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

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-17 | Mensuration |

Writer / Book | Understanding |

Topics | Solutions of MCQs |

Academic Session | 2024-2025 |

**Mensuration MCQs**

ML Aggarwal Class 10 ICSE Maths Solutions

**Choose the correct answer from the given four options (1 to 32) :**

**Question 1. ****If two cylinders of the same lateral surface have their radii in the ratio 4 : 9, then the ratio of their heights is**

(a) 2 : 3

(b) 3 : 2

(c) 4 : 9

(d) 9 : 4

**Answer :**

Ratio in two cylinder having same lateral surface area their radii is 4 : 9

Let r_{1} be the radius of the first and r_{2} be the second cylinder and h_{1}, h_{2} and their heights

Let r_{1} = 4x and r_{2} = 9x

∴ 2πr_{1}h_{1} = 2πr_{2}h_{2}

⇒ 2π**× **4x **×** h_{1 }= 2 × π × 9x **×**h_{2}

⇒ h_{1}/h_{2} = 9x/4x

= 9 : 4

Ratio in their heights = 9 : 4 (d)

**Question 2. ****The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is**

(a) 10 : 17

(b) 20 : 27

(c) 17 : 27

(d) 20 : 37

**Answer :**

Radii of two cylinder are in the ratio = 2 : 3

Let radius (r_{1}) = 2x

and radius (r_{2}) = 3x

Ratio in their height = 5 : 3

Let height of the first cylinder = 5y

and of second = 3y

Now, volume of the first cylinder

πr_{1}^{2}h = π(2x)^{2} **× **5h

= 20πx^{2}y

And volume of second = π(3x)^{2} **× **3y

**= **π **× **27x^{2}y

∴ Ratio is 20πx^{2}y : 27πx^{2}y

= 20 : 27

**Mensuration MCQs**

ML Aggarwal Class 10 ICSE Maths Solutions

**Page 436**

**Question 3. ****The total surface area of a cone whose radius is r/2 and slant height 2l is**

(a) 2πr (l + r)

(b) πr(l + r/4)

(c) πr(l + r)

(d) 2πrl

#### Answer :

Radius of a cone = r/2

and slant height = 2l

total surface area of a cone

πrl + πr^{2}

= (πr/2 **× **2l) + π(r/2)^{2}

**= **πrl + πr^{2}/4

= πr(l + r/4)

**Question 4. ****If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is**

(a) 60π cm^{2}

(b) 65π cm^{2}

(c) 90π cm^{2}

(d) 120π cm^{2}

**Answer :**

Diameter of the base of a cone = 10 cm

Radius (r) = 10/2 = 5 cm

and height (h) = 12 cm

**(ML Aggarwal Mensuration MCQs Class 10)**

**Question 5. ****If the radius of a hemisphere is 5 cm, then its volume is**

(a) 250/3 π cm^{3}

(b) 500/3 π cm^{3}

(c) 75π cm^{3}

(d) 125/3 π cm^{3}

**Answer :**

Radius of a hemisphere (r) = 5 cm

Volume = 2/3πr^{3}

= 2/3π (5)^{3} cm^{3}

= 250/3 π cm^{3}

**Question 6. ****If the ratio of the diameters of the two spheres is 3 : 5, then the ratio of their surface areas is**

(a) 3 : 5

(b) 5 : 3

(c) 27 : 125

(d) 9 : 25

**Answer :**

Ratio in the diameters of two spheres = 3 : 5

Let radius of the first sphere = 3x cm

and radius of the second sphere = 5x cm

Ratio in their surface area

**Question 7. ****The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is**

(a) 1 : 4

(b) 1 : 3

(c) 2 : 3

(d) 2 : 1

**Answer :**

Radius of balloon (hemispherical) in the original position = 6 cm

and in increased position = 12 cm

Ratio of their surface area = 4π(6)^{2} : 4π(12)^{2}

⇒ 6^{2} : 12^{2}

⇒ 36 : 144

⇒ 1 : 4

**Question 8. ****If two solid hemisphere of same base radius r are joined together along with their bases, then the curved surface of this new solid is**

(a) 4πr^{2}

(b) 6πr^{2}

(c) 3πr^{2}

(d) 8πr^{2}

**Answer :**

Radius of two solid hemispheres = r

These are joined together along with the bases

Curved surface area = 2π^{2} × 2 = 4πr^{2} (a)

**(ML Aggarwal Mensuration MCQs Class 10)**

**Question 9. ****If a solid of one shape is converted to another, then the surface area of the new solid**

(a) remains same

(b) increases

(c) decreases

(d) can’t say

**Answer :**

If a solid of one shape has conversed into another then

the surface area of the new solid will same or not same

i.e. can’t say. (d)

**Question 10. ****The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is**

(a) 9.7 cm^{3}

(b) 77.6 cm^{3}

(c) 58.2 cm^{3}

(d) 19.4 cm^{3}

**Answer :**

Edge of cube = 4.2 cm

Radius of largest cone cut out = 42.2/2 = 2.1 cm

And height = 4.2 cm

Volume = 1/3 πr^{2}h

= 1/3 × 22/7 × 2.1 × 2.1 × 4.2 cm^{3}

= 19.404

Therefore, 19.4 cm^{3}

**Question 11. ****The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is**

(a) 288 π cm^{3}

(b) 4/3π cm^{3}

(c) 6 π cm^{3}

(d) 4 π cm^{3}

**Answer :**

Radius of cylinder (r) = 1 cm

Height (h) = 6 cm

The largest sphere that can be cut off from the cylinder of radius 1 cm

**Question 12. ****The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is**

(a) 3 : 4

(b) 4 : 3

(c) 9 : 16

(d) 16 : 9

**Answer :**

Ratio in volumes of two spheres = 64 : 27

Ratio of their radii = r_{1}^{3}/r_{2}^{3} = 64/27

= (r_{1}/r_{2})^{3}

= (4/3)^{3}

⇒ r_{1}/r_{2} = 4/3

∴ Ratio in their surface area = 4πr_{1}^{2}/4πr_{2}^{2}

= r_{1}^{2}/r_{2}^{2}

= (4/3)^{2}

= 16/9

∴ Ratio is 16 : 9

**Question 13. ****If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is**

(a) 1 : 3 : 2

(b) 2 : 3 : 1

(c) 2 : 1 : 3

(d) 1 : 2 : 3

**Answer :**

If a cone, a hemisphere and a cylinder have equal bases = r (say)

and height = h in each case and r = h

Ratio in their volumes = 1/3πr^{2}h : 2/3πr^{3} : πr^{2}h

= 1/3 πr^{2}r : 2/3 πr^{3} : πr^{2}r

= 1/3 πr^{3} : 2/3 πr^{3} : πr^{3}

= 1/3 : 2/3 : 1

= 1 : 2 : 3

**Question 14. ****If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is**

(a) 1 : 2

(b) 2 : 1

(c) √π : √6

(d) √6 : √π

**Answer :**

**(d) √6 : √π**

A sphere and a cube have equal surface area

Let a be the edge of a cube and r be the radius of the sphere, then

4πr^{2} = 6a^{2}

⇒ π(2r)^{2} : 6a^{2} **(∵ d = 2r)**

⇒ d^{2}/a^{2} = 6/π

⇒ d/a = √6/√π

∴ Radii d : a = √6 : √π

**Question 15. ****A solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm is molded to form a sphere. The radius of the sphere is**

(a) 21 cm

(b) 23 cm

(c) 25 cm

(d) 19 cm

**Answer :**

Dimension of a cuboid = 49 cm × 33 cm × 24 cm

Volume of a cuboid = 49 × 33 × 24 cm^{3}

⇒ Volume of sphere = Volume of a cuboid

Volume of a sphere = 49 × 33 × 24 cm^{3}

∴ Radius = {Volume/(4/3.π)}^{1/3}

= [(49 **× **33 × 24 × 3 × 7)/(4 × 22)]^{1/3}

= (49 × 7 × 3 × 3)^{1/3}

= (7 × 7 × 7 × 3 × 3 × 3)^{1/3}

= 7 × 3

= 21 cm

**Question 16. ****If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is**

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cm

**Answer :**

Height of a circular cone (h) = 24 cm

and radius (r) = 6 cm

Volume of a cone = 1/3.πr^{2}h

= 1/3.π × 6 × 6 × 24 cm^{3}

Volume of sphere = Volume of a cone

Now volume of sphere = 1/3π × 36 × 24 cm^{3}

Let r be in radius of sphere

Then 4/3 πr^{3} = 1/3 π × 36 × 24

4r^{3} = 36 × 24

⇒ r^{3} = (36 × 24)/4

⇒ r^{3} = 3 × 3 × 3 × 2 × 2 × 2

⇒ r^{3 }= 3^{3} × 2^{3}

∴ r = 3 × 2

= 6 cm

**Question 17. ****If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is**

(a) 15 cm

(b) 10 cm

(c) 7.5 cm

(d) 5 cm

**Answer :**

Diameter of a cylinder = 15 cm

Radius = 15/2 cm

and height = 10 cm

∴ Volume = πr^{2}h

= π × 15/2 × 15/2 × 10 cm^{3}

= 1125π/2 cm^{3}

∴ Volume of sphere = 1125π/2 cm^{3}

∴ Radius of sphere = Volume/(4/3π)^{1/3}

= {(1125π **× **3)/(2 × 4π)}^{1/3}

= (3375/8)^{1/3}

= {(5^{3} × 3^{3})/(2^{3})}^{1/3}

= (5 × 3)/2 cm

= 15/2

= 7.5 cm

**(ML Aggarwal Mensuration MCQs Class 10)**

**Question 18. ****The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is**

(a) 100

(b) 1000

(c) 10000

(d) 100000

**Answer :**

Radius of sphere (R) = 10 cm

Volume of sphere = 4/3πR^{3}

= 4/3 π (10)^{3} cm^{3}

**= **4/3 π **× **1000 cm^{3}

And radius of one ball (r) = 1 cm

∴ Volume of one hall = 4/3 π(1)^{3} cm^{3}

= 4/3.π cm^{3}

∴ Number of balls = (4π × 1000 × 3)/(3 × 4 × π)

= 1000

**Mensuration MCQs**

ML Aggarwal Class 10 ICSE Maths Solutions

**Page 437**

**Question 19. ****A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is**

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 18 cm

**Answer :**

The internal diameter of the metallic shell = 4 cm

and external diameter = 8 cm

∴ Internal radius (r) = 4/2 = 2 cm

And external radius (R) = 8/2 = 4 cm

Volume of metal used = 4/3 π(R^{3} – r^{3})

= 4/3 π(4^{3} – 2^{3}) cm^{3}

= 4/3 × π(64 – 8) cm^{3}

= 4/3 π × 56 cm^{3}

Diameter of cone = 8 cm

∴ Radius of cone = 8/2 = 4 cm

⇒ Height = Volume/(1/3 πr^{2})

= (4π × 56 × 3)/(3 × 1 × π × 4 × 4)

= 14 cm

**Question 20. ****A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2 cm and height 7 cm up to its brim. The number of children who will get the ice cream cones is**

(a) 163

(b) 263

(c) 363

(d) 463

**Answer :**

Edge of a cubical ice cream brick = 22 cm

Volume = a^{3} = (22)^{3} = 10648 cm^{3}

Radius (r) of ice cream cone (r) = 2 cm

and height (h) = 7 cm

Volume of one cone = 1/3πr^{2}h

= 1/3 × 22/7 × 2 × 2 × 7 cm^{3}

= 88/3 cm^{3}

∴ Number of cones = (10648 × 3)/88

= 363

**Question 21. ****Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is**

(a) 4 cm

(b) 3 cm

(b) 2 cm

(d) 6 cm

**Answer :**

Diameter of cylinder = 2 cm

Radius = 2/2 = 1 cm

and height = 16 cm

Volume = πr^{2}h

= 22/7 × 1 × 1 × 16

= 352/7 cm^{3}

∴ Volume of 12 solid spheres so formed = 352/7 cm^{3}

∴ Volume of each sphere = 352/(7 × 12)

= 352/84 cm^{3}

∴ Radius of each sphere = (352 × 3 × 7)/(84 × 4 × 22)^{1/3}

= (1)^{1/3}

= 1 cm

∴ Diameter = 2 × 1

= 2 cm

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

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

Thanks

Please Share with Your Friends