ML Aggarwal Mensuration Chapter Test Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Chapter Test 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 Chapter Test Class 10 ICSE Maths Solutions**

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-17 | Mensuration |

Writer / Book | Understanding |

Topics | Solutions of Chapter Test |

Academic Session | 2024-2025 |

**Mensuration Chapter Test **

ML Aggarwal Class 10 ICSE Maths Solutions

**Question 1. ****A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m ^{2}, find the cost of the tin for making the container.**

**Answer :**

Height of container opened at the top (h) = 1 m = 100 cm

and diameter = 70 cm

∴Radius (r) = 70/2 = 35 cm

∴Total surface area = 2πrh + πr^{2}

= πr(2h + r)

= 22/7 × 35(2×100 + 35) cm^{2}

= 110(200 + 35)

= 110 × 235 cm^{2}

= (110 × 235)/(100 × 100)m^{2}

= 517/200 m^{2}

∴ Area of sheet required = 517/200 m^{2}

Cost of 1 m^{2} sheet **= **₹ 300

∴ Total cost = 517/200 × 300

**= **₹ 1551/2

= ₹ 775.50

**Question 2. ****A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.**

**Answer :**

Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm

Volume = 30 × 14 × 14 = 5880 cm^{3}

Largest size of cylinder cut out of the wooden cuboid will be of diameter = 14 cm

And height = 30 cm

∴ Radius of cylinder = 14/2= 7 cm

Volume of cylinder = πr^{2}h

= 22/7 × 7 × 7 × 30 cm^{3}

= 4620 cm^{3}

∴ Volume of wooden wasted = 5880 – 4620 = 1260 cm^{3}

**Question 3. ****Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.**

**Answer :**

Slant height of a cone (l) = 17 cm

Diameter of base = 30 cm

Radius (r) = 30/2 = 15 cm

**Question 4. ****Find the volume of a cone given that its height is 8 cm and the area of base 156 cm**^{2}.

^{2}.

**Answer :**

Height of a cone = 8 cm

Area of base = 156 cm

.’. Volume = 1/3 × area of base × height

= 1/3 × 156 × 8

= 1248/3 cm^{3}

= 416 cm^{3}

**Question 5. ****The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.**

**Answer :**

Circumference of the edge of bowl = 132 cm

Radius of a hemispherical bowl

= 132/2π = (132 × 7)/(2 × 22)

= 21 cm

Now volume of the bowl = 2/3πr^{3}

= 2/3 × 22/7 × (21)^{3} cm^{3}

= 2/3 × 2/7 × 9261 cm^{3}

= 19404 cm^{3}

**Question 6. ****The volume of a hemisphere is 2425.1/2 cm**^{2}. Find the curved surface area.

^{2}. Find the curved surface area.

**Answer :**

Volume of a hemisphere = 2425.1/2 cm^{3}

= 4851/2 cm^{3}

Let radius = r, then

2/3πr^{3 }= 4851/2

⇒ 2/3 × 22/7 × r^{3} = 4851/2

⇒ r^{3} = (4851 × 3 × 7)/(2 × 2 × 22)

= 9261/8

= (21/2)^{3}

∴ r = 21/2 cm

∴ Curved surface area = 2πr^{2}

= 2 × 22/7 × 21/2 × 21/2 cm^{2}

= 693 cm^{2}

**Question 7. ****A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy**

**Answer :**

A wooden solid toy is of a shape of a right circular cone

mounted on a hemisphere.

Radius of hemisphere (r) = 4.2 cm

Total height = 10.2 cm

**Mensuration Chapter Test **

ML Aggarwal Class 10 ICSE Maths Solutions

**Page 439**

**Question 8. ****A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.**

**Answer :**

Radius of cylinder = 7/2cm

and height of cylinder = 19 – 2 × 7/2 cm

= 19 – 7 = 12 cm

and radius of hemisphere = 7/2 cm

^{3}+ πr

^{2}h

= 4/3 × 22/7 × (7/2)^{3} + 22/7 × (7/2)^{2} × 12 cm^{3}

= 4/3 × 22/7 × (7×7×7)/(2×2×2) + 22/7 × 7/2 × 7/2 × 12 cm^{3}

= 539/3 + 462

= (539 + 1386)/3

= 1925/2 cm^{3}

= 641.2/3 cm^{3}

And total surface area of the solid = 2×2πr^{2 }+ 2πrh

= 4πr^{2} + 2πrh

= 2πr(2r + h)

= 2× 22/7 × 7/2(2×7/2 + 12)

= 22(7 + 12)

= 22 × 19 cm^{2}

= 418 cm^{2}

**Question 9. ****The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).**

**Answer :**

Let radius of cone (r) = 5x

then height (h) = 12x

Volume = 1/3 πr^{2}h

= 1/3(3.14)×(5x)^{2} × 12x

We know, volume = 2512 cm^{3}

⇒ 1/3(3.14)× 25x^{2 }× 12x = 2512

⇒ 1/3 × 3.14 × 300x^{3} = 2512

x^{3} = (2513×3)/(3.14×300)

= (2512×3×100)/(314×300)

= 8

= (2)^{3}

∴ x = 2

∴ Radius of cone (r) = 5 × 2 = 10 cm

And height (h) = 12 × 2 = 24 cm

= 26 cm

**Question 10. ****A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.**

**Answer :**

Let height of cone and cylinder = h

Diameter of the base of cone = 3x

Diameter of base of cylinder = 2x

Volume of cone = 1/3π(r_{1})^{2} h

= 1/3 π×(3x/2)^{2} × h

= 1/3 π× 9/4x^{2} ×h

= ¾ πx^{2}h

And volume of cylinder = πr^{2}h

= π(2x/2)^{2 }h

= πx^{2}h

= ¾ : 1

⇒ 3 : 4

**Question 11. ****A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.**

**Answer :**

Radius of the cone (r) = 9 cm

Height of the cone (h) = 10 cm

Volume of water filled in cone

= 1/3 πr^{2}h

= 1/3π(9)^{2} × 10 cm^{3}

= 810/3 π

= 270 π cm^{3}

Now radius of the cylindrical jar = 10 cm

Let h be the height of water in the jar

∴ πr^{2}h = 270π

⇒ π(10)^{2}h = 270π

⇒ 100 πh = 270π

⇒ h = 270π/100π

= 2.7 cm

**Question 12. ****An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.**

**Answer :**

Radius of the base of cone = 8 cm

And radius of cylinder = 8 cm

Height of cylindrical part (h_{1}) = 240 cm

And height of conical part (h_{2}) = 36 cm

Volume of the iron pillar = 1/3 πr^{2}h_{2} + πr^{2}h_{1}

= πr^{2}(1/3h_{2} + h_{1})

= 22/7 × 8 × 8 [1/3 ×36 + 240] cm^{3}

= 1408/7 [36/3 + 240] cm^{3}

= 1408/7 [252] cm^{3}

= 1408/7 × 252

= 1408 × 36

= 50688 cm^{3}

Weight of 1 cm^{3 }= 7.8 gm

∴ Total weight of the pillar = 50688 × 7.8 gm

= 395366.4 gm

= 395.3664 kg

**Question 13. ****A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.**

**Answer :**

Diameter of the cylindrical part = 126 m

Radius (r) = 126/2 = 63m

Height of cylindrical part = 5 m

Total height of the tent = 21 m

∴ Height of conical portion = 21 – 5

= 16

Surface area of the tent = 2πrh + πrl

= πr (2h + l)

= 22/7 × 63(2×5 + 65)

= 198 × (10 + 65)

= 198 × 75 m^{2}

= 14850 m^{2}

Cost of one 1 sq. m cloth = ₹ 36

∴ Total cost = Rs 14850 × 36

= ₹ 534600

**Question 14. ****The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their**

**(i) curved surfaces**

**(ii) volumes.**

**Answer :**

Radius of the base of a cone (r) = 3 cm

Height (h) = 4 cm

= 5

^{2}

= πr(l + r)

= 22/7 × 3(5 + 3) cm^{2}

= 66/7 × 8

= 528/7 cm^{2}

Diameter of cylinder = 4 cm

∴ Radius (r_{1}) = 4/2 = 2 cm

Total surface area = 528/7 cm^{2}

Let h be the height, then

∴ 2πr_{1}h_{1} + 2πr^{2} = 528/7

⇒ 2πr(h_{1} + r) = 528/7

⇒ 2 × 22/7 × 2(h1 + 2) = 528/7

⇒ h_{1 }+ 2 = 528/7 × 7/(2×22×2)

⇒ h_{1} + 2 = 6

⇒ h_{1} = 6 – 2 = 4 cm

**(i)** Ratio between curved surface of cone and cylinder = πrl : 2πr_{1}h_{1}

= πrl : 2πr_{1}h_{1}

= π×3×5 : 2×π×2×4

= 15 : 16

**(ii)** Ratio between their volumes

= 1/3 πr^{2}h : πr_{1}^{2}h_{1}

= 1/3 π×3×3×4 : π×2×4

= 3 : 4

**Question 15. ****A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.**

**Answer :**

Radius of base of a cone (r) = 2. 1 cm

and height (h) = 8.4 cm

Volume = 1/3πr^{2}h

= 1/3π × (2.1)^{2} × (8.4) cm^{3}

= π × 4.41 × 2.8 cm^{3}

= 12.348 π cm^{3}

∴ Volume of sphere = 12.348 π cm^{3}

Radius = Volume/(4/3.π)^{1/3}

= [12.348π × 3]/[4 × π]^{1/3}

= (9.261)^{1/3}

= (2.1 × 2.1 × 2.1)^{1/3}

= 2.1 cm

**Question 16. ****Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.**

**Answer :**

Radius of a cylinder (r) = 3 cm

Height (h) = 5 cm

Volume = πr^{2}h = π×3×3×5 = 45π cm^{2}

Diameter of a coins = 2.5 cm

∴ Radius (r_{1}) = 2.5/2 = 1.25 cm

And height (h_{1}) = 3 mm = 3/10 cm

∴ Volume of a coin = πr_{1}^{2}h_{1}

= π×1.25×1.25×3/10 cm^{3}

= 0.46875π cm^{3}

∴ Number of coins required = 45π/0.4687π

= 45/0.46875

= 96 coins

**Question 17. ****A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.**

**Answer :**

Radius of hemisphere = 8 cm

Volume = 2/3πr^{3} cm^{3}

= 2/3 π ×(8)^{3} cm^{3}

= 2/3 π ×512 cm^{3}

= 1024/3 π cm^{3}

∴ Volume of right circular cone = 1024/3 π cm^{3}

Radius = 6 cm

Let h be the height of the cone

∴ 1/3 πr^{2} h = 1024/3 π

⇒ 1/3 π×(6)^{2}h = 1024/3 π

⇒ 12 πh = 1024/3 π

⇒ h = 1024π/(3×12π)

= 256/9

= 28.44 cm

**Question 18. ****A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.**

**Answer :**

Radius of hemispherical bowl = 6 cm

.’. Volume of the water in the bowl

= 2/3 πr^{3}

= 2/3 π×(6)^{3} cm^{3}

= 144 π cm^{3}

∴ Volume of water in the cylinder = 144π cm^{3}

Radius of the cylinder = 4 cm

Let h be the height of water

∴ πr^{2}h = 144π

⇒ (4)^{2}h = 144

⇒ 16h = 144

∴ h = 144/16 = 9

Hence,

height of water in the cylinder = 9 cm

**Question 19. ****A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?**

**Answer :**

Radius of sphere = 6/2 = 3 cm

Let height of water raised = h cm

Now volume of sphere = 4/3 πr^{3}

= 4/3 π(3)^{3} cm^{3}

= 36 π cm^{3}

And volume of water in the cylinder = 36 π cm^{3}

∴ πr^{2}h = 36 π

⇒ (6)^{2}h = 36

⇒ 36 h = 36

⇒ h = 1

∴ Height of raised water = 1 cm

**Question 20. ****A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.**

**Answer :**

Radius of solid sphere = 6 cm

Volume of solid sphere

= 4/3 πr^{3}

= 4/3 × π × (6)^{3} cm^{3}

= 288π cm^{3}

∴ Volume of hollow cylinder = 288π cm^{3}

External radius of cylinder (R) = 5 cm

And height (h) = 32 cm

Let r be the inner radius

∴ Volume = π(R^{2 }– r^{2})h

∴ π(R^{2 }– r^{2})h = 288 π

⇒ [(5)^{2} – r^{2}] × 32 = 288

⇒ 25 – r^{2} = 288/32

⇒ 25 – r^{2} = 9

⇒ r^{2} = 25 – 9 = 16

⇒ r^{2 }= (4)^{2}

∴ r = 4

∴ Thickness of hollow cylinder = R – r

= 5 – 4

= 1 cm

–: End of ML Aggarwal Mensuration Chapter Test Class 10 ICSE Maths Solutions :–

Return to :- ML Aggarwal Solutions for ICSE Class-10

Thanks

Please Share with Your Friends

Thanks a lot