380 cm² = 10 × 8 × h

380 cm² = 80h

Q4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of 30 per m³.

Solution:

Length = 8m,  breadth = 6m,  height = 3m

Volume of cuboid = l × b × h

⇒ 8 × 6 × 3 m

⇒ 48 × 3 m

⇒ 144 m³

Cost of digging = 144 × 30

⇒ Rs. 4320

Q5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Solution:

1m³ = 1000 l

50000 l = 50 m³

Volume of cuboid = l × b × h

50m³ = 2.5 × 10 × h

50m³ = 25h

Q6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?

Solution:

Population = 400

Per head per day = 150 l

Total water require per day = 400 × 150

⇒ 600000 Litre

Volume of tank = 20 × 15 × 6 m

⇒ 1800 m³  =   1800000 l

Q7. A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.

Solution :

Volume of godown = l × b × h

⇒ 40 × 25 × 15 m

⇒ 15000 m³

Volume of crates = l × b × h

⇒ 1.5 × 1.25 × 0.5 m

⇒ 0.9375 m³

Q8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Solution:

Volume of cube = a³

⇒ 12 × 12 × 12

⇒ 1728 cm³

Q9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Solution:

Deep = 3m,  width = 40 m,  speed = 2 kmph

Volume of river = 2000 × 40 × 3

⇒ 24000 m³

EXERCISE 13.6

Assume π = 22/7, unless stated otherwise.

Q1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1 l)

Solution:

Circumference of cylinder = 132 cm

Height = 25 cm

2πr = 132 cm

Q2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.

Solution:

Inner radius = 24 cm

Outer radius = 28 cm

(i) volume of inner cylinder = πr²h

⇒ 110 × 196

⇒ 21560 cm³

Total volume = v₁ + v₂

⇒ 15840 + 21560 cm³

⇒ 37400 cm³

Mass = 37400 × 0.6

⇒ 22440

Q3. A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:

Length = 5cm, breadth = 4cm, height = 15cm

Diameter = 7cm, height = 10cm, r = 3.5cm

Volume of cuboid = l × b × h

⇒ 5 × 4 × 15 cm

⇒ 20 × 15 cm

⇒ 300 cm³

Volume of cylinder = πr²h

⇒  × 3.5 × 3.5 × 10

⇒ 22 × 5 × 3.5

⇒ 110 × 3.5

⇒ 385 cm³

Capacity of both

300 cm³  =  385 cm³

⇒ 385 – 300

⇒ 85 cm³

Q4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find

(i) radius of its base       (ii) its volume. (Use  = 3.14)

Solution:

Height = 5 cm

Lateral surface area of the cylinder = 2πrh

2 × 3.14 × r × 5 = 94.2 cm²

Q5. It costs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of 20 per m², find

(i) inner curved surface area of the vessel,

(ii) radius of the base,

(iii) capacity of the vessel.

Solution:

Cost = Rs. 2200

Height = 10 m

Q6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Solution:

Height = 1 m

Capacity = 15.4 l

Q7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:

Q8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Solution:

 

EXERCISE 13.7

Assume π = 22/7, unless stated otherwise.

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:

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:

(i) h² = 25² - 7²

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

Solution:

Height = 15 cm,  volume = 1570 cm³

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

Solution:

Height = 9 cm,  volume = 48π cm³

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

Solution:

Diameter = 3.5 cm ,  height = 12 cm ,  radius = 1.75

6. 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

Solution:

Volume = 9856 cm³ ,  diameter = 28 cm ,  radius = 14cm

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:

Height = 12 cm  ,  radius = 5 cm

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:

Height = 5 m  ,  radius = 12 m

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:

Diameter = 10.5 m,  height = 3 m ,  radius = 5.25

 

EXERCISE 13.8

Assume π =  , unless stated otherwise.

1. Find the volume of a sphere whose radius is

(i) 7 cm                            (ii) 0.63 m

Solution:

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

(i) 28 cm                                     (ii) 0.21 m

Solution:

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³ ?

Solution:

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 diameter of earth = x

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Solution:

Diameter = 10.5 ,  radius = 5.25

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Solution:

Radius = 1.01 m ,  radius = 1 m

7. Find the volume of a sphere whose surface area is 154 cm².

Solution:

Surface area = 154 cm²

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of 498.96. If the cost of white-washing is 2.00 per square metre, find the

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

Solution:

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S². Find the

(i) radius r² of the new sphere,                  (ii) ratio of S and S².

Solution:

R = r

10. 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?

Solution :

Diameter = 3.5 mm ,  r = 1.75 mm

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