Important Questions for CBSE Class 9 Maths Chapter 13 – Surface Areas and Volumes -CoolGyan

CBSE Class 9 Maths Chapter-13 Important Questions - Free PDF Download

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1 Marks Quetions

1. If the perimeter of one of the faces of a cube is 40 cm, them its volume is

(a) 6000 cu cm

(b) 1600 cu cm

(c) 1000 cu cm

(d) 600 cu cm

Ans. (c) 1000 cu cm

2. A cuboid having surface areas of 3 adjacent faces as a, b and c has the volume

(a)

(b)

(c) abc

(d)

Ans. (b)

3. The diameter of a right circular cylinder is 21 cm and its height is 8 cm. The Volume of the cylinder is

(a) 528 cu cm

(b) 1056 cu cm

(c) 1386 cu cm

(d) 2772 cu cm

Ans. (d) 2772 cu cm

4. Each edge of a cube is increased by 40%. The % increase in the surface area is.

(a) 40

(b) 96

(c) 160

(d) 240

Ans. (b) 96

5. Find the curved (lateral) surface area of each of the following right circular cylinders:

(a)

(b)

(c)

(d)None of these

Ans. (a)

6. The radius and height of a right circular cylinder are each increased by 20%. The volume of cylinder is increased by-

(a) 20%

(b) 40%

(c) 54%

(d) 72.8%

Ans. (d) 72.8%

7. A well of diameter 8 meters has been dug to the depth of 21 m. the volume of the earth dug out is

(a) 1056 cu m

(b) 352 cu m

(c) 1408 cu m

(d) 4224 cu m

Ans. (a) 1056 cu m

8. The radius of a cylinder is doubled and the height remains the same. The ratio between the volumes of the new cylinder and the original cylinder is

(a) 1:2

(b) 1:3

(c) 1:4

(d) 1:8

Ans. (c) 1:4

9. Length of diagonals of a cube of side a cm is

(i)

(ii)

(iii)

(iv) 1

Ans. (ii)

10. Surface area of sphere of diameter 14 cm is

(i)

(ii) 516

(iii) 400

(iv) 2244

Ans. (i)

11. Surface area of bowl of radius r cm is

(i)

(ii) 2

(iii) 3

(iv)

Ans. (iii) 3

12. Volume of a sphere whose radius 7 cm is

(i)

(ii)

(iii)

(iv)

Ans. (i)

13. The curved surface area of a right circular cylinder of height 14 cm is 88 find the diameter of the base of the cylinder

(i) 1 cm

(ii) 2 cm

(iii) 3 cm

(iv) 4 cm

Ans. (ii) 2 cm

14. Volume of spherical shell

(i)

(ii)

(iii)

(iv) none of these

Ans. (iii)

15. The area of the three adjacent faces of a cuboid are x,y,z. Its volume is V, then

(i)

(ii)

(iii)

(iv) none of these

Ans. (ii)

16. A conical tent is 10 m high and the radius of its base is 24 m then slant height of the tent is

(i) 26

(ii) 27

(iii) 28

(iv) 29

Ans. (i) 26

17. Volume of hollow cylinder

(i)

(ii)

(iii)

(iv)

Ans. (i)

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

(i) 155

(ii) 165

(iii) 150

(iv) none of these

Ans. (ii)

19. The surface area of a sphere of radius 5.6 cm is

(i) 96.8

(ii) 94.08

(iii) 90.08

(iv) none of these

Ans. (ii)

20. The height and the slant height of a cone are 21 cm and 28 cm respectively then volume of cone

(i) 7556

(ii) 7646

(iii) 7546

(iv) none of these

Ans. (c)

2 Marks Quetions

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 m² cost Rs. 20.

Ans. (i) Given: Length = 1.5 m, Breadth = 1.25 m and Depth = 65 cm = 0.65 m

Area of the sheet required for making the box open at the top =

= 3.575 + 1.875

(ii) Since, Cost of sheet = Rs. 20

Cost of sheet = Rs. 109

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

Ans. Given: Length = 5 m, Breadth = 4 m and Height = 3 m

Area of the four walls = Lateral surface area = =

Area of ceiling =

Total area of walls and ceiling of the room = 54 + 20

Now Cost of white washing for = Rs. 7.50

Cost of white washing for = Rs. 555

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

Ans. Given: Perimeter of rectangular wall = = 250 m ……….(i)

Now Area of the four walls of the room

= = 1500 ……….(ii)

Area of the four walls = Lateral surface area = = = 1500

[using eq. (i) and (ii)

= 6 m

Hence required height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to . How many bricks of dimensions can be painted out of this container?

Ans. Given: Length of the brick = 22.5 cm, Breadth = 10 cm and Height = 7.5 m

Surface area of the brick =

= 2 (225 + 75 + 468.75)

[1 cm = 0.01 m]

Now No. of bricks to be painted

= = 100

Hence 100 bricks can be painted.

5. A cubical box has each edge 10 cm and a cuboidal box is 10 cm wide, 12.5 cm long 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 how much?

Ans. (i) Lateral surface area of a cube

Lateral surface area of a cuboid =

Lateral surface area of cubical box is greater by (400 – 360)

(ii) Total surface area of a cube

Total surface area of cuboid =

= 2 (125 + 80 + 100)

Total surface area of cuboid box is greater by (610 – 600)

6. 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 simensions ?

Ans. Given: Length of base = 4 m, Breadth = 3 m and Height = 2.5 m

Tarpaulin required to make shelter = Surface area of 4 walls + Area of roof

= 35 + 12

Hence of the tarpaulin is required to make the shelter for the car.

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

Ans. Given: Height of cylinder = 14 cm, Curved Surface Area = 88 cm²

Let radius of base of right circular cylinder = cm

= 88

1 cm

Diameter of the base of the cylinder = = 2 cm

8. 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 meters of the sheet are required for the same?

Ans. Given: Diameter = 140 cm

Radius = 70 cm = 0.7 m

Height of the cylinder = 1 m

Total Surface Area of the cylinder = =

Hence metal sheet is required to make the close cylindrical tank.

9. 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.

Ans. Diameter of roller = 84 cm

Radius of the roller = 42 cm

Length (Height) of the roller = 120 cm

Curved surface area of the roller =

= =

Now area leveled by roller in one revolution =

Area leveled by roller in 500 revolutions = 1584.0000

10. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of white washing the curved surface of the pillar at the rate of Rs. 12.50 per .

Ans. Diameter of pillar = 50 cm

Radius of pillar = 25 cm = m

Height of the pillar = 3.5 m

Now, Curved surface area of the pillar = = =

Cost of white washing = Rs. 12.50

Cost of white washing = = Rs. 68.75

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

Ans. Curved surface area of the cylinder, Radius of cylinder = 0.7 m

Let height of the cylinder =

= 4.4

= 1 m

12. 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.

Ans. Inner diameter of circular well = 3.5 m

Inner radius of circular well = = 1.75 m

And Depth of the well = 10 m

(i) Inner surface area of the well =

(ii) Cost of plastering = Rs. 40

Cost of plastering = = Rs. 4400

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

Ans. The length (height) of the cylindrical pipe = 28 m

Diameter = 5 cm

Radius = cm

Curved surface area of the pipe = =

14. In the adjoining 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. [See fig.]

Ans. Height of each of the folding at the top and bottom = 2.5 cm

Height of the frame (H) = 30 cm

Diameter = 20 cm

Radius = 10 cm

Now cloth required for covering the lampshade

= CSA of top part + CSA of middle part + CSA of bottom part

15. The students of a Vidyalayawere 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?

Ans. Radius of a cylindrical pen holder = 3 cm

Height of the cylindrical pen holder = 10.5 cm

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

= =

Since Cardboard required for making 1 pen holder

Cardboard required for making 35 pen holders

(approx.)

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

Ans. Diameter = 10.5 cm

Radius = = cm

Slant height of cone = 10 cm

Curved surface area of cone = =

Total surface area of cone = =

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

Ans. Slant height of cone = 21 m

Diameter of cone = 24 m

Radius of cone = = 12 m

Total surface area of cone =

18. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per.

Ans. Slant height of conical tomb = 25 m, Diameter of tomb = 14 m

Radius of the tomb = 7 m

Curved surface are of tomb = =

Cost of white washing = Rs. 210

Cost of white washing =

Cost of white washing = = Rs. 1155

19. 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.

Ans. Radius of cap = 7 cm, Height of cap = 24 cm

Slant height of the cone =

= = = 25 cm

Area of sheet required to make a cap = CSA of cone =

Area of sheet required to make 10 caps

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

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

Ans. (i) Radius of sphere = 105 cm

Surface area of sphere = =

(ii) Radius of sphere = 5.6 m

Surface area of sphere = =

(iii) Radius of sphere = 14 cm

Surface area of sphere = =

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

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

Ans. (i) Diameter of sphere = 14 cm, Therefore Radius of sphere = = 7 cm

Surface area of sphere = =

(ii) Diameter of sphere = 21 cm

Radius of sphere = cm

Surface area of sphere = =

(iii) Diameter of sphere = 3.5 cm

Radius of sphere = = 1.75 cm

Surface area of sphere = =

22. Find the total surface area of a hemisphere of radius 10 cm.

Ans. Radius of hemisphere 10 cm

Total surface area of hemisphere =

Hence total surface area of hemisphere is.

23. Find the radius of a sphere whose surface area is.

Ans. Surface area of sphere =

= 154

= 3.5 cm

24. 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.

Ans. Inner radius of bowl = 5 cm

Thickness of steel = 0.25 cm

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

Outer curved surface area of bowl = =

25. A right circular cylinder just encloses a sphere of radius (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).

Ans. (i) Radius of sphere =

Surface area of sphere =

(ii) The cylinder just encloses the sphere in it.

The height of cylinder will be equal to diameter of sphere.

And The radius of cylinder will be equal to radius of sphere.

Curved surface area of cylinder = =

(iii)

Required ratio = 1: 1

26. A matchbox . What will be the volume a packet containing 12 such boxes?

Ans. Given: Length = 4 cm, Breadth = 2.5 cm, Height = 1.5 cm

Volume of a matchbox =

Volume of a packet containing 12 such

27. A cubical water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (= 1000 )

Ans. Volume of water in cuboidal tank

= 135000 liters

Hence tank can hold 135000 liters of water.

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

Ans. Let height of cuboidal vessel = m

Volume of liquid in cuboidal vessel

= 380

= = 4.75 m

Hence cuboidal vessel is 4.75 m high.

29. 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.

Ans. Volume of cuboidal pit

= 144

Cost of digging cuboidal pit = Rs. 30

Cost of digging cuboidal pit

= Rs. 4320

30. 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. (= 1000 )

Ans. Capacity of cuboidal tank = 50000 liters

= 50000 liters

= 2 m

Hence breadth of cuboidal tank is 2 m.

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

Ans. Since water flows at the rate of 2 km per hour, the water from 2 km of the river flows into the sea in one hour.

Therefore, the volume of water flowing into the sea in one hour = Volume of a cuboid

[1 km = 1000 m]

Now, Volume of water flowing into sea in 1 hour (in 60 minutes)

Volume of water flowing into sea in 1 minute =

32. Find the length of a wooden plank of width 2.5 m, thickness 0.025 m and volume 0.25 m³.

Ans. Given: Volume of wooden plank

= 0.25

= 4 m

Hence required length of wooden plank is 4 m.

33. If the lateral surface of a cylinder is and its height is 5 cm, then (i) radius of its base (ii) volume of the cylinder.

Ans. Height of the cylinder = 5 cm

Lateral surface area of the cylinder

= 94.2

= 3 cm

Volume of cylinder = =

34. A bag of grain contains of grain. How many bags are needed to fill a drum of radius 4.2 m and height 5 m?

Ans. Radius of drum = 4.2 m and Height of drum = 5 m

Volume of a drum = =

Now, Number of bags =

= 99

Hence 99 bags are needed to fill the drum.

35. 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 diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the columns of the wood and that of the graphite.

Ans. Diameter of graphite = 1 mm

Radius of drum = 0.5 mm = 0.05 cm

Height of graphite = 14 cm

Volume of graphite = = =

Diameter of pencil = 7 mm

Radius of pencil (R) = 3.5 mm = 0.35 cm

Volume of pencil = = =

Now, Volume of wood = Volume of pencil – Volume of graphite

= 5.39 – 0.11 =

3 Marks Quetions

1. A small indoor green house (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 surface area of the glass?

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

Ans. (i) Given: Length of glass herbarium = 30 cm,

Breadth = 25 cm and Height = 25 m

Total surface area of the glass =

= 2 (750 + 625 + 750)

Hence 4250 cm² of the glass is required to make a herbarium.

(ii) Tape is used at 12 edges.

Tape is used at 4 lengths, 4 breadths and 4 heights.

Total length of the tape = = 2 (30 + 25 + 25) = 320 cm

Hence 320 cm of the tape if needed to fix 12 edges of herbarium.

2. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. [See fig.]. Find its:

(i) Inner curved surface area

(ii) Outer curved surface area

(iii) Total surface area

Ans. (i) Length of the pipe = 77 cm, Inner diameter of cross-section = 4 cm

Inner radius of cross-section = 2 cm

Inner curved surface area of pipe = =

(ii) Length of pipe = 77 cm, Outer diameter of pipe = 4.4 cm

Outer radius of the pipe = 2.2 cm

Outer surface area of the pipe = =

(iii) Now there are two circles of radii 2 cm and 2.2 cm at both the ends of the pipe.

Area of two edges of the pipe = 2 (Area of outer circle – area of inner circle)

= =

Total surface area of pipe

= Inner curved surface area + Outer curved surface area + Area of two edges

= 968 + 1064.8 + 5.28 =

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

Ans. (i) Slant height of cone = 14 cm

Curved surface area of cone =

= 308

= 7 cm

(ii) Total surface area of the cone = Curved surface area + Area of circular base

= 462 cm²

4. A conical tent is 10 m high and the radius of its base is 24 m. Find:

(i) slant height of the tent.

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

Ans. Height of the conical tent = 10 m

Radius of the conical tent = 24 m

(i) Slant height of the tent

= = 26 m

(ii) Canvas required to make the tent = Curved surface area of the tent

= = =

Cost of canvas = Rs. 70

Cost of canvas = = Rs. 137280

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.

Ans. Height of the conical tent = 8 m and Radius of the conical tent = 6 m

Slant height of the tent

= = 10 m

Area of tarpaulin = Curved surface area of tent =

Width of tarpaulin = 3 m

Let Length of tarpaulin = L

Area of tarpaulin = = 3L

Now According to question, 3L = 188.4

L = = 62.8 m

The extra length of the material required for stitching margins and cutting is 20 cm = 0.2 m.

So the total length of tarpaulin bought is (62.8 + 0.2) m = 63 m

6. 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, what will be the cost of painting all these cones?

Ans. Diameter of cone = 40 cm

Radius of cone

= 20 cm

= m

= 0.2 m

Height of cone = 1 m

Slant height of cone

= = m

Curved surface area of cone = =

Cost of painting of a cone = Rs. 12

Cost of painting of a cone = = Rs. 7.68672

Cost of painting of 50 such cones = = Rs. 384.34 (approx.)

7. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans. I case: Radius of balloon = 7 cm

Surface area of balloon = = ……….(i)

II case: Radius of balloon (R) = 14 cm

Surface area of balloon = = ……….(ii)

Now, Ratio [from eq. (i) and (ii)],

= =

Hence, required ratio = 1: 4

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

Ans. Capacity of cuboidal tank = = = =

= 1800000 liters

Water required by her head per day = 150 liters

Water required by 4000 persons per day = = 600000 liters

Number of days the water will last =

= = 3

Hence water of the given tank will last for 3 days.

9. A godown measures . Find the maximum number of wooden crates each measuring that can be stored in the godown.

Ans. Capacity of cuboidal godown = =

Capacity of wooden crate = =

Maximum number of crates that can be stored in the godown =

= = 16000

Hence maximum 16000 crates can be stored in the godown.

10. Find the minimum number of bricks each measuring required to construct a wall 10 m long, 6 m high and 1.5 m thick.

Ans. Volume of one cuboidal brick =

Volume of cuboidal wall

Minimum number of bricks required =

= =

= = 46376.81

= 46377 [Since bricks cannot be in fraction]

11. 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 )

Ans. Height of vessel = = 25 cm

Circumference of base of vessel = 132 cm

= 132

= 21 cm

Now, Volume of cylindrical vessel = = =

= liters [= 1 liter]

= 34.65 liters

12. The inner diameter of a cylindrical wooden pipe is 24 cm and its out diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if of wood has a mass of 0.5 g.

Ans. Inner diameter of pipe = 28 cm

Inner radius of pipe = = 12 cm

And Outer diameter of pipe = 28 cm

Outer radius of pipe (R) = = 14 m

Length of pipe = 35 cm

Volume of wood = Volume of outer cylinder – Volume of inner cylinder

= =

Weight of of wood = 0.6 g

Weight of of wood =

= 3432 g = 3.432 kg

13. 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 height of 15 cm (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and how much?

Ans. I case: Length of tin = 5 cm, Width of tin = 4 cm

and Height of tin = 15 cm

Then, Capacity of tin =

II case: Diameter of base of cylinder = 7 cm

Radius of base of cylinder = cm

Height of cylinder = 10 cm

Capacity of cylinder = =

From the cases I and II, we observed that cylindrical container has greater capacity by (385 – 300).

14. 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 m², find:

(i) inner curved surface area of the vessel.

(ii) radius of the base.

(iii) capacity of the vessel.

Ans. Total cost to paint inner curved surface area of the vessel = Rs. 2200

Rate = Rs. 20 per square meter

(i) Inner curved surface area of vessel = = =

(ii) Depth of the vessel = 10 m

Now, Inner surface area of vessel =

= 110

= 1.75 m

(iii) Since = 1.75 m and = 10 m

Capacity of vessel = Volume of cylinder =

= =

= 96.25 kl [ 1 m³ = 1 kl]

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

Ans. Height of the vessel = 1 m

Capacity of vessel = 15.4 liters

= kilo liters

= 0.0154 [ 1 = 1 kl]

= 0.0154

= 0.0048

= 0.07 m

Now, Area of metal sheet required = TSA of cylindrical vessel

16. Find the capacity of a conical vessel with:

(i) Radius 7 cm, Slant height 25 cm

(ii) Height 12 cm, Slant height 13 cm

Ans. (i) Given: = 7 cm, = 25 cm

= = 24 cm

Capacity of conical vessel =

= =

= 1.232 liters [= 1 liter]

(ii) Given: = 12 cm, = 13 cm

= =

= = 5 cm

Capacity of conical vessel =

= =

= liters [= 1liter]

= liter

17. 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 volume of the two solids obtained.

Ans. When right angled triangle ABC is revolved about side 5 cm, then the solid formed is a cone.

In that cone, Height = 5 cm

And radius = 12 cm

Therefore, Volume of cone =

Now, = =

Required ratio = 5: 12

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

Ans. Let diameter of earth be

Radius of earth =

Now, Volume of earth = [ Earth is considered to be a sphere]

= = ………..(i)

According to question, Diameter of moon = Diameter of earth=

Radius of moon (R) =

Now, Volume of Moon = [ Moon is considered to be a sphere]

= =

= = Volume of Earth [From eq. (i)]

Volume of moon is the volume of earth.

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

Ans. Diameter of hemispherical bowl = 10.5 cm

Radius of hemispherical bowl = = 5.25 cm

Volume of milk in hemispherical bowl =

= = =

= liters

= 0.303187 liters = 0.303 liters (approx.)

20. Find the volume of a sphere whose surface area is.

Ans. Surface area of sphere =

= 154

= cm

Now, Volume of sphere = =

= = =

21. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm [See fig.]. The thickness of the planks 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 and the rate of painting is 10 paise per , find the total expenses required for polishing and painting the surface of the bookshelf

Ans. External faces to be polished

= Area of six faces of cuboidal bookshelf – 3 (Area of open portion ABCD)

[ AB = 85 – 5 – 5 = 75 cm and AD = = 30 cm]

= 2 (2750 + 2125 + 9350)

= 28450 – 6750

Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise.

= Rs. 0.20 per = Rs. = Rs. 4340

Here, three equal five sides inner faces.

Therefore, total surface area = [ Depth = 25 – 5 = 20 cm]

= 3 [ 4200 + 2250]

= 19350

Now, cost of painting inner faces at the rate of 10 paise i.e. Rs. 0.10 per .

= Rs. 1935

22. If diameter of a sphere is decreased by 25% then what percent does its curved surface area decrease?

Ans. Diameter of original sphere = D = 2R R =

Curved surface area of original sphere = = =

According to the question, Decreased diameter = 25% of D = =

Diameter of new sphere = =

Radius of new sphere =

Now, curved surface area of new sphere = = =

Change in curved surface area =

Percent change in the curved surface area =

= = = 43.75%

23. The surface area of cuboids is; its dimensions are in the ratio 4:3:2. Find the volume of the cuboid.

Ans. Let the dimensions of the cuboid be 4x, 3x and2x meters

Surface area of the cuboid = 2

Given surface area = 3328sq m

From (i) and (ii) we get

or x =8

and 2x = 16

Thus the dimensions of the cuboid are 32m, 24m and 16m

Volume of the cuboid=

= 12288cu m

24. The volume of a rectangular slower of stone is 10368 and is dimensions are in the ratio of 3:2:1. (i) Find the dimensions (ii) Find the cost of polishing its entire surface @ Rs. 2 per.

Ans. Let the length of the block be 3xdm

Width = and height = x dm

Volume of the block = 10368

=1728

=12

also 2x = 24 and 3x = 36

Thus dimensions of the block are 36dm, 24dm and 12dm

Surface area of the block =

= 2(864+288+432)

Cost of polishing the surface =

= Rs. 6336

25. In a cylindrical drum of radius 4.2 m and height 3.5 m, how many full bags of wheat can be emptied if the space required for each bag is 2.1 cu m.

Ans. Radius of the drum = 4.2m =

Height of the drum =3.5m =

Volume of the drum

….(i)

Volume of wheat in each bags = 2.1cu m =

=92

Hence the number of full bags is 92

26. The inner diameter of a cylindrical wooden tripe is 24 cm. and its outer diameter is 28 cm. the length of wooden tripe is 35 cm. find the mass of the tripe, if 1 cu cm of wood has a mass of 0.6 g.

Ans. Inside diameter of the pipe = 24cm

Outside diameter of the pipe = 28cm

Length of the pipe = 35cm = (h says)

Outside radius of the pipe =

Volume of the wood = External volume – Internal volume

cu cm

= 5720 cu cm

Mass of 1cu cm = 0.6g

Mass of the pipe =

= 3432g

= 3.432kg

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

Ans. Diameter of the bowl = 7 cm.

Radius of the bowl =

Height up to which soup is filled (h) = 4 cm.

Volume of the soup in one bowl =

=154 cu cm

soup given to one patient =154 cu cm.

Soup given to 250 patients

=38500cu cm.

[ltrs=1000cu cm]

=38.5 ltrs.

Hence the hospital has to prepare 38.5 litre daily to serve 250 patients.

28. The diameter of a roller is 84cm and its length is 120cm. It takes 500 complete revolutions to move once over to level a playground.

(a) Find the area of playground in sq m.

(b) Determine the cost of leveling the playground at the rate of Rs 1.75 per sq m.

Ans. (a) R= Radius of the roller = Area =42 cm. as – 0.42 m.

H= length of the roller = 120 cm. = 1.2 m.

Area covered in the revolution =

Area covered in 500 revolutions =5003.168 sq m.

= 1584 sq m.

thus area of playground = 1584 sq m.

(b) cost of leveling 1 sq m. of playground =Rs 1.75

Cost of total leveling = Rs (15841.75)

= Rs 2772

29. A metal cube of edge 12 cm is melted and formed into three similar cubes. If the edge of two smaller cubes is 6cm and 8cm, find the edge of the third smaller cube (Assuming that there is no loss of metal during melting).

Ans. Volume of cube with edge 12cm

=1728cu cm. …..(i)

Volume of the first smaller cube with edge 6cm

=216cu cm. …..(ii)

Volume of the second smaller cube with edge 8cm.

=512cu cm. …..(iii)

Let the edge of the third smaller cube be a cm.

Volume of the third smaller cube =

=1728 [using (i) and (ii)]

By the given condition.

Area

Thus the edge of the third required cube is 10 cm.

30. How many bricks, each measuring 18cm by 12cm by 10cm will be required to build a wall 15m long 6dm wide and 6.5m high when of its volumes occupied by mastar? Please find the cost of the bricks to the nearest rupees, at Rs 1100 per 1000 bricks.

Ans. Length of the wall =15 m. = 1500 cm.

Width of the wall = 6 dm. =60 cm.

Height of the wall = 6.5 m. = 650 cm.

Volume of the wall = (150060650) cu cm.

= 58500000 cu cm. (I)

Volume occupied by mastar = ( ) cu cm.

= 5850000 cu cm.

Volume occupied by bricks = (i) - (ii)

= (58500000 – 5850000) cu cm.

= 52650000 cu cm. (iii)

Volume of a brick = (181210) cu cm.

= 2160 cu cm. (iv)

No of brick required = (iii) (iv)

= 24375

cost of 1000 bricks = Rs 1100

Total cost = Rs

= Rs 26812.50

= Rs 26813.

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

Ans. Depth of river = 3m

Water of the river = 40m

Rate of flow of water = 2km/hr = 2000m/hr

Volume of water flowing in one hour

Hence Volume of water flowing in one minute =

32. If the lateral surface of a cylinder is and its height is 5 cm. then find

(i) radius of its base (ii) its volume

Ans. Given lateral surface of cylinder =

H = 5cm

R=3cm

(ii) Volume of cylinder =

33. A shot put is a metallic sphere of radius 4.9 cm If the density of the metal is 7.8 g per Find the mass of the shot put.

Ans. Volume of sphere = and radius r =4.9 cm

Mass of of metal is 7.8g

Mass of the shot put = volume density

34. The capacity of a hemispherical tank is 155.232. Find its radius.

Ans. Capacity of tank = Its Volume =

Hence radius of tank = 42cm

35. What length of tarpaulin 3 m wide will 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

Ans. Here h=8m and r=6m

Curved surface area =

Length of tarpaulin required

Extra length required for wastage = 20cm=0.2m

Hence, total length required = 62.8+0.2

= 63m

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

Ans. Given radius of capsule =

Amount of medicine = Volume of capsule

37. A wall of length 10 m was to be built across an open ground. The height of wall is 4 m and thickness of the wall is 34 cm. If this wall is to be built up with bricks whose dimensions are How many bricks would be required

Ans. Length of wall =10m =1000cm

Thickness = 24cm

Height = 4m = 400cm

Volume of wall = length thickness height

Now each brick is a cuboid with length = 24cm

Breadth = 12cm and height = 8cm

Volume of each brick = = 24128

Number of bricks required =

The wall requires 4167 bricks.

38. The pillars of a temple are cylindrically shaped if each pillar has a circular base of radius 20cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?

Ans. Radius of base of cylinder = 20cm

Height of pillar = 10m = 1000cm

Volume of each cylinder =

Volume of 14 pillars = volume of each cylinder 14

So 14 pillars would need of concrete mixture.

39. A right triangle ABC with sides 5 cm, 12cm, and 13 cm is revolved about the side 12 cm, find the volume of the solid so obtained

Ans. The solid obtained by revolving the given right triangle is a right circular cone with radius = 5cm.

And height = 12cm

Volume of solid

40. The inner diameter of a circular well is 3.5 cm. 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

Ans. Given Inner diameter of well =3.5m

Inner radius

and depth

(i) Inner surface area

(ii) The cost of plastering is Rs 40 per

Cost of plastering this surface area

41. A Godown measures Find the maximum number of wooden crates each measuring that can be stored in the go down

Ans. Dimensions of Go down

Volume of Go down

volume of wooden carts

No. of wooden crates

Hence, 800 wooden crates are required.

42. The volume of a right circular cylinder is and radius of its base is 8 cm. Find the total surface area of the cylinder.

Ans. Volume of cylinder

Volume of cylinder

H=9cm

Total surface area

=854.989 cm

4 Marks Quetions

1. 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 by 20 cm by 5 cm and the smaller of dimensions 15 cm by 12 cm by 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of the card board is Rs. 4 for, find the cost of cardboard required for supplying 250 boxes of each kind.

Ans. Given: Length of bigger cardboard box (L) = 25 cm

Breadth (B) = 20 cm and Height (H) = 5 cm

Total surface area of bigger cardboard box

= 2 (LB + BH + HL)

= 2 (500 + 100 + 125)

5% extra surface of total surface area is required for all the overlaps.

5% of 1450 =

Now, total surface area of bigger cardboard box with extra overlaps

= 1450 + 72.5

Total surface area with extra overlaps of 250 such boxes

Since, Cost of the cardboard for 1000 cm² = Rs. 4

Cost of the cardboard for 1cm² = Rs.

Cost of the cardboard for 380625 cm² = Rs. = Rs. 1522.50

Now length of the smaller box = 15 cm,

Breadth = 12 cm and Height = 5 cm

Total surface area of the smaller cardboard box

= 2 (180 + 60 +75)

5% of extra surface of total surface area is required for all the overlaps.
5% of 630 = =
Total surface area with extra overlaps = 630 + 31.5 =

Now Total surface area with extra overlaps of 250 such smaller boxes

Cost of the cardboard for = Rs. 4

Cost of the cardboard for = Rs.

Cost of the cardboard for = Rs. = Rs. 661.50

Total cost of the cardboard required for supplying 250 boxes of each kind

= Total cost of bigger boxes + Total cost of smaller boxes

= Rs. 1522.50 + Rs. 661.50

= Rs. 2184

2. Find:

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

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

Ans. (i) Diameter of cylindrical petrol tank = 4.2 m

Radius of the cylindrical petrol tank = = 2.1 m

And Height of the tank = 4.5 m

Curved surface area of the cylindrical tank = = =

(ii) Let the actual area of steel used be meters

Since of the actual steel used was wasted, the area of steel which has gone into the tank.

= =

= 59.4

Hence steel actually used is .

3. 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.

Ans. Inner diameter of bowl = 10.5 cm

Inner radius of bowl = 5.25 cm

Now, Inner surface area of bowl =

= =

Cost of tin-plating per = Rs. 16

Cost of tin-plating per =

Cost of tin-plating per = = Rs. 27.72

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

Ans. Let diameter of Earth =
Radius of Earth =
Surface area of Earth = =

Now, Diameter of Moon = of diameter of Earth =

Radius of Moon =

Surface area of Moon = = =

Now, Ratio = = = =

Required ratio = 1: 16

5. 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.

Ans. Volume of solid cube = = =
According to question, Volume of each new cube = (Volume of original cube)

= =

Side of new cube = = 6 cm

Now, Surface area of original solid cube =

Now according to the question,

= =

Hence required ration between surface area of original cube to that of new cube = 4: 1.

6. The volume of a right circular cone is. If the diameter of the base if 28 cm, find:

(i) Height of the cone

(ii) Slant height of the cone

(iii) Curved surface area of the cone.

Ans. (i) Diameter of cone = 28 cm

Radius of cone = 14 cm

Volume of cone =

= 9856

= 48 cm

(ii) Slant height of cone =

= =

= = 50 cm

(iii) Curved surface area of cone =

= =

7. 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 he cost of paint required if silver paint costs 25 paise per and black paint costs 5 paise per .

Ans. Diameter of a wooden sphere = 21 cm.

Radius of wooden sphere = cm

And Radius of the cylinder = 1.5 cm

Surface area of silver painted part = Surface area of sphere - Upper part of cylinder for support

= =

Surface area of such type of 8 spherical part

Cost of silver paint over = Rs. 0.25

Cost of silver paint over = = Rs. 2757.85

Now, curved surface area of a cylindrical support =

= =

Curved surface area of 8 such cylindrical supports

Cost of black paint over of cylindrical support = Rs. 0.50

Cost of black paint over of cylindrical support

= Rs. 26.40

Total cost of paint required = Rs. 2757.85 + Rs. 26.4

= Rs. 2784.25

8. The difference between outside and inside surface of a cylindrical metallic tripe 14 cm. long is 44 sq cm. if the tripe is made of 99 cu cm. of metal, find the outer and inner radius of the tripe.

Ans. Let r₁cm and r₂cm can be the inner and outer radii respectively of the pipe

Area of the outside surface =