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### Transcript of ML Aggarwal Solutions for Class 10 Maths Chapter 17 ... ... Chapter 17: Mensuration Exercise 17.1...

• ML Aggarwal Solutions Class 10 Maths

Chapter 17: Mensuration

Exercise 17.1

Take π = 22/7 unless stated otherwise.

1. Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in

terms of π.

Solution:

Given radius of the cylinder, r = 5 cm

Height of the cylinder, h = 10 cm

Total surface area = 2r(r+h)

= 2×5(5+10)

= 2×5×15

= 150 cm2

Hence the total surface area of the solid cylinder is 150 cm2.

2. An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2m. Neglecting the

thickness of its walls, calculate

(i) its outer lateral surface area,

(ii) its capacity in litres.

Solution:

Given diameter of the cylinder, d = 35 cm

radius, r = d/2 = 35/2 = 17.5 cm

Height of the cylinder, h = 1.2 m = 120 cm

(i)Outer lateral surface area = 2rh

= 2×(22/7)×17.5×120

= 13200 cm2

(ii)Capacity of the cylinder = r2h

= (22/7)×17.52×120

= 115500 cm3

= 115.5 litres [∵1000 cm3 = 1 litre]

Hence the outer lateral surface area and the capacity of the cylinder is 13200 cm2 and 115.5 litres respectively.

3. A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled

with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.

Solution:

Given diameter of the cylindrical glass, d = 7 cm

Radius, r = d/2 = 7/2 = 3.5 cm

Height of the cylindrical glass, h = 12 cm

Volume of the cylindrical glass, V = r2h

= (22/7)×3.52×12

= 462 cm3

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• ML Aggarwal Solutions Class 10 Maths

Chapter: Mensuration

Number of students = 1600

Milk needed for 1600 students = 1600×462

= 739200 cm3

= 739.2 litres [∵1000 cm3 = 1 litre]

Hence the amount of milk needed to serve 1600 students is 739.2 litres.

4. In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of

height 16 cm. Find the volume of the cylinder.

Solution:

Given height of the cylinder, h = 16 cm

When rectangular foil of length 22 cm is folded to form cylinder, the base circumference of the cylinder is 22cm

2r = 22

r = 22/2

= 3.5 cm

Volume of the cylinder, V = r2h

= (22/7)×3.52×16

= 616 cm3

Hence the volume of the cylinder is 616 cm3.

5. (i) How many cubic metres of soil must be dug out to make a well 20 metres deep and 2 metres in

diameter?

(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 50 per m2, find

the cost of plastering.

Solution:

(i) Given diameter of the well, d = 2 m

Radius, r = d/2 = 2/2 = 1 m

Depth of the well, h = 20 m

Volume of the well, V = r2h

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• ML Aggarwal Solutions Class 10 Maths

Chapter: Mensuration

= (22/7)×12×20

= 62.85 m3

Hence the amount of soil dug out to make the well is 62.85 m3.

(ii)Curved surface area of the well = 2rh

= 2×(22/7)×1×20

= 880/7 m2

Cost of plastering the well per m2 = Rs. 50

Total cost of plastering the well = 50×(880/7)

= Rs. 6285.71

Hence the total cost of plastering is Rs. 6285.71.

6. A road roller (in the shape of a cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least

number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.

Solution:

Given diameter of the road roller, d = 0.7 m

Radius, r = d/2 = 0.7/2 = 0.35 m

Width, h = 1.2 m

Curved surface area of the road roller = 2rh

=2 ×(22/7)×0.35×1.2

= 2.64 m2

Area of the play ground = l×b

= 120×44

= 5280 m2

Number of revolutions = Area of play ground / Curved surface area

= 5280/2.64

= 2000

Hence the road roller must take 2000 revolutions to level the ground.

7. If the volume of a cylinder of height 7 cm is 448 π cm3, find its lateral surface area and total surface area.

Solution:

Given Height of the cylinder, h = 7 cm

Volume of the cylinder, V = 448  cm3

r2h = 448 

×r2×7 = 448 

 r2 = 448/7 = 64

 r = 8

Lateral surface area = 2rh

= 2××8×7

= 112  cm2

Total surface area = 2r(r+h)

= 2××8×(8+7)

= 2××8×15

= 240 cm2

Hence the lateral surface area and the total surface area of the cylinder are 112  cm2 and 240 cm2.

8. A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m3.

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• ML Aggarwal Solutions Class 10 Maths

Chapter: Mensuration

Solution:

Given height of the pole, h = 7 m

Diameter of the pole, d = 20 cm

radius, r = d/2 = 20/2 = 10 cm = 0.1m

Volume of the pole = r2h

= (22/7)×0.12×7

= 0.22 m3

Weight of wood per m3 = 225 kg

weight of 0.22 m3 wood = 225×0.22 = 49.5 kg

Hence the weight of the wood is 49.5 kg.

9. The area of the curved surface of a cylinder is 4400 cm2, and the circumference of its base is 110 cm. Find

(i) the height of the cylinder.

(ii) the volume of the cylinder.

Solution:

Given curved surface area of a cylinder = 4400 cm2

Circumference of its base = 110 cm

2r = 110

r = 110/2 = (110×7)/2×22 = 17.5 cm

(i) Curved surface area of a cylinder, 2rh = 4400

2×(22/7)×17.5×h = 4400

h = (4400×7)/2×22×17.5

= 40 cm

Hence the height of the cylinder is 40 cm.

(ii)Volume of the cylinder, V = r2h

= (22/7)×17.52×40

= 38500 cm3

Hence the volume of the cylinder is 38500 cm3.

10. A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm2. Find

(i) the height of the cylinder correct to one decimal place.

(ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)

Solution:

(i) Given diameter of the cylinder, d = 20 cm

Radius, r = d/2 = 20/2 = 10 cm

Curved surface area = 1000 cm2

2rh = 1000

2×3.14×10×h = 1000

62.8h = 1000

h = 1000/62.8

= 15.9 cm

Hence the height of the cylinder is 15.9 cm.

(ii)Volume of the cylinder, V = r2h

= 3.14×102×15.9

= 4992.6 cm3

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• ML Aggarwal Solutions Class 10 Maths

Chapter: Mensuration

Hence the volume of the cylinder is 4992.6 cm3.

11. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of

ink in the pen will be used up when writing 310 words on an average. How many words would use up a

bottle of ink containing one-fifth of a litre?

Answer correct to the nearest. 100 words.

Solution:

Height of the barrel of a pen, h = 7 cm

Diameter, d = 5mm = 0.5 cm

Radius, r = d/2 = 0.5/2 = 0.25 cm

Volume of the barrel of pen, V = r2h

= (22/7)×0.252×7

= 1.375 cm3

Ink in the bottle, = one fifth of a litre

= (1/5)×1000 = 200 ml

Number of words written using full barrel of ink = 310

Number of words written by using this ink = (200/1.375)×310 = 45090.90 words

Round off to nearest hundred, we get 45100 words.

Hence the number of words written using the ink is 45100 words.

12. Find the ratio between the total surface area of a cylinder to its curved surface area given that its height

and radius are 7.5 cm and 3.5 cm.

Solution:

Given radius of the cylinder, r = 3.5 cm

Height of the cylinder, h = 7.5 cm

Total surface area = 2r(r+h)

Curved surface area = 2rh

Ratio of Total surface area to curved surface area = 2r(r+h)/ 2rh

= (r+h)/h

= (3.5+7.5)/7.5

= 11/7.5

= 22/15

Hence the required ratio is 22:15.

13. The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio

of the volume of the new cylinder to that of the original cylinder?

Solution:

Let the radius of the base of a right circular cylinder be r and height be h.

Volume of the cylinder, V1 = r 2h

The radius of the base of a right circular cylinder is halved and the height is doubled.

So radius of new cylinder = r/2

Height of new cylinder = 2h

Volume of the new cylinder, V2 = r 2h

= (r/2)2 ×2h

= ½ r2h

So ratio of volume of new cylinder to the original cylinder , V2/V1= ½ r 2h/r2h = ½

Hence the requ