Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes

 Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes

Ex 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 opened at the top. Ignoring the thickness of the plastic sheet, determine
1. The area of the shut required for making the box.
2. The cost of sheet for it, if a sheet measuring 1 m2 casts ₹ 20
Solution:
1. l = 1.5 m
b = 1.25 m2
and h = 65 cm = 0.65 m
∴ The area of the sheet required for making the box = lb + 2(bh + hl)
= (1.5) (1.25) + 2{(1.25) (0.65) + (0.65) (1.5)}
= 1.875 + 2{0.8125 + 0.975}
= 1.875 + 2(1.7875)
= 1.875 + 3.575
= 5.45 m2

2. The cost of sheet for it = ₹ 5.45 x 20 = ₹ 109

Question 2.
The length, breadth, and height of a room are 5 m, 4 m, and 3m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of ₹ 7.50 perm?
Solution:
l = 5m, b = 4m and h = 3 m
Area of the walls of the room
= 2(1 + b)h = 2(5 + 4)3 = 54 m2
Area of ceiling = lb
= (5)(4) = 20 m2
Total area of the walls of the room and the ceiling
= 54 m2 + 20 m2 = 74 m2
∴ Cost of white washing the walls of the room and the ceiling
= 74 x ₹ 7.50 = ₹ 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 ₹ 10 per m2 is ₹ 15000, find the height of the hall.
[Hint. Area of the four walls = Lateral surface area]
Solution:
Let the length, breadth, and height of the rectangular hall be l m, bm, and hm respectively.
Perimeter = 250 m
2(l + b) = 250 ⇒ l + b = 125

∴ 2(l+ b)h = 1500
(l + b)h = 750
125 h = 750
h = 750/125
h = 6m
Hence, the height of the hall is 6 m.

Question 4.
The paint in a certain container is sufficient to paint on an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?
Solution:
For a brick, l = 22.5 cm, b = 10 cm and h = 7.5 cm
∴ Total surface area of a brick = 2(lb + bh + hl)
= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2(468.75) = 937.5 cm2 = 0.9375 m2
∴ Number of bricks that can be painted out

= 9.375/0.9375 = 10

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.
1. Which box has the greater lateral surface area and by how much?
2. Which box has the smaller total surface area and by how much?
Solution:
1. Each edge of the cubical box (a) = 10 cm
∴ Lateral surface area of the cubical box
= 4a2 = 4(10)2 = 400 cm2
For cuboidal box
l = 12.5 cm, b = 10 cm and h = 8 cm
∴ Lateral surface area of the cuboidal box
= 2(l + b)h
= 2(12.5 + 10)(8) = 360 cm2
∴ Cubical box has the greater lateral surface area than the cuboidal box by (400 – 360) cm2 , i.e., 40 cm2 .

2. Total surface area of the cubical box = 6a2
= 6(10)2 = 600 cm2
Total surface area of the cuboidal box
= 2 (lb + bh + hl)
= 2[(12.5) (10) + (10) (8) + (8) (12.5)]
= 2[125 + 80 + 100] = 610 cm2
∴ Cubical box has the smaller total surface area than the cuboidal box by (610 – 600) cm2, i.e., 10 cm2.

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.
1. What is the area of the glass?
2. How much of tape is needed for all the 12 edges?
Solution:
1. For herbarium
l = 30 cm
b = 25 cm
h = 25 cm
∴ Area of the glass = 2(lb + bh + hl)
= 2 [(30) (25) + (25) (25) + (25) (30)]
= 2 [750 + 625 + 750] = 4250 cm2

2. The tape needed for all the 12 edges
= 4 (l + b + h)
= 4(30 + 25 + 25)
= 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 x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 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:
For bigger box
l = 25 cm
b = 20 cm
h = 5 cm
∴ Total surface area of the bigger box = 2 (lb + bh + hi)
= 2[(25) (20) + (20) (5) + (5) (25)]
= 2 [500 + 100 + 125] = 1450 cm2 Cardboard required for all the overlaps
= 1450 x 5/100 = 72.5 cm2
∴ Net surface area of the bigger box = 1450 cm2 + 72.5 cm2 = 1522.5 cm2
∴ Net surface area of 250 bigger boxes = 1522.5 x 250 = 380625 cm2
∴ Cost of cardboard
= 4/1000 x 380625 = ₹ 1522.50
For smaller box
l = 15 cm
b = 12 cm
h = 5 cm
∴Total surface area of the smaller box
= 2 (lb + bh + hl)
= 2[(15) (12) + (12) (5) + (5) (15)]
= 2 [180 + 60 + 75] = 630 cm2

Cardboard required for all the overlaps
= 630 x 5/100 = 31.5 cm2
∴ Net surface area of the smaller box
= 630 cm2 + 31.5 cm2 = 661.5 cm2
∴ Net surface area of 250 smaller boxes
= 661.5 x 250 = 165375 cm2
Cost of cardboard
4/1000 x 165375 = ₹ 661.50
∴ Cost of cardboard required for supplying 250 boxes of each kind
= ₹ 1522.50 + ₹ 661.50 = ₹ 2184

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with a 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 many tarpaulins would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Sol. For shelter
l = 4 m
b = 3 m
h = 2.5 m
∴ Total surface area of the shelter
= lb + 2(bh + hl)
= (4) (3) + 2[(3) (2.5) + (2.5) (4)]
= 12 + 2 [7.5 + 10] = 47 m2
Hence, 47 m2 of tarpaulin will be required.

Ex 13.2

Question 1.
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:
Let the radius of the base of the cylinder be r cm.
h = 14 cm
Curved surface area = 88 cm2
2πrh = 88
= 2 x 22/7 x r x 14 = 88
r = 88×7/2×22×14
r = 1 ⇒ 2r = 2
Hence the diameter of the base of the cylinder is 2 cm.

Question 2.
It is required to make a closed cylindrical tank of height 1 m and a base diameter 140 cm from a metal sheet. How many square meters of the sheet is required for the same?
Solution:
h = 1 m = 100 cm
2r = 140 cm
r = 140/2 cm = 70 cm
∴ Total surface area of the closed cylindrical tank
= 2πr(h + r)
= 2 x 22/7 x 70 (100 + 70)
= 74800 cm2 = 74800/100x100 m2 = 7.48 m2
Hence, 7.48 square metres of the sheet are 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. Find its
1. inner curved surface area,
2. outer curved surface area,
3. the total surface area.

Solution:
h = 77 cm
2r = 4 cm
⇒ r = 2 cm
2R = 4.4 cm
⇒ R = 2.2 cm

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 m2.
Solution:
2r = 84 cm
r = 42 cm
h = 120 cm
∴ Area of the playground levelled in taking 1 complete revolution
= 2πrh = 2 x 22/7 x 42 x 120
= 31680 cm2
∴ Area of the playground
= 31680 x 500 = 15840000 cm2
= 15840000/100×100 m2 = 1584 m2
Hence, the area of the playground is 1584 m2.

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 ₹ 12.50 per m2.
Solution:
2r = 50 cm
∴ r = 25 cm = 0.25 m
h = 3.5 m
∴ Curved surface area of the pillar
= 2πrh = 2 x 22/7 x 0.25 x 3.5
= 5.5 m2
∴ Cost of painting the curved surface area of the pillar at the rate of ₹ 12.50 per m2
= ₹ (5.5 x 12.50) = ₹ 68.75

Question 6.
The 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:
Let the height of the right circular cylinder be h m.
r = 0.7 m
Curved surface area = 4.4 m2
2πrh = 4.4
2 x 22/7 x 0.7 x h = 4.4
4.4h = 4.4 = h = 1m
Hence, the height of the right circular cylinder is 1 m.

Question 7.
The inner diameter of a circular wall is 3.5 m. It is 10 m deep. Find
1 its inner curved surface area,
2. the cost of plastering this curved surface at the rate of 40 per m2.
Solution:
1. 2r = 3.5 m
r = 3.5/2 m
= r = l.75 m
h = 10m
∴ Inner curved surface area of the circular wall = 2πrh
= 2 x 22/7 x 1.75 x 10 = 110 m2.

2. Cost of plastering the curved surface at the rate of ₹ 40 per m2
= ₹ (110 x 40) = ₹ 4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and a diameter 5 cm. Find the total radiating surface in the system.
Solution:
h = 28m
2r = 5 cm

Question 9.
Find:
1. 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.
2. how much steel was actually used, if of the steel actually used was wasted in making the tank.
Solution:
1. 2r = 4.2 m

Question 10.
In the figure given below, 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:
2r = 20 cm r = 10cm
h = 30cm
∴ Cloth required = 2,tr (h + 2.5 + 2.5)
= 2πr (h + 5)
= 2 x 22/7 x 10 x (30+5)
= 2200 cm2

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?
r = 3 cm
h = 10.5 cm
∴ Cardboard required for 1 competitor
= 2πrh + πr2

Hence, 7920 cm2 of cardboard was required to be bought for the competition.

Ex 13.3

Question 1.
The diameter of the base of the cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
∴ Diameter of the base = 10.5 cm
∴ Radius of the base (r) = 10.5/7 cm = 5.25 cm
Slant height (l) = 10 cm
∴ Curved surface area of the cone = πrl
= 22/7 x 5.25 x 10 = 165 cm2

Question 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.
Solution:
Slant height (l) = 21 m
Diameter of base = 24 m
  

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 20 cm. (Use it = 3.14)
Solution:
For conical tent
h = 8m
r = 6 m

Question 6.
The slant height and a base diameter of a conical tomb is 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2.
Solution:
Slant height (l) = 25 m
Base diameter (d) = 14 m
∴ Base radius (r) = 14/2 m = 7m
∴ Curved surface area of the tomb
= πrl = 22/7 x 7 x 25 = 550 m2
∴ Cost of white-washing the curved surface area of the tomb at the rate of 210 per 100 m2
= ₹ 210/100 x 550 = ₹ 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.
Solution:
Base radius (r) = 7 cm
Height (h) = 24 cm

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 a 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?

∴ Curved surface area = πrl
= 3.14 x 0.2 x 1.02 = 0.64056 m2
∴ Curved surface area of 50 cones
= 0.64056 x 50 m2 = 32.028 m2
∴ Cost of painting all these cones at ₹ 12 per m2
= ₹ 32.028 x 12
= ₹ 384.336 = ₹ 384.34 (approximately)

Ex 13.4

Question 1.
Find the surface area of a sphere of radius:
1. 10.5 cm
2. 5.6 cm
3. 14 cm
Solution:
1. r = 10.5 cm
∴ Surface area = 41πr2
= 4 x 22/7 x (10.5)2
= 1386 cm2

2. r = 5.6 cm
∴ Surface area = 4m2
= 4x 22/7 x (5.6)2
= 394.24 cm2

3. r = 14 cm
∴ Surface area = 4πr2
= 4 x 22/7 x (14)2 = 2464 cm2

Question 2.
Find the surface area of a sphere of diameter:
1. 14 cm
2. 21 cm
3. 3.5 m
Solution:
1. Diameter = 14 cm
∴ Radius (r) = 14/2 cm = 7 cm
∴ Surface area = 4πr2
= 4 x 22/7 x (7)2 = 616 cm2

2. Diameter = 21 cm
∴ Radius (r) = 21/2 cm
∴ Surface area = 4πr2
= 4 x 22/7 x (22/7)2
= 1386 cm2

3. Diameter = 3.5 m
∴ Radius (r) = 3.5/2 m = 1.75 m
∴ Surface area = 4πr2
= 4 x 22/7 x (1.75)2
= 38.5 m2

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
r = 10 cm
∴ Total surface area of the hemisphere
= 3πr2 = 3 x 3.14 x (10)2
= 942 cm2

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.
Solution:
Case I:
r = 7 cm
Surface area = 4πr2
= 4 x  (7) 2 = 616 cm2

Case II:
r = 14 cm
Surface area = 4πr2
= 4 x 22/7 x (14)2 = 2464 cm2
∴ Ratio of surface area of the balloon
= 616:2464 = 616/2464 = 1/4 = 1.4

Question 5.
A hemispherical bowl is 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:
Inner diameter = 10.5 cm
∴ Inner radius (r) = 10.5/2 cm = 5.25 cm
∴ Inner surface area = 2πr2
= 2 x 22/7 x (5.25)2
= 173.25 cm2
∴ Cost of tin-plating at the rate of ₹ 16 per 100 cm2
= 16/100 x 173. 25
= ₹ 27.72

Question 6.
E Find the radius of a sphere whose surface area is 154 cm2
Solution:
Let the radius of the sphere be r cm.
Surface area = 154 cm2

Question 7.
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 the earth be 2r.

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.
Solution:
The inner radius of the bowl = 5 cm
Thickness of steel = 0.25 cm
∴ The outer radius of the bowl
= 5 + 0.25 = 5.25 cm
∴ The outer curved surface area of the bowl
4πr = 4 x 22/7 x (5.25)2
= 346.5 cm2

Question 9.
A right circular cylinder just encloses a sphere of radius r. Find
1. the surface area of the sphere,
2. the curved surface area of the cylinder,
3. the ratio of the areas obtained in (i) and (ii).

Solution:
1. The surface area of the sphere = 4πr2

2. For cylinder:
The radius of the base = r
Height = 2r
∴ The curved surface area of the cylinder
= 2π(r) (2r) = 4πr2

3. Ratio of the areas obtained in (i) and (ii)

Ex 13.5

Question 1.
A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Solution:
Volume of a matchbox = 4 x 2.5 x 1.5 cm3 = 15 cm3
∴ Volume of a packet containing 12 such boxes = 15 x 12 cm3 = 180 cm3

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 m3 = 1000 L)
Solution:
Capacity of the tank = 6 x 5 x 4.5 m3 = 135 m3
∴ The volume of water it can hold = 135 m3
= 135 x 1000 L = 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 a liquid?
Solution:
Let the height of the cuboidal vessel be h m.
l = 10 m
b = 8 m
Capacity of the cuboidal vessel = 380 m3
lbh = 380
⇒ (10) (8) = 380 ⇒ h = 380/(10)(8)4.75 m
⇒ h = 19/4 ⇒ h = 4.75 m
Hence, the cuboidal vessel must be made 4.75 m high.

Question 4.
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:
l = 8 m, b = 6 m and h = 3 m
∴ Volume of the cuboidal pit = Ibh = 8 x 6 x 3 m3 = 144 m3
∴ Cost of digging the cuboidal pit @ ₹ 30 per m3 = ₹ (144 x 30) = ₹ 4320

Question 5.
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:
Let the breadth of the cuboidal tank be b m.
l = 2.5 m
h = 10 m
Capacity of the cuboidal tank = 50000 litres
= 50000/1000 m3 = 50 m3
lbh = 50
⇒ 2.5 x b x 10 = 50
⇒ 25b = 50
⇒ b = 50/25 = 2m
Hence, the breadth of the cuboidal tank is 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 x 15 m x 6 m. For how many days will the water of this tank last?
Solution:
Requirement of water per head per day = 150 lires
∴ Requirement of water for the total population of the village per day
= 150 x 4000 litres = 600000 litres
= 600000/1000 m3 = 600 m3

For tank:
l = 20 m, b = 15 m and h = 6 m
∴ Capacity of the tank
= 20 x 15 x 6 m3 = 1800 m3
∴ Number of days for which the water of this tank last

= 1800/600 = 3
Hence, the water of this tank will last for 3 days.

Question 7.
A godown measures 40 m x 25 m x 15 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown.
Solution:
For godown:
l = 40 m
b = 25 m
h = 15 m
∴ Capacity of the godown = lbh
= 40 x 25 x 15 m3 = 15000 m3
For a wooden crate:
l = 1.5 m
b = 1.25 m
h = 0.5m
∴ Capacity of a wooden crate = lbh
= 1.5 x 1.25 x 0.5 m3 = 0.9375 m3

We have,
15000/0.9375 = 16000
Hence, the maximum number of wooden crates that can be stored in the godown 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.
Solution:
Side of the solid cube (a) = 12 cm
∴ Volume of the solid cube = a3
= (12)3 = 12 x 12 x 12 cm3 = 1728 cm3
∴ It is cut into eight cubes of equal volume.
∴ Volume of a new cube
= 1728/8 cm3
= 216 cm3
Let the side of the new cube be x cm.
Then, volume of the new cube = x3 cm3
According to the question,
x3 = 216
⇒ x = (216)1/3
⇒ x = (6 x 6 x 0)1/3
⇒ x = 6 cm
Hence, the side of the new cube will be 6 cm.
Surface area of the original cube
= 6a2 = 6(12)2 cm2
Surface area of the new cube = 6x2 = 6(6)2 cm2
∴ Ratio between their surface areas

Question 9.
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:
In one hour
l = 2 km = 2 x 1000 m = 2000 m
b = 40 m
h = 3 m
∴ Water fall into the sea in one hour = Ibh
= 2000 x 40 x 3 m3
∴ Water fall into the sea in a minute
2000×40×3/60 m3 = 400 m3
Hence, 4000 m3 of water will fall into the sea in the minute.

Ex 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 cm3 = 1L)
Solution:
Let the base radius of the cylindrical vessel be r cm. Then, the circumference of the base of the cylindrical vessel = 2πr cm.
According to the question,
2πr = 132

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 cm3 of wood has a mass of 0.6 g.
Solution:
∴ Inner diameter = 24 cm

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 a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) For tin can
1 = 5 cm
b = 4 cm
h = 15 cm
Capacity = l x b x h
= 5 x 4 x 15 cm3 = 300 cm3

(ii) For plastic cylinder
Diameter = 7 cm
∴ Radius (r) = 7/2 cm
Height (h) = 10 cm
∴ Capacity = πr2h
= 22/7 x (7/2)2x 10 = 385 cm2
Clearly the second container i.e., a plastic cylinder has greater capacity than the first container i.e., a tin can by 385 – 300 = 85 cm3.

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)
Solution:
(i) Let the radius of the base of the cylinder be r cm.
h = 5 cm
Lateral surface = 94.2 cm2
⇒ 2πrh = 94.2
⇒ 2 x 3.14 x r x 5 = 94.2

(ii) r = 3 cm
h = 5 cm
∴ Volume of the cylinder = πr2h
= 3.14 x (3)2 x 5 = 141.3 cm3

Question 5.
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:
(i) Inner curved surface area of the vessel
= 2200/20 = 110 m2

(iii) r = 1.75 m
h = 10 m
∴ Capacity of the vessel = πr2h
= 22/7 x (1.75)2 x 10
= 96.25 m3
Hence, the capacity of the vessel is 96.25 m3 (or 96.25 kL).

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of the metal sheet would be needed to make it?
Solution:

∴ Total surface area = 2πrh + 2πr2
= 2 x 22/7 x 0.07 x 1 + 2 x 22/7 x (0.07)2
= 0.44 + 0.0308 = 0.4708 m2
Hence, 0.4708 m2 of the metal sheet should 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.
Solution:

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of a 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:
Diameter = 7 cm
∴ Radius (r) = 7/2 cm
Height (h) = 4 cm
∴ Volume of soup in the cylindrical bowl = πr2h
= 22/7 x (7/2)2 x 4 cm3 = 154 cm3
∴ Volume of soup to be prepared daily to serve 250 patients
= 154 x 250 cm3 = 38500 cm3 (or 38.5 L)
Hence, the hospital has to prepare 38500 cm3 (or 38.5 L) of soup daily to serve 250 patients.

Ex 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
Solution:
(i) r = 6 cm
h = l cm

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
Solution:
(i) r = 7 cm
l = 25 cm
r2 + h2 = l2
⇒ (7)2 + h2 = (25)2
⇒ h2 = (25)2 – (7)2
⇒ h2 = 625 – 49
⇒ h2= 576

Question 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:
Let the radius of the base of the cone be r cm.
h = 15 cm
Volume = 1570 cm3

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Solution:
Let the radius of the base of the right circular
cone be r cm.
h = 9 cm
Volume = 48π cm3

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

Question 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:

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.
Solution:
The solid obtained will be a right circular cone whose radius of the base is 5 cm and height is 12 cm.

Question 8.
If the triangle ABC in 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:
The solid obtained will be a right circular cone whose radius of the base is 12 cm and the height is 5 cm.
∴ r = 12 cm
h = 5 cm

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.
Solution:

Ex 13.8

Question 1.
Find the volume of a sphere whose radius is ∝

Question 2.
Find the amount of water displaced by a solid
spherical ball of diameter
1. 28 cm
2. 0.21m
Solution:

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 cm3?
Solution:

2. Density = 8.9 g per cm3
∴ Mass of the ball = Volume x Density
= 38.808 x 8.9
= 345.39 g (approx.)

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?
Solution:

Question 5.
How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:

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.
Solution:
Inner radius (r) = 1 m
Thickness of iron sheet = 1 cm = 0.01 m
∴ Outer radius (R) = Inner radius (r) + Thickness of iron sheet
= 1 m + 0.01 m = 1.01 m
∴ Volume of the iron used to make the tank

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

Question 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 whitewashing is 2.00 per square meter, find the
1. inside surface area of the dome,
2. the volume of the air inside the dome.
Solution:

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
1. radius r’ of the new sphere,
2. the ratio of S and S’.
Solution:

Question 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:

Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows:
Height = 110 cm,
Depth = 25 cm,
Breadth = 85 cm (See figure).

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 cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Solution:
Surface area to be polished
= [(110 x 85) + 2(110 x 25) + 2(85 x 25) + 2(110 X 5) + 4(75 x 5)1
= (9350 + 5500 + 4250 + 1100 + 1500) cm2 = 21700 cm2
∴ Expenses required for polishing @ 20 paise per cm2
= 21700 x 20 paise
= ₹ 21700×20/100 = ₹ 4340

Surface area to be painted
= [2(20 x 90) + 6(75 x 20) + (75 x 90)1
= (3600 + 9000 + 6750) cm2 = 19350 cm2
∴ Expenses required for painting @ 10 paise per cm2
= 19350 x lo paise
= ₹ 19350×10/100 = ₹ 1935
∴ Total expenses required for polishing and painting the surface of the bookshelf
= ₹4340 + ₹1935 = ₹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 figure. 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 cm2 and black paint costs 5 paise per cm2.

Solution:
For a wooden sphere
Diameter = 21 cm

Question 3
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution: