Surface Areas and Volumes-NCERT Solutions

Class IX Math
NCERT Solution for Surface Areas and Volumes

NCERT TEXTBOOK QUESTIONS SOLVED
EXERCISE 13.1 (Page 213)

Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It the top. Ignoring the thickness of the plastic sheet, determine:

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

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

Sol: (i) Here, l = 1.5 m, b = 1.25 m

∵ It is open from the top.

∵ Its surface area = [Lateral surface area] + [Base area]

= [2(l + b)h] + [l × b]

= [2(1.50 + 1.25)0.65 m²] + [1.50 × 1.25 m²]

= [2 × 2.75 × 0.65 m²] + [1.875 m²]

= 3.575 m² + 1.875 m² = 5.45 m²

∵ The total surface area of the box = 5.45 m²

∴ Area of the sheet required for making the box = 5.45 m²

(ii) Rate of sheet = Rs. 20 per m²

Cost of 5.45 m² = Rs. 20 × 5.45

⇒ Cost of the required sheet = Rs.109

Q2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate off 7.50 per m².

Sol: Length of the room (l) = 5m

Breadth of the room (b) = 4 m

Height of the room (h) = 3 m

The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.

∴ Area for white washing = [Lateal surface area] + [Area of the ceiling]

= [2(1 + b)h] + [1 × b]

= [2(5 + 4) } 3 m²] + [5 × 4 m²]

= [54m²] + [20m²] = 74m²

Cost of white washing:

Cost of white washing for 1 m² = Rs. 7.50

∴ Cost of white washing for 74 m² = Rs. 7.50 × 74

The required cost of white washing is Z555.

Q3. 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 Rs. 15000, find the height of the hall.

Sol: Note: Area of four walls = Lateral surface area.

A rectangular hall means a cubiod.

Let the length and breadth of the hall be ‘l’ and ‘b’ respectively.

∵[Perimeter of the floor] = 2(l + b)

⇒ [2(l + b)] = 250 m.

∵Area of four walls = lateral surface area

⇒ [2(l + b)] × h [where ’h’ is the height of hall.]

Cost of painting the four walls = Rs. 10 × 250 h = Rs. 2500 h

⇒ Rs. 2500 h = Rs. 15000

Thus, the required height of the hall 6 m

Q4. 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?

Sol: Total area that can be painted = 9.375 m², since a brick is like a cuboid

∴ Total surface area of a brick = 2[lb + bh + hl]

= 2[(22.5 × 10) + (10 × 7.5) + (7.5 × 22.5)] cm²

= 2[(225) + (75) + (168.75)] cm²

Thus, the required number of bricks = 100

Q5. 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?

Sol: For the cubical box

∵Edge of the cubical box = 10 cm

∴ Lateral surface area = 4a²

= 4 × 102 cm²

= 4 × 100 cm²

= 400 cm²

Total surface area = 6a²

= 6 × 102 = 6 × 100 cm²

= 600 cm²

∵For the cuboidal box, l = 12.5 cm, b = 10 cm, h = 8 cm

∴ Lateral surface area = 2[(l + b)] × h

= 2[12.5 + 10] × 8 cm²

= 2[22.5 × 8] cm²

= 360 cm²

Total surface area = 2[lb + bh + hl]

= 2[(12.5 × 10) + (10 × 8) + (8 × 12.5)] cm²

= 2[125 + 80 + 100] cm²

= 2[305] cm² = 610 cm²

Obviously,

(i) ∵ 400 cm² > 360 cm² and 400 – 360 = 40

∴ The cubical box, has greater lateral surface area by 40 m².

(ii) ∵ 610 cm² > 600 cm² and 610 – 600 = 10

∴The cuboidal box has greater total surface area by 10 m².

Q6. 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: The herbarium is like a cuboid

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

(i) ∵ Area of a cuboid = 2[lb + bh + hl]

∴ Surface area of the herbarium (glass) = 2[(30 × 25) + (25 × 25) + (25 × 30)] cm²

= 2[750 + 625 + 750] cm²

= 2[2125] cm²

= 4250 cm²

Thus, the required area of glass is 4250 cm².

(ii) Total length of 12 edges = 4l + 4b + 4h

= 4(1 + b + h) = 4(30 + 25 + 25) cm

= 4 × 80 cm = 320 cm

Thus, length of tape needed = 320 cm

Q7. 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 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Sol: For bigger box:

Length (1) = 25 cm, Breadth (b) = 20 cm, Height (h) = 5 cm

∵ The box is like a cuboid and total surface area of a cuboid = 2(lb + bh + hl)

Area of a box = 2([25 × 20) + (20 × 5) + (5 × 25)] cm²

= 2[500 + 100 + 125] cm²

= 2[725] cm² = 1450 cm²

Total surface area of 250 boxes = 250 × 1450 cm² = 362500 cm²

For smaller box:

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

Total surface area of a box = 2[lb + bh + hl]

= 2[(15 × 12) + (12 × 5) + (5 × 15)] cm²

= 2[180 + 60 + 75] cm²

= 2[315] cm² = 630 cm²

⇒ Total surface area of 250 boxes = 250 × 630 cm²

= 157500 cm²

Now, total surface area of both kinds of boxes

= 362500 cm² + 157500 cm²

= 5,20,000 cm²

Area for overlaps = 5% of [ total surface area]

∴ Total area of the cardboard required = [Total area of 250 boxes] + [5% of total surface area]

= 520000 cm² + 26000 cm²

= 546000 cm²

Cost of cardboard:

∵ Cost of 1000 cm² = Rs. 4

Q8. 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 he rolled up). Assuming that the stitching ma,gins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 in, with base dimensions 4m × 3m?

Sol: Here, height (h) = 2.5 m

Base dimension = 4 m × 3 m

⇒ Length (1) = 4 m and Breadth (b) = 3 m

∵The structure is like a cuboid.

∴ The surface area of the cuboid, excluding the base.

= [Lateral surface area] + [Area of ceiling]

= [2 (l + b)h] + [lb]

= [2 (4 + 3) × 2.5] + [4 × 3] m²

= [35] + [12] m²

= 47 m²

Thus, 47 m² tarpaulin would be required.

EXERCISE 13.2 (Page 216)

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

Sol: Let 'r' be the radius of the cylinder.

Here, height (h) = 14 cm

and curved surface area = 88 cm²

Curved surface area of a cylinder = 2πrh

∴ 2πrh = 88

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?

Sol: Here, height (h) = 1 m

∵ Diameter of the base = 140 cm = 1.40 m

Hence, the required sheet is 7.48 m²

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 figure) Find its.

(i) inner curved surface area,

(ii) outer curved surface area,

(iii) total surface area.

Sol: Length of the metal pipe = 77 cm

∵ It is in the form of a cylinder.

∴ Height (h) of the cylinder = 77 cm

Inner diameter = 4 cm

(iii) Total surface area

= [Inner curved surface] + [outer curved surface area] + [Two base circular lamina]

= [27πrh] + [2πRh] + [2π(R² – r²)]

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.

Sol: The roller is in the form of a cylinder diameter of the roller = 84 cm

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 Rs. 12.50 per m².

Sol: Diameter of the pillar = 50 cm

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.

Sol: Radius (r) = 0.7 m

Let height of the cylinder be 'h' metres.

∴ Curved surface area = 2πrh

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 W.

Sol: Inner diameter of the well = 3.5 m

Depth of the well (= height of the cylinder) h = 10 m

(i) Inner curved surface area = 2πrh

(ii) Cost of plastering

∵ Rate of plastering = Rs. 40 per m²

∴ Total cost of plastering 110 m² = Rs.110 × 40 = Rs. 4400

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.

Sol: Length of the cylindrical pipe = 28 m

⇒ h = 28m

Diameter of the pipe = 5 cm

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

Sol: The storage tank is in the form of a cylinder, and diameter of the tank = 4.2 m

Thus, the required area of the steel that was actually used is 95.04 m²

Q10. In the figure you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has 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.

Sol: The lampshade is in the form of a cylinder, where

∵A margin of 2.5 cm is to be added to top and bottom.

∴ Total height of the cylinder

h = 30 cm + 2.5 cm + 2.5 cm

= 35 cm

Now, curved surface area = 2πrh

= 2 × 22 × 10 × 5 cm² =2200 cm²

Thus, the required area of the cloth = 2200 cm²

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 Vdyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Sol: Here, the penholders are in the form of cylinders

Radius of a cylinder (r) = 3 cm

Height of a cylinder (h) = 10.5 cm

Since, a penholder must be open from the top.

∴ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]

EXERCISE 13.3 (Page 221)

Q1. Diameter of the be of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Sol: Here, diameter of the base = 10.5 cm

Q2. Find the total surfce area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Q3. Curved surface area of a cone is 308 cm' and its slant height is 14 cm. Find: (i) radius of the base and (ii) total surface area of the cone.

Sol: Here, curved surface area = 308 cm² Slant height (1) = 14 cm

(i) Let the radius of the base be 'r' cm.

∴ πrl = 308

and curved surface area = 308 cm²

∴ Total surface area = [Curved surface area] + [Base area]

= 308 cm² + 154 cm²

= 462 cm²

Q4. A conical tent is 10 m high and the radius of its base

(i) slppnt height of the tent.

(ii) cost, of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70.

Sol: Here, height of the tent (h) = 10 m

Radius of the base (r) = 24 m

Q5. 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).

Sol: Here, Base radius (r) = 6 m

Height (h) = 8 m

∴ Area of the canvas (tarpaulin) required to make the tent = 188.4 m²

Let the length of the tarpaulin = 'L' m

∴ Length × Breadth = 188.4

⇒ L × 3 = 188.4

Q6. 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 m².

Q7. 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.

Sol: Here,

Radius of the base (r) = 7 cm

height (h) = 24 cm

Lateral surface area of 10 caps = 10 × 550 cm² = 5500 cm²

Thus, the required area of the sheet = 5500 cm²

Q8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is f 12 per m², what will be the cost of painting ail the cones? (Use π = 3.14 and take

Sol: Here,

∵ Diameter of the base = 40 cm

EXERCISE 13.4 (Page 225)

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

(i) 10.5 cm (ii) 5.6 cm

Sol: (i) Here r = 10.5 cm

∴ Surface area of the sphere = 4πr²

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

(i) 14 cm (ii) 21 cm

Sol: (i) Here, Diameter = 14 cm

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

Sol: Here, radius (r) = 10 cm

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.

Sol: Case I: When radius (r₁) = 7 cm

Q5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm².

Sol: Inner diameter of the hemisphere = 10.5 m

∵ Curved surface area of a hemisphere = 2πr²

∴ Inner curved surface area of hemispherical bowl

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

Sol: Let the radius of the sphere be 'r' cm.

∴ Surface area = 4πr²

⇒ 4πr² = 154

Thus, the required radius of the sphere is 3.5 cm

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

Sol: Let the radius of the earth = r

∵ Surface area of a sphere = 4πr²

Since, the earth as well as the moon are considered to be spheres.

∴ Surface area of the earth = 4πr²

or [Surface area of the moon] : [Surface area of the earth] = 1 : 16

Thus, the required ratio = 1 : 16

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.

Sol: Inner radius (r) = 5cm

Thickness = 0.25 cm

∴ Outer radius (R) = [5.00 + 0.25] cm = 5.25 cm

∴ Outer surface area of the bowl = 2πR²

Q9. A right circular cylinder just encloses a sphere of radius r (see the 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).

Sol: (i) For the sphere:

Radius = r

∴ Surface area of the sphere = 4 πr²

(ii) For the right circular cylinder:

∵ Radius of the cylinder = Radius of the sphere

∴ Radius of the cylinder = r

Height of the cylinder = Diameter of the sphere

⇒ Height of the cylinder (h) = 2r

Since, curved surface area of a cylinder = 2πrh = 2πr(2r) = 4πr²

EXERCISE 13.5 (Page 228)

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

Sol: 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

∴ Volume of matchbox = (l × b) × h

= [4 cm × 2.5 cm] × 1.5 cm³

= 15 cm³

⇒ Volume of 12 boxes = 12 × 15 cm³ = 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?

(l m³ = 1000 l)

Sol: Here, Length (l) = 6 m

Breadth (b) = 5 m

Depth (h) = 4.5 m

Capacity = l × b × h = 6 × 5 × 4.5 m³

∵1 m³ can hold 1000 l.

∴135 m³ can hold (135 × 1000 l = 135000 l) of water.

∴ The required amount of water in the tank = 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 liquid?

Sol: Length (l) = 10 m

Breadth (b) = 8 m

Volume (v) = 380 m³

Let height of the cuboid be 'h'.

Since, volume of a cuboid = 1 × b × h

∴ Volume of the cuboidal vessel = 10 × 8 × h m³ = 80h m³

⇒ 80h = 380

Thus, the required height of the liquid = 4.75 m

Q4. 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 m³.

Sol: Length (l) = 8 m

Breadth (b) = 6 m

Depth (h) = 3 m

∴ Volume of the cuboidal pit = l × b × h = 8 × 6 × 3 m³ = 144 m³

Since, rate of digging the pit is Rs. 30 per m³.

Cost of digging = Rs. 30 × 144 = Rs. 4320

Q5. 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.

Sol: Length of the tank (l) = 2.5 m

Depth of the tank (h) = 10 m

Let breadth of the tank = b m

∴ Volume (capacity) of the tank= l × b × h = 2.5 × b × 10 m³

But the capacity of the tank = 50000 l = 50 m³

∴ 25b = 50 m³

Thus, the depth of the tank = 2 m

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?

Sol: 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 m³ = 1800 m³

Since 1 m³ = 1000 l

∴ Capacity of the tank = 1800 × 1000 l = 1800000 l

Village population = 4000

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

∴ Amount of water is required per day = 150 × 4000 l.

Let the required number of days = x

∴ 4000 × 150 × x = 1800000

Thus, the required number of days is 3.

Q7. 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.

Sol: Volume of the godown = 60 × 25 × 10 m³

Volume of a crate = 1.5 × 1.25 × 0.5 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.

Sol: Side of the given cube = 12 cm

∴ Volume of the given cube = (side)3 = (12)3 cm³

Side of the smaller cube:

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

⇒ Volume of smaller cube = n³

⇒ Volume of 8 smaller cubes = 8n³

∴ 8n³ = (12)³ = 12 × 12 × 12

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 × 122 cm² = 6 × 12 × 12 cm²

Surface area of one smaller cube = 6 × (side)2 = 6 × 62 cm²

= 6 × 6 × 6 cm²

∴ Surface area of 8 smaller cubes = 8 × 6 × 6 × 6 cm²

Thus, the required ratio = 1 : 4

Q9. 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?

Sol: 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

∴ Water volume (volume of the cuboid so formed)

= l × b × h = 2000 × 40 × 3 m³

Now, volume of water fallowing in 1 hr (= 60 minutes)

= 2000 × 40 × 3 m³

Volume of water that will fall in 1 minute

= [2000 × 40 × 3] ÷ 60 m³

EXERCISE 13.6 (Page 230)

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)

Sol: Let the base radius of the cylindrical vessel be 'r' cm.

∴ Circumference = 2πr

⇒ 2πr = 132 [∵Circumference = 132 cm]

∵ Capacity of the vessel = Volume of the vessel

∴ Capacity of cylindrical vessel = 34650 cm³

Since 1000 cm³ = 1 litre

Thus, the vessel can hold 34.65 l of water.

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 l cm³ of wood has a mass of 0.6g.

Sol: Here,

Inner diameter of the cylindrical pipe = 24 cm

Length of the pipe (h) = 35 cm

∵ Inner volume of the pipe = πr²h

Outer volume of the pipe = πR²h

∴ Amount of wood (volume) in the pipe = Outer volume – Inner volume

= πR²h – πr²h

= πh(R² – r)

= πh(R + r)(R – r) [∵ a² – b² = (a + b)(a – b)]

Mass of the wood in the pipe

= [Mass of wood in 1 m³ of wood] × [Volume of wood in the pipe]

= [0.6g] × [22 × 5 × 26 × 2] cm³

= 6 × 22 × 10 × 26g

= 6 × 22 × 26 g = 3432 g

Thus, the required mass of the pipe is 3.432 kg.

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?

Sol: For rectangular pack:

Length (l) = 5 cm

Breadth (b) = 4 cm

Height (h) = 15 cm

∴ Volume = l × b × h = 5 × 4 × 15 cm³ = 300 cm³

⇒ Capacity of the rectangular pack = 300 cm³ ...(1)

For cylindrical pack:

Base diameter = 7 cm

Q4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find: (1) radius of its base (ii) its volume. (Use π = 3.14)

Sol: Height of the cylinder (h) = 5 cm

Let the base radius of the cylinder be ‘r’.

(i) Since lateral surface of the cylinder = 2 πrh

But lateral surface of the cylinder = 94.2 cm²

⇒ 2πrh = 94.2

Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = πr²h

Volume of the given cylinder = 3.14 × (3)2 × 5 cm³

Thus, the required volume = 141.3 cm³

Q5. It costs Rs. 2200 to paint the inner curved surface of cylindrical vessel 10 m the cost of painting is at the rate of Rs. 20 per m²; find:

(i) inner curved surface of the vessel (ii) radius of the base

(iii) capacity of the vessel.

Sol: (i) To find inner curved surface

Total cost of painting = Rs. 2200

Rate of painting = Rs. 20 per m²

Inner curved surface of the vessel = 110 m²

(ii) To find radius of the base

Let the base radius of the cylindrical vessel.

∵ Curved surface of a cylinder = 2 πrh

∴ 2πrh = 110

The required radius of the base = 1.75 m

(iii) To find the capacity of the vessel

Since, volume of a cylinder = πr²h

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

Sol: Capacity of the cylindrical vessel = 15.4 l

= 15.4 × 1000 cm³

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 graphite is I mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Thus, the required volume of the graphite = 0.11 cm³

For the pencil

Volume of the wood

Volume of the wood = [Volume of the pencil] – [Volume of the graphite]

= 5.39 cm³ – 0.11 cm³

= 5.28 cm³

Thus, the required volume of the wood is 5.28 cm³.

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?

Sol: The bowl is cylindrical.

Diameter of the base = 7 cm

Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.

EXERCISE 13.7 (Page 233)

Q1. Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm.

Sol: (i) Here, radius of the cone r = 6 cm

Height (h) = 7 cm

Q2. 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

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

Q3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base.

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

Volume of the cone (v) = 1570 cm³

Let the radius of the base be 'r' cm.

Q4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Sol: Volume of the cone = 48 π cm³

Height of the cone (h) = 9 cm

Let 'r' be its base radius.

∵Diameter = 2 × Radius

∴Diameter of the base of the cone = 2 × 4 = 8 cm

Q5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Sol: Here, diameter of the conical pit = 3.5 m

= 38.5 m³

∴ 1000 cm³ = 1 l and 1000000 cm³ = l m³

∴ 1000 × 1000 cm³ = 1000 l = 1 kl

Also 1000 × 1000 cm³ = 1 m³

⇒¹ m³ = 1 kl

⇒³8.5 m³ = 38.5 kl

Thus, the capacityof the conical pit is 38.5 kl.

Q6. 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.

Sol: Volume of the cone (v) = 9856 cm³

Diameter of the base = 28 cm

Thus, the required height is 48 cm.

(ii) To find the slant height

Let the slant height be 'l' cm.

∵ (Slant height)² = (Radius)² + (Height)²

∴ l² = 14² + 48² = 196 + 2304 = 2500 = (50)²

⇒ l = 50

Thus, the required height = 50 cm.

(iii) To find the curved surface area

∵The curved surface area of a cone is given by nrl

Thus, the curved surface area of the cone is 2200 cm².

Q7. 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.

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

∵The right angled triangle is revolved about the 12 cm side.

∴ 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

= π × 100 cm³

= 100π cm³

Thus, the required volume of the cone is 100π cm³.

Q8. 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.

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

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

Radius of the cone (r) = 12 cm

Q9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

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

Base diameter = 10.5 m

Thus, the required volume = 86.625 m³

Area of the canvas

∵The area of the canvas to cover the heap must be equal to the curved surface area of the conical heap.

Thus, the required area of the canvas is 99.825 m².

EXERCISE 13.8 (Page 236)

Q1. Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

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

Q2. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

Sol: (i) Diameter of the ball = 28 cm

Q3. 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?

Sol: Diameter of the metallic ball = 4.2 cm

Q4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Sol: Let diameter of the earth be 2r.

Q5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Sol: Diameter of the hemisphere = 10.5 cm

Q6. A hemispherical tank is made up of an iron sheet 1 then find the volume of the iron used to make the tank.

Sol: Inner radius (r) = 1m

Q7. Find the volume of a sphere whose surface area is 154 cm²

Sol: Let 'r' be the radius of the sphere.

Its surface area = 4πr²

Q8. 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.

Sol: Total cost of white washing = Rs. 498.96

Rate of white washing = Rs. 2 per m

(i) To find the inside surface area of the dome:

∵Radius of the hemisphere (r) = 6.3 m

Surface area of a hemisphere = 2πr² m²

Thus, the required surface area of the dome = 249.48 m²

(ii) To find the volume of air in the dome:

Thus, the required volume of air inside the dome is 523.9 m³ (approx.).

Q9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r' of the sphere, (ii) ratio of S and S’.

Sol: (i) To find r':

∵Radius of a small sphere = r

Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mmj) is needed to fill this capsule?

Sol: Diameter of the spherical capsule = 3.5 mm

Thus, the required quantity of medicine = 22.46 mm³ (approx).

EXERCISE 13.9 (OPTIONAL)

Q1. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the figure). The thickness of the plank is 5 cm every-where. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 poise per cm² and the rate of painting is 10 poise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.

Sol: Here, Height = 110 cm

Depth = 25 cm

Breadth = 85 cm

Thickness of the plank = 5 cm

Cost of polishing

Area to be polished = [(110 × 85)] + [(85 × 25) × 2] + [(110 × 25) × 2] +[(5 × 110) × 2] + [(75 × 5) × 4] cm²

= [9350] + [4250] + [5500] + [1100] + [1500] cm² = 21700 cm²

Cost of polishing at the rate 20 poise per cm²

Cost of painting

∵Area to be painted = [(75 × 20) × 6] + [(90 × 20) × 2] + [90 × 75] cm²

= [9000 + 3600 + 6750] cm²

= 19350 cm²

∴Cost of painting at the rate of 10 paise per cm²

Total Expenses

Total expenses = (Cost of polishing) + (Cost of painting)

= Rs. 4340 + Rs.1935

= Rs. 6275

Thus, the total required expense = Rs. 6275.

Q2. 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' and black paint costs 5 paise per cm².

Sol: For spherical part

Surface area of a sphere = 4πr²