Class 9 - Maths - Surface Areas Volumes

**Exercise 13.1**

**Question 1:**

A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open 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 1 m^{2} costs Rs 20.

Answer:

Here, l = 1.5 m, b = 1.25 m, h = 65 cm = 0.65 m.

Since the box is open at the top, it has only five faces.

(i) So, surface area of the box = lb + 2(bh + hl)

= 1.5 * 1.25 + 2 (1.25 * 0.65 + 0.65 * 1.5)

= 1.875 + 2 (1.7875)

= (1.875 + 3.575)

= 5.45 m^{2}

Hence, 5.45 m^{2} of sheet is required.

(ii) Cost of 1 m^{2} of the sheet = Rs 20

So, cost of 5.45 m^{2} of the sheet = Rs 20 * 5.45 = Rs 109

**Question 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 Rs 7.50 per m^{2}.

Answer:

Here, l = 5 m, b = 4 m, h = 3 m

Surface area of the walls of the room and the ceiling = 2h (l + b) + lb

= [2 * 3 (5 + 4) + 5 * 4]

= (6 * 9 + 20)

= 74 m^{2}

Cost of white washing = Rs 7.50 per m^{2}

So, total cost of white washing the walls and the ceiling of the room = Rs 74 * 7.50 = Rs 555

**Question 3:**

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m^{2} is Rs 15000, find the height of the hall.

Answer:

Let length, breadth and height of the hall be l, b and h respectively.

Perimeter of the floor of the hall = 2(l + b) = 250 m

Area of the four walls of the hall = 2h(l + b) .............1

Also, area of the four walls of the hall = 15000/10 = 1500 m^{2} .............2

From equation 1 and 2, we have,

2h (l + b) = 1500

=> h * 250 = 1500 [Since 2(l + b) = 250]

=> h = 1500/250

=> h = 6

Hence, height of the hall is 6 m.

**Question 4:**

The paint in a certain container is sufficient to paint an area equal to 9.375 m^{2}. How many bricks of dimensions 22.5 cm * 10 cm * 7.5 cm can be painted out of this container?

Answer:

Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm.

Total surface area of 1 brick = 2(lb + bh + hl)

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

= 2(225 + 75 + 168.75) cm^{2}

= 937.5 cm^{2}

= 9375/(100 * 100) m^{2}

= 9375/10000

= 0.09375 m^{2}

So, required number of bricks = 9.375/0. 09375 = 100

**Question 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?

Answer:

Here, a = 10 cm, l = 12.5 cm, b = 10 cm, h = 8 cm

(i) Lateral surface area of the cubical box = 4a^{2}

= 4 * (10)^{2}

= 4 * 100

= 400 cm^{2}

Lateral surface area of the cuboidal box = 2h(l + b)

= 2 * 8 (12.5 + 10)

= 16 × 22.5

= 360 cm^{2}

Difference in the lateral surface areas of the two boxes = (400 – 360) = 40 cm^{2}.

Hence, the cubical box has greater lateral surface area by 40 cm^{2}.

(ii) Total surface area of the cubical box = 6a^{2}

= 6 * (10)^{2}

= 6 * 100

= 600 cm^{2}

Total surface area of the cuboidal box = 2(lb + bh + hl)

= 2(12.5 * 10 + 10 * 8 + 8 * 12.5)

= 2(125 + 80 + 100)

= 2 × 305

= 610 cm^{2}

Difference in the total surface areas of the two boxes = (610 – 600) = 10 cm^{2}

Hence, the cubical box has smaller total surface area by 10 cm^{2}.

**Question 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?

Sol. Here, l = 30 cm, b = 25 cm, h = 25 cm.

(i) Total surface area of the herbarium = 2(lb + bh + hl)

= 2(30 × 25 + 25 × 25 + 25 × 30)

= 2(750 + 625 + 750)

= 2 × 2125

= 4250 cm^{2}

Hence, area of the glass = 4250 cm^{2}

(ii) A cuboid has 12 edges. These consist of 4 lengths, 4 breadths and 4 heights.

Hence, length of the tape required = 4l + 4b + 4h

= (4 × 30 + 4 × 25 + 4 × 25)

= (120 + 100 + 100) cm

= 320 cm

**Question 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 Rs 4 for 1000 cm^{2},

find the cost of cardboard required for supplying 250 boxes of each kind.

Answer:

**For bigger boxes:**

l = 25 cm, b = 20 cm, h = 5 cm

Total surface area of 1 bigger box = 2(lb + bh + hl)

= 2(25 * 20 + 20 * 5 + 5 * 25)

= 2 (500 + 100 + 125)

= 1450 cm^{2}

Area of cardboard required for overlaps = 5% of 1450

= (5/100) * 1450

= 7250/100

= 72.5 cm^{2}

Total area of cardboard needed for 1 bigger box = 1450 + 72.5 = 1522.5 cm^{2}

Total area of cardboard needed for 250 bigger boxes = 1522.5 * 250 = 380625 cm^{2}.

** **

**For smaller boxes:**

l = 15 cm, b = 12 cm, h = 5 cm

Total surface area of 1 smaller box = 2 (lb + bh + hl)

= 2(15 * 12 + 12 * 5 + 5 * 15)

= 2 (180 + 60 + 75)

= 630 cm^{2}

Area of cardboard required for overlaps = 5% of 630

= (5/100) * 630

= 3150/100

= 31.5 cm^{2}

Total area of cardboard needed for 1 smaller box = 630 + 31.5

= 661.5 cm^{2}

Total area of cardboard needed for 250 smaller boxes = 661.5 * 250 = 165375 cm^{2}

Now, total area of cardboard needed for 500 boxes (250 bigger and 250 smaller boxes)

= 380625 + 165375 = 546000 cm^{2}

Cost of 1000 cm^{2} of cardboard = Rs 4

So, Cost of 546000 cm^{2} of cardboard = Rs (4/1000) * 546000

= 4 * 546

= Rs 2184

**Question 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?

Answer:

Here, l = 4 m, b = 3 m, h = 2.5 m

The tarpaulin is needed to cover 5 faces only (excluding the floor) Surface area of the shelter

= lb + 2(bh + hl)

= 4 * 3 + 2(3 * 2.5 + 2.5 * 4)

= 12 + 2(7.5 + 10)

= 12 + 35

= 47 m^{2}

Hence, 47 m^{2} of tarpaulin is required to make the shelter.

**Exercise 13.2**

**Question 1:**

The curved surface area of a right circular cylinder of height 14 cm is 88 cm^{2}. Find the diameter of the base of the cylinder.

Answer:

Here, h = 14 cm, curved, surface area = 88 cm^{2}, r = ?

Curved surface area of the cylinder = 2πrh

=> 88 = 2 * (22/7) * r * 14

=> 88 = 44 × 2 × r

=> 88 = 88 * r

=> r = 88/88

=> r = 1 cm

Hence, the base diameter of cylinder = 1 * 2 = 2 cm

**Question 2:**

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?

Answer:

Here, h = 1 m, r = 140/2 cm = 70 cm = 0.7 m

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

= 2 * (22/7) * 0.7 (1 + 0.7)

= 44 * 0.1 * 1.7

= 7.48 m^{2}

Hence, 7.48 m^{2} of sheet is required.

**Question 3:**

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 figure).

Find its (i) inner curved surface area (ii) outer curved surface area (iii) total surface area.

Answer:

Here, h = 77 cm,

Outer radius (R) = 4.4/2 cm = 2.2 cm

Inner radius (r) = 4/2 cm = 2 cm

(i) Inner curved surface area of the pipe = 2πrh

= 2 * (22/7) * 2 * 77

= 2 * 22 * 22

= 968 cm^{2}

(ii) Outer curved surface area of the pipe = 2πRh

= 2 * (22/7) * 2.2 * 77

= 44 * 24.2

= 1064.80 cm^{2}

(iii) Total surface area of the pipe = inner curved surface area + outer curved surface area

+ areas of the two base rings.

= 2πrh + 2πRh + 2π (R^{2} – r^{2})

= 968 + 1064.80 + 2 * (22/7)[(2.2)^{2} – 2^{2}]

= 2032.80 + 5.28

= 2038.08 cm^{2}

**Question 4:**

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 m^{2}.

Answer:

Radius of the roller (r) = 84/2 = 42 cm

Length of the roller (h) = 120 cm

Curved surface area of the roller = 2πrh

= 2 * (22/7) * 42 * 120

= 44 * 6 * 120

= 44 * 720

= 31680 cm^{2 }

Since area covered by the roller in 1 revolution = 31680 cm^{2}

So, area covered by the roller in 500 revolutions = 31680 × 500 cm^{2}

= 15840000 cm^{2}

Hence, area of the playground = 15840000/(100 * 100)

= 15840000/10000

= 1584 cm^{2}

**Question 5:**

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 Rs 12.50 per m^{2}.

Answer:

Here, r = 50/2 = 25 cm = 0.25 m, h = 3.5 m

Curved surface area of the pillar = 2πrh

= 2 * (22/7) * 0.25 * 3.5

= 44 * 0.25 * 0.5

= 5.5 m^{2}

Cost of painting 1 m^{2} = Rs 12.50

So, total cost of painting the curved surface of the pillar = Rs 12.50 * 5.5 = Rs 68.75

**Question 6:**

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.

Answer:

Curved surface area of the cylinder = 4.4 m^{2}, r = 0.7 m, h = ?

Curved surface area of the cylinder = 2πrh

=> 4.4 = 2 * (22/7) * 0.7 * h

=> 4.4 = 44 * 0.1 * h

=> 4.4 = 4.4 * h

=> h = 4.4/4.4

=> h = 1

Hence, height of the cylinder is 1 m.

**Question 7:**

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 Rs 40 per m^{2}.

Sol. Here, r = 3.5/2 m , h = 10 m

(i) Inner curved surface area of the well = 2πrh

= 2 * (22/7) * (3.5/2) * 10

= 22 * 5 = 110 m^{2}

(ii) Cost of plastering 1 m^{2} = Rs 40

So, Cost of plastering the curved surface area of the well = Rs 110 * 40 = Rs 4400

Question 8:

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.

Answer:

Here, r = 5/2 cm = 2.5 cm = 0.025 m, h = 28 m.

Total radiating surface in the system = total surface area of the cylinder

= 2πr(h + r)

= 2 * (22/7) * 0.025 (28 + 0.025)

= (44 * 0.025 * 28.025)/7

= 4.4 m^{2} (approx)

**Question 9:**

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 1/12 of the steel actually used was wasted in making the tank.

Answer:

Here, r = 4.2/2 = 2.1 m, h = 4.5 m

(i) Curved surface area of the storage tank = 2πrh

= 2 * (22/7) * 2.1 * 4.5

= 44 * 0.3 * 4.5

= 59.4 m^{2}

(ii) Total surface area of the tank = 2πr(h + r)

= 2 * (22/7) * 2.1 4.5 + 2.1)

= 44 * 0.3 * 6.6

= 87.12 m^{2}

Let the actual area of steel used be x m^{2}.

Area of steel wasted = 1/12 of x

= x/12 m^{2}

So, area of the steel used in the tank = (x – x/12) = 11x/12 m^{2}

=> 87.12 = 11x/12

=> x = (87.12 * 12)/11

=> x = 7.92 * 12

=> x = 95.04 m^{2}

Hence, 95.04 m^{2} of steel was actually used.

**Question 10:**

In the figure, 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.

Answer:

Here, r = 20/2 cm = 10 cm, Height = 30 cm

Circumference of the base of the frame = 2πr

= 2π * 10 = 20π cm

Height of the frame = 30 cm

Height of the cloth needed for covering the frame (including the margin)

= (30 + 2.5 + 2.5) cm = 35 cm

Also, breadth of the cloth = circumference of the base of the frame.

So, area of the cloth required for covering the lampshade = length * breadth

= 35 * 20π

= 35 * 20 * (22/7)

= 5 * 20 * 22

= 2200 cm^{2}

**Question 11:**

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?

Answer:

Here, r = 3 cm, h = 10.5 cm

The penholders have only one base i.e., these are open at one end.

Total surface area of 1 penholder = 2πrh + πr^{2}

= πr(2h + r)

= (22/7) * 3 (2 * 10.5 + 3)

= (22/7) * 3 * 24

Total surface area of 35 penholders = (22/7) * 3 * 24 * 35

= 22 * 3 * 24 * 5

= 7920 cm^{2}

Hence, 7920 cm^{2} of cardboard is needed.

**Exercise 13.3**

**Question 1:**

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

Answer:

Here, r = 10.5/2 cm = 5.25 cm, l = 10 cm.

Curved surface area of the cone = πrl

= (22/7) * 5.25 * 10

= 22 * 7.5 * 10

= 165 cm^{2}

**Question 2:**

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

Answer:

Here, l = 21 m, r = 24/2 = 12 m

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

= (22/7) * 12 (21 + 12)

= (22/7) * 12 * 33

= 8712/7

= 1244.57 m^{2}

**Question 3:**

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.

Answer:

Here, l = 14 cm, curved surface area = 308 cm^{2}, r = ?

(i) Curved surface area of the cone = πrl

=> 308 = (22/7) * r * 14

=> 308 = 22 * r * 7

=> 308 = 154 * r

=> r = 308/154

=> r = 2

Hence, base radius of the cone = 7 cm.

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

= (22/7) * 7(14 + 7)

= 22 * 21

= 462 cm^{2}

**Question 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 m^{2} canvas is Rs 70.

Answer:

Here, h = 10 m, r = 24 m

(i) We have, l^{2} = h^{2} + r^{2}

= (10)^{2} + (24)^{2}

= 100 + 576

= 676

=> l = √676

=> l = 26 m

(ii) Curved surface area of the tent = πrl

= (22/7) * 24 * 26

Cost of 1 m^{2} canvas = Rs 70

So, Cost of (22/7) * 24 * 26 m^{2} of canvas = Rs 70 * (22/7) * 24 * 26

= Rs 10 * 22 * 24 * 260

= Rs 137280

**Question 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 29 cm (use π = 3.14)

Answer:

Here h = 6 m, r = 8 m

We have, l = √(r^{2} + h^{2})

= √(6^{2} + 8^{2})

= √(36 + 64)

= √100

=> l = 10 m

Curved surface area of the tent = πrl

= 3.14 * 6 * 10

So, required length of tarpaulin = (3.14 * 6 * 10)/3 m + 20 cm

= (3.14 * 2 * 10) m + 20 cm

= 62.8 + 0.2 m

= 63 m

**Question 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 Rs 210 per 100 m^{2}.

Answer:

Here, l = 25 m, r = 14/2 m = 7 m

Curved surface area of the tomb = πrl

= (22/7) * 7 * 25

= 22 * 25

= 550 m^{2}

Cost of white washing 100 m^{2} = Rs 210

So, Cost of white washing 550 m^{2} = Rs (210/100) * 550

= 21 * 55

= Rs 1155

**Question 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.

Answer:

Here, r = 7 cm, h = 24 cm

We have, l = √(h^{2} + r^{2})

= √(24^{2} + 7^{2})

= √(576 + 49)

= √625

=> l = 25 cm

Total curved surface area of 1 cap = πrl

= (22/7) * 7 * 25

= 22 * 25

= 550 cm^{2}

Area of sheet required to make 10 such caps = 10 * 550 = 5500 cm^{2}

**Question 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 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)

Answer:

Here, r = 40/2 cm = 20 cm = 0.20 m, h = 1 m

l = √(h^{2} + r^{2})

= √{1^{2} + (0.2)^{2}}

= √(1 + 0.4)

= √1.04

=> l = 1.02 m

Curved surface area of 1 cone = πrl

Curved surface area of 50 cones = 50 * 3.14 * 0.2 * 1.02

= 32.028 m^{2}

Cost of painting an area of 1 m^{2} = Rs 12

So, Cost of painting an area of 32.028 m^{2} = Rs 12 * 32.028

= Rs 384.34 (approx)

**Exercise 13.4**

**Question 1:**

Find the surface area of a sphere of radius:

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

Answer:

(i) r = 10.5 cm

Surface area of the sphere = 4πr^{2}

= 4 * (22/7) * (10.5)^{2}

= 4 * (22/7) * 10.5 * 10.5

= 4 * 22 * 10.5 * 1.5

= 1386 cm^{2}

(ii) r = 5.6 cm

Surface area of the sphere = 4πr^{2}

= 4 * (22/7) * (5.6)^{2}

= 4 * (22/7) * 5.6 * 5.6

= 4 * 22 * 5.6 * 0.8

= 394.24 cm^{2}

(iii) r = 14 cm

Surface area of the sphere = 4πr^{2}

= 4 * (22/7) * (14)^{2}

= 4 * (22/7) * 14 * 14

= 4 * 22 * 14 * 2

= 2464 cm^{2}

**Question 2:**

Find the surface area of sphere of a diameter:

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

Answer:

(i) r = 14/2 cm = 7 cm

Surface area of the sphere = 4πr^{2}

= 4 * (22/7) * 7^{2}

= 4 * (22/7) * 7 * 7

= 4 * 22 * 7

= 616 cm^{2}

(ii) r = 21/2 = 10.5 cm

Surface area of the sphere = 4πr^{2}

= 4 * (22/7) * (10.5)^{2}

= 4 * (22/7) * 10.5 * 10.5

= 4 * 22 * 10.5 * 1.5

= 1386 cm^{2}

(iii) r = 3.5/2 = 1.75 m

Surface area of the sphere = 4πr^{2}

= 4 * (22/7) * (1.75)^{2}

= 4 * (22/7) * 1.75 * 1.75

= 4 * 22 * 1.75 * 0.25

= 38.5 cm^{2}

**Question 3:**

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

Answer:

Given, radius r = 10 cm

Total surface area of the hemisphere = 3πr^{2}

= 3 * 3.14 * (10)^{2}

= 3 * 3.14 * 100

= 942 cm^{2}

**Question 4:**

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.

Answer:

When r = 7 cm:

Surface area of the balloon = 4πr^{2}

= 4 * π * 7 * 7 cm^{2}

When R = 14 cm:

Surface area of the balloon = 4πr^{2}

= 4 * π * 14 * 14 cm^{2}

Required ratio of the surface areas of the balloon = (4 * π * 7 * 7)/( 4 * π * 14 * 14)

= 1/4

= 1 : 4

**Question 5:**

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}.

Answer:

Here r = 10.5/2 = 5.25 cm

Inner surface area of the bowl = 2πr^{2}

= 2 * (22/7) * (5.25)^{2}

= 2 * (22/7) * 5.25 * 5.25

= 44 * 0.75 * 5.25

= 173.25 cm^{2}

Cost of tin plating 100 cm^{2} = Rs 16

Cost of tin plating 173.25 cm^{2} = Rs (16/100) * 173.25

= Rs 0.16 * 173.25

= Rs 27.72

**Question 6:**

Find the radius of a sphere whose surface area is 154 cm^{2}.

Answer:

Surface area of the sphere = 4πr^{2}

=> 154 = 4 * (22/7) * r^{2}

=> r^{2} = (154 * 7)/(4 * 22)

=> r^{2} = (7 * 7)/4

=> r^{2} = 49/4

=> r = √(49/4) => r = 7/2 => r = 3.5

Hence, radius of the sphere = 3.5 cm.

**Question 7:**

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

Answer:

Let diameter of the earth = 2r

Then radius of the earth = r

So, Diameter of the moon = 2r/4 = r/2

Radius of the moon = (r/2)/2 = r/4

Now, surface area of the moon = 4π(r/4)^{2}

= πr^{2}/4 ……………..1

Surface area of the earth = 4πr^{2} ……....2

So, Required ratio = (πr^{2}/4)/( 4πr^{2})

= 1/16

= 1 : 16

**Question 8:**

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.

Answer:

Inner radius of the bowl (r) = 5 cm

Thickness of the steel = 0.25 cm

So, Outer radius of the bowl (R) = 5 + 0.25 = 5.25 cm

Outer curved surface area of the bowl = 2πR^{2}

= 2 * (22/7) * (5.25)^{2}

= 2 * (22/7) * 5.25 * 5.25

= 2 * 22 * 5.25 * 0.75

= 173.25 cm^{2}

**Question 9:**

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

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

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

Answer:

Here, radius of the sphere = r

Radius of the cylinder = r

And, height of the cylinder = 2r

(i) Surface area of the sphere = 4πr^{2}

(ii) Curved surface area of the cylinder = 2πrh

= 2π * r * 2r

= 4πr^{2}

(iii) Required ratio = (4πr^{2})/( 4πr^{2}) = 1/1 = 1 : 1

**Exercise 13.5**

**Question 1:**

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

Answer:

Measures of matchbox (cuboid) is 4 cm * 2.5 cm * 1.5 cm

=> l = 4 cm, b = 2.5 cm and h = 1.5 cm

Now, Volume of matchbox = l * b * h

= 4 * 2.5 * 1.5

= 15 cm^{3}

So, volume of 12 boxes = 12 * 15 = 180 cm^{3}

**Question 2:**

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 m^{3} = 1000 l)

Answer:

Here, Length (l) = 6 m, Breadth (b) = 5 m, Depth (h) = 4.5 m

Capacity = l * b * h = 6 * 5 * 4.5 m^{3}

Since 1 m^{3} can hold 1000 l.

So, 135 m^{3} can hold 135 * 1000 l = 135000 l of water.

Hence, the required amount of water in the tank = 135000 l.

**Question 3:**

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

Answer:

Given, Length (l) = 10 m, Breadth (b) = 8 m, Volume (v) = 380 m^{3}

Let height of the cuboid be h.

Since, volume of a cuboid = l × b × h

=> 10 * 8 * h = 380

=> 80h = 380

=> h = 380/80 = 4.75

Thus, the required height of the liquid = 4.75 m

**Question 4:**

Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m^{3}.

Answer:

Given, Length (l) = 8 m, Breadth (b) = 6 m, Depth (h) = 3 m

Now, Volume of the cuboidal pit = l * b * h

= 8 * 6 * 3

= 144 m^{3}

Since, rate of digging the pit is Rs. 30 per m^{3}.

So, cost of digging = Rs. 30 * 144 = Rs. 4320

**Question 5:**

The capacity of 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 in.

Answer:

Length of the tank (l) = 2.5 m, Depth of the tank (h) = 10 m

Let breadth of the tank = b m

Now, Volume (capacity) of the tank= l * b * h = 2.5 * b * 10 = 25b m^{3}

But the capacity of the tank = 50000 l = 50000/1000 m^{3} = 50 m^{3}

So, 25b = 50

=> b = 50/25

=> b = 2

Thus, the depth of the tank = 2 m

**Question 6:**

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?

Answer:

Length of the tank (l) = 20 m, Breadth of the tank (b) = 15 m, Height of the tank (h) = 6 m

Volume of the tank = l * b * h = 20 * 15 * 6 = 1800 m^{3}

Since 1 m^{3} = 1000 l

So, Capacity of the tank = 1800 * 1000 = 1800000 l

Village population = 4000

Since, 150 l of water is required per head per day.

So, the amount of water is required per day = 150 * 4000 l.

Let the required number of days = x

=> 4000 * 150 * x = 1800000

=> x = (1800000)/(4000 * 150)

=> x = 3

Thus, the required number of days is 3

**Question 7:**

A godown measures 60 m * 25 m * 10 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.

Answer:

Volume of the godown = 60 * 25 * 10 m^{3}

Volume of a crate = 1.5 * 1.25 * 0.5 m^{3}

Let the required number of wooden create be n

So, n * 1.5 * 1.25 * 0.5 = 60 * 25 * 10

=> n = (60 * 25 * 10)/( 1.5 * 1.25 * 0.5)

=> n = (60 * 25 * 10 * 10 * 100 * 10)/(15 * 125 * 5)

=> n = (4 * 10 * 2 * 100 * 10)/5

=> n = 4 * 10 * 2 * 100 * 2

=> n = 16000

Hence, the maximum number of wooden crates is 16000.

**Question 8:**

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.

Answer:

Side of the given cube = 12 cm

So, volume of the given cube = (side)^{3} = (12)^{3} cm^{3}

Side of the smaller cube:

Let the side of the new (smaller) cube = n

=> Volume of smaller cube = n^{3}

=> Volume of 8 smaller cubes = 8n^{3}

So, 8n^{3} = (12)^{3}

=> n^{3} = (12)^{3} /8

=> n^{3} = (12/2)^{3}

=> n = 12/2

=> n = 6

Thus, the required side of the new (smaller) cube is 6 cm.

Ratio between surface areas:

Surface area of the given cube = 6 * (side)^{2}

= 6 * 12^{2}

= 6 * 12 * 12 cm^{2}

Surface area of one smaller cube = 6 * (side)^{2}

= 6 * 6^{2}

= 6 * 6 * 6 cm^{2}

So, surface area of 8 smaller cubes = 8 * 6 * 6 * 6 cm^{2}

Now, ratio = Area of the given cube/area of 8 new cubes

= (6 * 12 * 12)/(8 * 6 * 6 * 6)

= 1/4

Thus, the required ratio = 1 : 4

**Question 9:**

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

Answer:

The water flowing in a river can be considered in the form of a cuboid.

Such that Length (l) = 2 km = 2000 m, Breadth (b) = 40 m, Depth (h) = 3 m

So, Water volume (volume of the cuboid so formed) = l * b * h

= 2000 * 40 * 3 m^{3}

Now, volume of water fallowing in 1 hr (= 60 minutes) = 2000 * 40 * 3 m^{3}

So, volume of water that will fall in 1 minute = (2000 * 40 * 3)/60

= (6000 * 40)/60

= 1000 * 40

= 4000 m^{3}

**Exercise 13.6**

**Question 1:**

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 cm^{3} = 1l)

Answer:

Here, h = 25 cm, 2πr = 132 cm.

Given, the circumference of the base of a cylindrical vessel is 132 cm

=> 2πr = 132

=> 2 * (22/7) * r = 132

=> r = (132 * 7)/(2 * 22)

=> r = (132 * 7)/44

=> r = 3 * 7

=> r = 21 cm

Now, volume of the cylinder = πr^{2} h

= (22/7) * (21)^{2} * 25

= (22/7) * 21 * 21 * 25

= 22/ * 3 * 21 * 25

= 34650 cm^{3}

= 34650/1000 litres [Since 1000 cm^{3} = 1l]

= 34.65 litres

**Question 2:**

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 cm^{3} of wood has a mass of 0.6 g.

Answer:

Here, inner radius (r) = 24/2 = 12 cm

Outer radius (R) = 28/2 = 14 cm, h = 35 cm

Volume of the wood used in the pipe = π(R^{2} – r^{2}) h

= (22/7) * (14^{2} – 12^{2}) * 35

= 22 * (196 – 144) * 5

= 22 * 52 * 5

= 5720 cm^{3}

Mass of 1 cm^{3} of wood = 0.6 g

So, Mass of 5720 cm^{3} of wood = 0.6 * 5720 g

= 3432 g

= 3432/1000 kg [Since 1000 g = 1 kg]

= 3.432 kg

**Question 3:**

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?

Answer:

For tin can with rectangular 6 base.

l = 5 cm, b = 4 cm, h = 15 cm

Volume of the tin can = lbh = 5 * 4 * 15 = 300 cm^{3}

For plastic cylinder with circular base.

r = 7/2 cm = 3.5 cm, h = 10 cm

Volume of the plastic cylinder = πr^{2} h

= (22/7) * 3.5 * 3.5 × 10

= 22 * 3.5 * 0.5 * 10

= 385 cm^{3}

Difference in the capacities of the two containers = 385 – 300 = 85 cm^{3}

Hence, the plastic cylinder with circular base has greater capacity by 85 cm^{3}

**Question 4:**

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 π = 3.14)

Answer:

Here, h = 5 cm, 2πrh = 94.2 cm^{2}

(i) 2πrh = 94.2

=> 2 * 3.14 * r * 5 = 94.2

=> r = 94.2/(2 * 3.14 * 5)

=> r = 94.2/(10 * 3.14)

=> r = 94.2/31.4

=> r = 3

Hence, base radius of the cylinder = 3 cm

(ii) Volume of the cylinder = πr^{2} h

= 3.14 * 3 * 3 * 5

= 141.3 cm^{3}

**Question 5:**

It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per m^{2}, find

(i) Inner curved surface area of the vessel (ii) radius of the base (iii) capacity of the vessel.

Answer:

Here, h = 10 m

(i) Inner curved surface area = Total cost of painting per m^{2}

= 2200/20

= 110 m^{2}

(ii) We have, 2πrh = 110

=> 2 * (22/7) * r * 10 = 110

=> r = (110 * 7)/(2 * 22 * 10)

=> r = 770/440

=> r = 1.75 m

(iii) Capacity of the vessel = πr^{2} h

= (22/7) * 1.75 * 1.75 * 10

= 22 * 1.75 * 0.25 * 10

= 96.25 m^{3}

= 96.25 kl [Since 1 m^{3} = 1 kl]

**Question 6:**

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?

Answer:

Here, h = 1 m,

Given, volume = 15.4 litres

= 15.4/1000 m^{3}

= 0.0154 m^{3}

Also, volume of the cylindrical vessel = πr^{2} h

=> 0.0154 = (22/7) * r^{2} * 1

=> r^{2} = (0.0154 * 7)/22

=> r^{2} = 0.0049

=> r = √(0.0049)

⇒ r = 0.07 m

So, total surface area of the cylinder = 2πr (h + r)

= 2 * (22/7) * 0.07 (1 + 0.07)

= 44 * 0.01 * 1.07

= 0.4708 m^{2}

Hence, 0.4708 m^{2} of metal sheet would be needed.

**Question 7:**

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.

Answer:

Here, h = 14 cm.

Radius of the pencil (R) = 7/2 mm = 0.35 mm.

Radius of the graphite (r) = 1/2 mm = 0.05 mm.

Volume of the graphite = πr^{2} h

= (22/7) * 0.05 * 0.05 * 14

= 22 * 0.05 * 0.05 * 2

= 0.11 mm^{3}

Volume of the wood = π(R^{2} – r^{2})h

= (22/7) * [(0.35)^{2} – (0.05)^{2}] * 14

= 22 * [0.1225 – 0.0025] * 2

= 22 * 0.12 * 2

= 5.28 cm^{3}

Hence, volume of the wood = 5.28 cm^{3} and volume of the graphite = 0.11 cm^{3}

**Question 8:**

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?

Answer:

Here, r = 7/2 cm = 3.5 cm, h = 4 cm

Capacity of 1 cylindrical bowl = πr^{2} h

= (22/7) * 3.5 * 3.5 * 4

= 22 * 3.5 * 0.5 * 4

= 154 cm^{3}

Hence, soup consumed by 250 patients per day = 250 * 154 = 38500 cm^{3}

**Exercise 13.7**

**Question 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.

Answer:

(i) Here, radius of the cone (r) = 6 cm, Height (h) = 7 cm

So, volume = πr^{2} h/3

= (1/3) * (22/7) * 6 * 6 * 7

= (1/3) * 22 * 6 * 6

= 22 * 6 * 2

= 264 cm^{3}

(ii) Here, radius of cone (r) = 3.5 cm, Height (h) = 12 cm

So, volume = πr^{2} h/3

= (1/3) * (22/7) * 3.5 * 3.5 * 12

= (1/3) * 22 * 3.5 * 0.5 * 12

= 22 * 3.5 * 0.5 * 4

= 154 cm^{3}

**Question 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

Answer:

(i) Here, r = 7 and I = 25 cm

Now, radius (r) = √(I^{2} – h^{2})

= √(13^{2} – 12^{2})

= √(169 – 144)

= √25

=> r = 5 cm

Now, volume of conical vessel = πr^{2} h/3

= (1/3) * (22/7) * 5 * 5 * 12

= (22/7) * 5 * 5 * 4

= 2200/7 cm^{3}

Now, capacity of the conical vessel = (2200/7) * (1/1000) l

= 22/70 l

= 11/35 l

Thus, the required capacity of the conical vessel is 11/35 l.

**Question 3:**

The height of a cone is 15 cm. If its volume is 1570 cm^{3}, find the radius of the base.

Answer:

Here, height of the cone (h) = 15 cm

Volume of the cone (v) = 1570 cm^{3}

Let the radius of the base be r cm.

=> πr^{2} h/3 = 1570

=> (3.14 * r^{2} * 15)/3 = 1570

=> 3.14 * r^{2} * 5 = 1570

=> 15.70 * r^{2} = 1570

=> r^{2} = 1570/15.70

=> r^{2} = (1570 * 100)/1570

=> r^{2} = 100

=> r = √100

=> r = 10

Thus, the required radius of the base is 10 cm.

**Question 4:**

If the volume of a right circular cone of height 9 cm is 48π cm^{3}, find the diameter of its base.

Answer:

Volume of the cone = 48 π cm^{3}

Height of the cone (h) = 9 cm

Let r be its base radius.

=> πr^{2} h/3 = 48π

=> (r^{2} * 9)/3 = 48

=> 3r^{2} = 48

=> r^{2} = 48/3

=> r^{2} = 16

=> r = √16

=> r = 4

Now, Diameter = 2 * r = 2 * 4 = 8

So, diameter of the base of the cone = 8 cm

**Question 5:**

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

Answer:

Here, diameter of the conical pit = 3.5 m

Radius (r) = 3.5/2 m, Depth (h) = 12 m

Volume (capacity) = πr^{2} h/3

= (22/7) * (3.5/2) * (3.5/2) * 12 * (1/3)

= 22 * (0.5/2) * (3.5/2) * 4

= 22 * 0.5 * 3.5

= 38.5 m^{3}

Since 1000 cm^{3} = 1 l

and 1000000 cm^{3} = l m^{3}

So, 1000 * 1000 cm^{3} = 1000 l = 1 kl

Also, 1000 * 1000 cm^{3} = 1 m^{3}

=> 1 m3 = 1 kl

=> 38.5 m^{3} = 38.5 kl

Thus, the capacity of the conical pit is 38.5 kl.

**Question 6:**

The volume of right circular cone is 9856 cm3. If the diameter of the be is 28 cm, find

(i) height of the cone (ii) slant height of the cone

(iii) curved surface area of the cone.

Answer:

Volume of the cone (v) = 9856 cm^{3}

Diameter of the base = 28 cm

Radius of the base = 28/2 = 14 cm

Volume (capacity) = πr^{2} h/3

= (1/3) * (22/7) * 14 * 14 * h

= (1/3) * 22 * 2 * 14 * h

Now, (1/3) * 22 * 2 * 14 * h = 9856

=> h = (9856 * 3)/(22 * 2 * 14)

=> h = 16 * 3

=> h = 48 cm

Thus, the required height is 48 cm.

(ii) Let the slant height be l cm.

(Slant height)^{2} = (Radius)^{2} + (Height)^{2}

=> l^{2} = 14^{2} + 48^{2}

=> l^{2} = 196 + 2304

=> l^{2} = 2500

=> l^{2} = √2500

=> l = 50

Thus, the required height = 50 cm.

(iii) To find the curved surface area

The curved surface area of a cone = πrl

= (22/7) * 14 * 50

= 22 * 2 * 50

= 2200

Thus, the curved surface area of the cone is 2200 cm^{2}.

**Question 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.

Answer:

Sides of the right triangle are 5 cm, 12 cm and 13 cm.

Since the right angled triangle is revolved about the 12 cm side.

So, Its height is 12 cm and base is 5 cm.

Thus, we have Radius of the base of the cone so formed (r) = 5 cm

Height (h) = 12 cm, Slant height = 13 cm

Now, volume of the cone so obtained = πr^{2} h/3

= (1/3) * π * 5 * 5 * 12

= π * 5 * 5 * 4

= 100π cm^{3}

Thus, the required volume of the cone is 100π cm^{3}

**Question 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.

Answer:

Since the right triangle is revolved about the side 5 cm.

So, Height of the cone so obtained (h) = 5 cm

And radius of the cone (r) = 12 cm

Now, volume of the cone so obtained = πr^{2} h/3

= (1/3) * π * 12 * 12 * 5

= π * 4 * 12 * 5

= 240π cm^{3}

Now, volume of cone having radius 5 cm/ volume of cone having radius 12 cm

= 100π/240π

= 5/12

= 5 : 12

Thus, the required ratio is 5 : 12

**Question 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.

Answer:

Here the heap of wheat is in the form of a cone such that

Base diameter = 10.5 m

Base radius (r) = 10.5/2 = 5.25 m

Height (h) = 3 m

Now, volume of heap = πr^{2} h/3

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

= 22 * 5.25 * 0.75

= 86.625 m^{3}

Thus, the required volume = 86.625 m^{3}

Since, the area of the canvas to cover the heap must be equal to the curved surface area of the

conical heap.

Area of the canvas = πrl

Now, l^{2} = r^{2} + h^{2}

=> l^{2} = (5.25)^{2} + 3^{2}

=> l^{2} = 27.5625 + 9

=> l^{2} = 36.5625

=> l^{2} = √36.5625

=> l = 6.046

=> l = 6.05 (approx.)

Now, Area of the canvas = πrl

= (22/7) * 5.25 * 6.05

= 22 * 0.75 * 6.05

= 99.825 m^{2}

Thus, the required area of the canvas is 99.825 m^{2}.

**Exercise 13.8**

**Question 1:**

Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

Answer:

(i) Here, radius (r) = 7 cm

Volume of the sphere = 4πr^{3} /3

= (4/3) * (22/7) * 7 * 7 * 7

= (4/3) * 22 * 7 * 7

= 4312/3 cm^{3}

(ii) Here, radius (r) = 0.63 m

Volume of the sphere = 4πr^{3} /3

= (4/3) * (22/7) * 0.63 * 0.63 * 0.63

= (4/3) * 0.09 * 0.63 * 0.63

= 4 * 0.09 * 0.21 * 0.63

= 1.05 m^{3 } (approx.)

**Question 2:**

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

(i) 28 cm (ii) 0.21 m

Answer:

(i) Diameter of the ball = 28 cm

Radius of the ball = 28/2 = 14 cm

Now, volume of the spherical ball = 4πr^{3} /3

= (4/3) * (22/7) * 14 * 14 * 14

= (4/3) * 22 * 14 * 14 * 2

= 34496/3 cm^{3}

(ii) Diameter of the ball = 0.21 m

Radius of the ball = 0.21/2 m

Now, volume of the spherical ball = 4πr^{3} /3

= (4/3) * (22/7) * (0.21/2) * (0.21/2) * (0.21/2)

= (4/3) * 22 * (0.03/2) * (0.21/2) * (0.21/2)

= 11 * 0.03 * 0.07 * 0.21

= 0.004851cm^{3}

**Question 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 cm^{3}?

Answer:

Diameter of the metallic ball = 4.2 cm

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

Now, volume of the ball = 4πr^{3} /3

= (4/3) * (22/7) * 2.1 * 2.1 * 2.1

= 4 * 22 * 0.7 * 0.3 * 2.1

= 38.808 cm^{3}

Density of the metal = 8.9 g/cm^{3}

So, mass of the ball = 8.9 * 38.808 = 345.39 g

**Question 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?

Answer:

Let diameter of the earth be 2r.

Then radius of the earth = 2r/2 = r

So, the diameter of the moon = 2r/4 = r/2

Radius of the moon = (r/2)/2 = r/4

Volume of the earth = 4πr^{3} /3 ………..1

Volume of the moon = {4π(r/4)^{3} }/3 ………..2

Now, Volume of the earth/ Volume of the moon = {4πr^{3} /3}/{4π(r/4)^{3} }/3

= r^{3} /(r^{3} /64)

= 64

=> Volume of the moon = (1/64) * Volume of the earth

Hence, volume of the moon is 1/164 of volume of the earth.

**Question 5:**

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

Answer:

Diameter of the hemisphere = 10.5 cm

Radius (r) of the hemisphere = 10.5/2 = 5.25 cm

Volume of the hemisphere = 2πr^{3} /3

= (2/3) * (22/7) * 5.25 * 5.25 * 5.25

= 2 * 22 * 1.75 * 5.25 * 0.75

= 303 cm^{3}

= 303/1000 l

= 0.303 l

Hence, the hemisphere bowl can hold 0.303 litres of milk.

**Question 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.

Answer:

Inner radius (r) of the tank = 1m

Thickness of the iron sheet = 1 cm = 1/100 m = 0.01 m

External radius of the tank (R) = 1 + 0.01 = 1.01 m

Volume of the iron used to make the tank = 2π(R^{3} - r^{3})/3

= (2/3) * (22/7) * [(1.01)^{3} - 1^{3}]

= (2/3) * (22/7) * 0.030301

= 0.06348 m^{3}

**Question 7:**

Find the volume of a sphere whose surface area is 154 cm^{2}.

Answer:

Let r be the radius of the sphere.

Its surface area = 4πr^{2}

=> 4πr^{2} = 154

=> r^{2} = 154/4π

=> r^{2} = 154/(4 * 22/7)

=> r^{2} = (154 * 7)/(4 * 22)

=> r^{2} = (7 * 7)/4

=> r^{2} = 49/4

=> r = √(49/4)

=> r = 7/2

=> r = 3.5

Now, volume of the sphere = 4πr^{3} /3

= (4/3) * (22/7) * 3.5 * 3.5 * 3.5

= (4/3) * 22 * 0.5 * 3.5 * 3.5

= 539/3 cm^{3}

**Question 8:**

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96.

If the cost of white-washing is Rs 2.00 per square metre, find the

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

Answer:

(i) Inner surface of the dome = Total cost/Cost of white per m^{2}

= 498.96/2

= 249.48 m^{2}

(ii) Let radius of the dome be r m.

Then, 2πr^{2} = 249.48

=> r^{2} = 249.48/2π

=> r^{2} = 249.48/(2 * 22/7)

=> r^{2} = (249.48 * 7)/(2 * 22)

=> r^{2} = (11.34 * 7)/2

=> r^{2} = 5.67 * 7

=> r^{2} = 39.69

=> r = √39.69

=> r = 6.3 cm

So, volume of the air inside the dome = 2πr^{3} /3

= (2/3) * (22/7) * 6.3 * 6.3 * 6.3

= 2 * 22 * 0.9 * 2.1 * 6.3

= 523.9 cm^{3}

**Question 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 S and S′.

Answer:

(i) Volume of a sphere of radius r = 4πr^{3} /3

Volume of 27 such sphere = 27 * 4πr^{3} /3

= 9 * 4πr^{3}

= 36πr^{3}

Volume of a sphere of radius r’ = 4π(r’)^{3} /3

=> 4π(r’)^{3} /3 = 36πr^{3}

=> (r’)^{3} /3 = 9r^{3}

=> (r’)^{3} = 9 * 3 * r^{3}

=> (r’)^{3} = 27r^{3}

=> (r’)^{3} = (3r)^{3}

=> r’ = 3r

(ii) Surface are (S) of the sphere with radius r = 4πr^{2}

Surface are (S’) of the sphere with radius r = 4π(r’)^{2}

= 4π(3r)^{2}

= 4π * 9r^{2}

= 36πr^{2}

Now, S/S’ = 4πr^{2} /36πr^{2}

=> S/S’ = 1/9

=> S : S’ = 1 : 9

**Question 10:**

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm^{3}) is needed to fill this capsule?

Answer:

Given, diameter of sphere = 3.5 mm

Radius of the sphere = 3.5/2 = 1.75 mm

Now, volume of the capsule = 4πr^{3} /3

= (4/3) * (22/7) * 1.75 * 1.75 * 1.75

= (4/3) * 22 * 0.25 * 1.75 * 1.75

= 67.375/3

= 22.46 mm^{3} (approx.)

Hence, 22.46 mm^{3} of medicine is needed to fill the capsule.

**Exercise 13.9 (Optional)**

**Question 1:**

A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see Fig.).

The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted.

If the rate of polishing is 20 paise per cm^{2} and the rate of painting is 10 paise per cm^{2}, find the total expenses required for polishing and painting the surface of the bookshelf.

Answer:

Here, Height = 110 cm, Depth = 25 cm

Breadth = 85 cm, Thickness of the plank = 5 cm

Area to be polished = [(110 * 85)] + [(85 * 25) * 2] + [(110 * 25) * 2]

+ [(5 * 110) * 2] + [(75 * 5) * 4]

= 9350 + 4250 + 5500 + 1100 + 1500

= 21700 cm^{2}

Cost of polishing at the rate 20 poise per cm^{2} = Rs (20/100) * 21700

= Rs 21700/5

= Rs 4340

Area to be painted = [(75 * 20) * 6] + [(90 * 20) * 2] + [90 * 75]

= 9000 + 3600 + 6750

= 19350 cm^{2}

So, cost of painting at the rate of 10 paise per cm^{2} = Rs (10/100) * 19350

= Rs 19350/10

= Rs 1935

Total Expenses = Cost of polishing + Cost of painting

= Rs. 4340 + Rs.1935

= Rs. 6275

Thus, the total required expense = Rs. 6275.

**Question 2:**

The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig.

Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black.

Find the cost of paint required if silver paint costs 25 paise per cm^{2} and black paint costs 5 paise per cm^{2}.

Answer:

Radius of sphere = 21/2 = 10.5 cm

Surface are of a sphere = 4πr^{2}

= 4 * (22/7) * 10.5 * 10.5

= 4 * 22 * 1.5 * 10.5

= 1386 cm^{2}

Area of the base of the cylinder (support) = πR^{2}

= (22/7) * (1.5)^{2}

= (22/7) * 1.5 * 1.5

= 49.5/7

= 7.07 cm^{2}

Area of a sphere to painted silver = 1386 – 7.07 = 1378.93 cm^{2}

Area of spheres to be painted silver = 8 * 1378.93 cm^{2}

So, cost of painting the spheres = Rs (8 * 1378.93 * 25)/100 = Rs 2757.86

Curved surface area of a cylinder (support) = 2 * (22/7) * 1.5 * 7

= 2 * 22 * 1.5

= 66 cm^{2}

Curved surface area of 8 supports = 8 * 66 = 528 cm^{2}

Cost of painting the supports = Rs 528 * (5/100)

= Rs 528/20

= Rs 26.40

Hence, total cost required of paint = Rs (2757.86 + 26.40) = Rs 2784.26

**Question 3:**

The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Answer:

Let originally the diameter of the sphere be 2r.

Then, radius of the sphere = 2r/2 = r

Surface area of the sphere = 4πr^{2}

New diameter of the sphere = 2r- 2r * (25/100)

= 2r – 2r/4

= 2r - r/2

= 3r/2

New radius of the sphere = (3r/2)/2 = 3r/4

Surface area of the new sphere = 4π * (3r/4)^{2}

= 4π * (9r^{2} /16)

= 9πr^{2}/4

Decrease in surface area = 4πr^{2} - 9πr^{2}/4 = 7πr^{2}/4

Percent decrease = (7πr^{2}/4 * 100)/ 4πr^{2}

= (7/16) * 100

= 175/4

= 43.75

Hence, the surface area decreases by 43.75%

.