ML Aggarwal Mensuration Exe-17.1 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-17.1 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Mensuration Exe-17.1 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-17 | Mensuration |

Writer / Book | Understanding |

Topics | Solutions of Exe-17.1 |

Academic Session | 2024-2025 |

**Mensuration Exe-17.1**

ML Aggarwal Class 10 ICSE Maths Solutions

**Take π = 22/7 unless stated otherwise.**

**Question 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.**

**Answer :**

Radius of the cylinder (r) = 5 cm

Height (h) = 10 cm

Total surface area = 2πr (h + r)

so = 2π x 5(10 + 5) cm^{2}

therefore = 10 x 15π

Hence = 150π cm^{2}

**Question 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.**

**Answer :**

Diameter of cylindrical geyser = 35 cm

Radius (r) = 35/2 cm

Height = 1.2 m = 120 cm

**(i)** Outer lateral surface area = 2πrh

= 2 × (22/7) × 17.5 × 120

= 13200 cm^{2}

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

= (22/7) × 17.5^{2 }× 120

= 115500 cm^{3}

= 115.5 litres **[∵ 1000 cm ^{3} = 1 litre]**

**Question 3. ****A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk up to a height of 12 cm, find how many litres of milk is needed to serve 1600 students.**

**Answer :**

Number of students = 1600

Diameter of cylindrical glasses = 7 cm

Radius (r) = 7/2 cm

Height of the cylindrical glass, h = 12 cm

Volume of the cylindrical glass, V = πr^{2}h

= (22/7) × 3.5^{2 }× 12

= 462 cm^{3}

Number of students = 1600

Milk needed for 1600 students = 1600 × 462

= 739200 cm^{3}

= 739.2 litres **[∵ 1000 cm ^{3} = 1 litre]**

Hence,

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

**Question 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.**

**Answer :**

Length of rectangular tin foil (l) = 22 cm

and breadth (b) = 16 cm

By folding lengthwise, the radius of the cylinder

2r = 22

⇒ r = 22/2 = 3.5 cm

Volume of the cylinder, V = πr^{2}h

= (22/7) × 3.5^{2 }× 16

= 616 cm^{3}

Hence, the volume of the cylinder is 616 cm^{3}.

**Question 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 m ^{2}, find the cost of plastering.**

**Answer :**

##### (i) Depth of well (h) = 20 m

and diameter = 2 m

Radius, r = d/2 = 2/2 = 1 m

Depth of the well, h = 20 m

Volume of the well, V = πr^{2}h

= (22/7) × 1^{2 }× 20

= 62.85 m^{3}

Hence the amount of soil dug out to make the well is 62.85 m^{3}.

**(ii)** Curved surface area of the well = 2πrh

= 2 × (22/7) × 1 × 20

= 880/7 m^{2}

Cost of plastering the well per m^{2} = Rs. 50

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

= Rs. 6285.71

Hence the total cost of plastering is Rs. 6285.71.

**Question 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.**

**Answer :**

Diameter of a road roller = 0.7 m

Radius (r) = 0.7/2 = 0.35 m

and width (h) = 1.2 m

Curved surface area of the road roller = 2πrh

= 2 × (22/7) × 0.35 × 1.2

= 2.64 m^{2}

Area of the play ground = *l × b*

*= *120 × 44

= 5280 m^{2}

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.

**Question 7.**

**(i) If the volume of a cylinder of height 7 cm is 448 π cm ^{3}, find its lateral surface area and total surface area.**

**(ii) A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m ^{3}.**

**Answer :**

##### (i) Volume of a cylinder = 448 π cm^{3}

Height (h) = 7 cm

Volume of the cylinder, V = 448 cm^{3}

πr^{2}h = 448π

⇒ πr^{2 }× 7 = 448π

⇒ r^{2} = 448/7 = 64

⇒ r = 8

Lateral surface area = 2πrh

= 2 × π× 8 × 7

= 112π cm^{2}

Total surface area = 2πr(r + h)

= 2 ×π × 8 × (8 + 7)

= 2 ×π × 8 × 15

= 240π cm^{2}

Hence, the lateral surface area and the total surface area of the cylinder are 112π cm^{2} and 240π cm^{2}.

##### (ii) Height of a wooden pole (h) = 7 m

Diameter = 20 cm

**Mensuration Exe-17.1**

ML Aggarwal Class 10 ICSE Maths Solutions

Page 398

**Question 8. **The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the :

(i) radius of the cylinder

(ii) volume of cylinder

**Answer :**

Let r be the radius of the base of cylindrical vessel and ft = 25 cm be its height.

Now, circumference of the base = 132 cm

2πr = 132

Hence, the radius of the cylinder is 21 cm and volume of the cylinder is 34650 cm^{3}

**Question 9. ****The area of the curved surface of a cylinder is 4400 cm**^{2}, and the circumference of its base is 110 cm. Find:

^{2}, and the circumference of its base is 110 cm. Find:

**(i) the height of the cylinder.**

**(ii) the volume of the cylinder.**

**Answer :**

Area of the curved surface of a cylinder = 4400 cm^{2}

Circumference of base = 110 cm

2πr = 110

⇒ πr = 110/2

⇒ r = (110 × 7)/(2 × 22) = 17.5 cm

**(i)** Curved surface area of a cylinder,

2πrh = 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 = πr^{2}h

= (22/7) × 17.5^{2 }× 40

= 38500 cm^{3}

Hence the volume of the cylinder is 38500 cm^{3}.

**Question 10. ****A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm**^{2}. Find

^{2}. 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)**

**Answer 10**

Diameter of a cylinder = 20 cm

Radius (r) = 20/2 = 10 cm

Curved surface area = 1000 cm^{2}

Curved surface area = 1000 cm^{2}

2πrh = 1000

⇒ 2 × 3.14 × 10 × h = 1000

⇒ 62.8 h = 1000

⇒ h = 1000/62.8 = 15.9 cm

Hence, the height of the cylinder is 15.9 cm.

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

= 3.14 × 10^{2 }× 15.9

= 4992.6 cm^{3}

Hence the volume of the cylinder is 4992.6 cm^{3}.

**Question 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.**

**Answer :**

##### Height of cylindrical barrel of a pen (h) = 7 cm

Diameter = 5 mm

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

Volume of the barrel of pen,

V = πr^{2}h

= (22/7) × 0.25^{2 }× 7

= 1.375 cm^{3}

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.

**Question 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.**

**Answer :**

Radius of a cylinder (r) = 3.5 cm

and height (h) = 7.5 cm

Total surface area = 2πr(r + h)

Curved surface area = 2πrh

Ratio of Total surface area to curved surface area = 2πr(r + h)/2πrh

= (r + h)/h

= (3.5 + 7.5)/7.5

= 11/7.5

= 22/15

Hence, the required ratio is 22:15.

**Question 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?**

**Answer :**

Let the radius of the base of a right circular cylinder = r

and height (h) = h

Volume = πr^{2}h

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, V_{2} = πr^{2}h

= π(r/2)^{2 }× 2h

= ½ πr^{2}h

So ratio of volume of new cylinder to the original cylinder, V_{2}/V_{1}= ½ πr^{2}h/πr^{2}h = ½

Hence, the required ratio is 1:2.

**Question 14.**

**(i) The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm ^{2}. Find the height and the volume of the cylinder.**

**(ii) The total surface area of a cylinder is 352 cm**

^{2}. If its height is 10 cm, then find the diameter of the base.**Answer :**

**(i)** Let r be the radius and h be the height of the cylinder.

Given the sum of radius and height of the cylinder, r + h = 37 cm

Total surface area of the cylinder = 1628 cm^{2}

2πr(r + h) = 1628

⇒ 2 × (22/7) × r × 37 = 1628

⇒ r = (1628 × 7)/(2 × 22 × 37)

⇒ r = 7 cm

We have,

r + h = 37

⇒ 7 + h = 37

⇒ h = 37 – 7 = 30 cm

Volume of the cylinder, V = πr^{2}h

= (22/7) × 7^{2 }× 30

= (22/7) × 49 × 30

= 4620 cm^{3}

Hence, the height and volume of the cylinder is 30 cm and 4620 cm^{3} respectively.

**(ii)** Total surface area of the cylinder = 353 cm^{2}

Height, h = 10 cm

2πr(r + h) = 352

⇒ 2 × (22/7) × r × (r + 10) = 352

⇒ r^{2 }+ 10r = (352 × 7)/2 × 22

⇒ r^{2}+ 10r = 56

⇒ r^{2 }+ 10r – 56 = 0

⇒ (r + 14)(r – 4) = 0

⇒ r + 14 = 0 or r – 4 = 0

⇒ r = – 14 or r = 4

Radius cannot be negative. So r = 4.

Diameter = 2 × r = 2 × 4 = 8 cm.

Hence, the diameter of the base is 8 cm.

**Question 15. ****The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm**^{2}.

^{2}.

**Answer :**

Ratio between curved surface area and total surface area = 1 : 2

Total surface area = 616 cm^{2}

Curved surface area = 616/2 = 308 cm^{2}

2πrh = 308

⇒ πrh = 308/2

= (308 × 7)/(2 × 22)

⇒ rh = 49 **…(i)**

Total surface area, 2πrh + 2πr^{2} = 616

308 + 2πr^{2} = 616

⇒ 2πr^{2} = 616 – 308

⇒ 2πr^{2} = 308

⇒ πr^{2 }= 308/2 = 154

⇒ r^{2 }= 154 × 7/22 = 49

Taking square root on both sides

r = 7

Substitute r in (i)

rh = 49

⇒ 7 × h = 49

⇒ h = 49/7 = 7 cm

Volume of the cylinder, V = πr^{2}h

= (22/7) × 7^{2 }× 7

= 1078 cm^{3}

Hence, the volume of the cylinder is 1078 cm^{3}.

**Question 16. ****Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.**

**Answer :**

Volume of two cylinders is the same

Diameter of both cylinder are in the ratio = 3 : 4

Then the ratio of radius r_{1}:r_{2} = 3 : 4

Given volume of both jars are same.

πr_{1}^{2}h_{1} = πr_{2}^{2}h_{2}

⇒ h_{1}/h_{2} = πr_{2}^{2}/ πr_{1}^{2}

⇒ h_{1}/h_{2} = 4^{2}/3^{2} = 16/9

Hence, the ratio of the heights are 16 : 9.

**Question 17. ****A rectangular sheet of tin foil of size 30 cm x 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinders thus formed.**

**Answer :**

##### Size of the sheet = 30 cm × 18 cm

(i) By rolling lengthwise,

The circumference of the cylinder = 2πr = 30

2 × (22/7)r = 30

⇒ r = 30 × 7/2 × 22 = 210/44 = 105/22 cm

Height, h = 18 cm

Volume of the cylinder, V_{1} = πr^{2}h

= (22/7) × (105/22)^{2 }× 18

= 15 × 105 × 9/11

If we roll it breadthwise, base circumference, 2r = 18

2 × (22/7)r = 18

⇒ r = 18 × 7/2 × 22 = 126/44 = 63/22 cm

Height, h = 30 cm

Volume of the cylinder, V_{2} = πr^{2}h

= (22/7) × (63/22)^{2 }× 30

= 9 × 63 × 15/11

V_{1}/V_{2} = (15×105× 9/11) ÷ (9×63× 15/11)

= (15 × 105 × 9/11) × (11/9×63×15)

= 105/63

= 15/9

= 5/3

Ratio of the volumes of two cylinders is 5 : 3.

**Question 18. ****A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.**

**Answer :**

Given internal diameter of the tube = 11.2 cm

Internal radius, r = d/2 = 11.2/2 = 5.6 cm

Length of the tube, h = 21 cm

Thickness = 0.4 cm

Outer radius, R = 5.6 + 0.4 = 6 cm

Volume of the metal = πR^{2}h – πr^{2}h

= πh(R^{2 }– r^{2})

= (22/7) × 21 × (6^{2 }– 5.6^{2})

= 66 × (6 + 5.6)(6 – 5.6)

= 66 × 11.6 × 0.4

= 306.24 cm^{3}

Hence,

the volume of the metal is 306.24 cm^{3}.

**Question 19. ****A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.**

**Answer:**

Diameter of the pencil = 7 mm

Radius (R) = 7/2 mm = 7/20 cm

Diameter of graphite (lead) = 1 mm

Radius of graphite, r = ½ mm = 1/20cm

Volume of graphite = πr^{2}h

= (22/7) × (1/20)^{2 }× 14

= 11/100

= 0.11 cm^{3}

Hence the volume of the graphite is 0.11 cm^{3}

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

= (22/7) × [(7/20)^{2 }– (1/20)^{2}]14

= (22/7) × [(49/400) – (1/400)]14

= (22/7) × (48/400) × 14

= 11 × 12/25

= 5.28 cm^{3}

Hence

the volume of the wood is 5.28 cm^{3}

**Question 20. ****A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weighs 8 g.**

**Answer :**

Length of cylindrical roller (h) = 2 m = 200 cm

Diameter = 35 cm

Inner radius = 35/2 cm

Thickness = 7 cm

Outer radius, R = (35/2) + 7

= (35 + 14)/2

= 49/2 cm

Volume of the iron in roller = π(R^{2 }– r^{2})h

= (22/7)[(49/2)^{2 }– (35/2)^{2}]200

= (22/7)[(49^{2 }– 35^{2}/4]200

= (22/7) × 50(49^{2 }– 35^{2})

= (22/7) × 50(49^{2 }– 35^{2})

= (22/7) × 50(2401 – 1225)

= (22/7) × 50 × 1176

= 184800 cm^{3}

Given 1 cm³ of iron weighs 8 g.

Weight of the roller = 184800 × 8 = 1478400 g

= 1478.4 kg **[1 kg = 1000 g]**

Hence the weight of the roller is 1478.4 kg.

-: End of ML Aggarwal Mensuration Exe-17.1 Class 10 ICSE Maths Solutions :–

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