# Mensuration 3D

Question 1 |

How many liters of water can filled in a tank of length, breadth and height of 4m,3 m and 1.5 m respectively?

20000 lt. | |

18000 lt. | |

22000 lt. | |

16000 lt. |

**Mensuration 3D**

**Discuss it**

Question 1 Explanation:

4m * 3m * 1.5m = 18000 litres

Question 2 |

A sheet of paper in square shape is rolled along its length to make it a cylinder. What is the ratio of the base radius to the side of the square?

5/2π | |

3/2π | |

2/π | |

1/2π |

**Mensuration 3D**

**Discuss it**

Question 3 |

A 5 cm cube is cut into as many 1 cm cubes as possible. What is the ratio of the surface area of the original cube to that of the sum of the surface areas of all the smaller cubes?

2:3 | |

1:5 | |

2:7 | |

1:8 |

**Mensuration 3D**

**Discuss it**

Question 3 Explanation:

The volume of the original cube = 5

The volume of a smaller cubes = 1

we will be getting total cubes = 125

The surface area of the larger cube = 6*a

The surface area of each of the smaller cubes = 6 (1

Therefore, surface area of all of the 125, 1 cm

Therefore, the required ratio = 150 : 750 = 1 : 5

^{3}= 125 cm^{3}.The volume of a smaller cubes = 1

^{3}= 1 cm^{3}.we will be getting total cubes = 125

The surface area of the larger cube = 6*a

^{2}= 6(5^{2}) = 6 * 25 = 150The surface area of each of the smaller cubes = 6 (1

^{2}) = 6.Therefore, surface area of all of the 125, 1 cm

^{3}cubes = 125 * 6 = 750.Therefore, the required ratio = 150 : 750 = 1 : 5

Question 4 |

A cube with sides of 5 cm is painted on all its faces. If sliced in 1 cm cubes, how many 1 cm cubes will have exactly one of their faces painted?

54 | |

76 | |

38 | |

40 |

**Mensuration 3D**

**Discuss it**

Question 5 |

A drum is full of water. Diameter of the drum is 35cm. The level of water will be dropped by how much, if 11 litres of water is taken out:

40/7 | |

80/7 | |

70/9 | |

13/8 |

**Mensuration 3D**

**Discuss it**

Question 5 Explanation:

Volume of cylinder= πr

=>22/7∗35/2∗35/2∗h=11000 {11 lt = 11000 mlt}

=>h=(11000∗7∗4)/(22∗35∗35)cm=80/7cm

^{2}h=>22/7∗35/2∗35/2∗h=11000 {11 lt = 11000 mlt}

=>h=(11000∗7∗4)/(22∗35∗35)cm=80/7cm

Question 6 |

A hemispherical bowl is filled to the brim with a beverage. The contents of the bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder, the volume of the beverage in the cylindrical vessel is:

66.66% | |

78.50% | |

100% | |

More than 100% (i.e., some liquid will be left in the bowl) |

**Mensuration 3D**

**Discuss it**

Question 6 Explanation:

Let the height of the vessel be h. Then,
Radius of the bowl = h/2
Radius of the vessel = h/2
And,
Volume of the bowl = 2/3 * Pi * (h/2)^3 = 1/12 * Pi * h^3
Volume of the vessel = Pi * (h/2)^2 * h = 1/4 * Pi * h^3
As the volume of the vessel is 3 times more than that of the bowl, it can contain 100% of the beverage.

Question 7 |

A metallic hemisphere is melted and recast in the shape of a cone with the same base radius (R) as that of the hemisphere. If H is the height of the cone, then:

H = 2R | |

H = 3R | |

H = 2/3R | |

H = 3/2R |

**Mensuration 3D**

**Discuss it**

Question 7 Explanation:

Volume of the Hemisphere = 2/3 * Pi * R^3
Volume of the Cone = 1/3 * Pi * R^2 * H
As the two volumes are same, we have
2/3 * Pi * R^3 = 1/3 * Pi * R^2 * H
Therefore, H = 2R.

Question 8 |

A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surfaces will be:

1:2 | |

2:1 | |

1:2^0.5 | |

2^0.5:1 |

**Mensuration 3D**

**Discuss it**

Question 8 Explanation:

Let r be the radius of the hemisphere and the cone. Given,
Height of Cone = Radius of Hemisphere = r
Slant height of Cone = √(r²+ r²) = √2r
Ratio of their Curved Surfaces = Hemisphere/Cone
= 2 * π * r² / π * r * √2r
= √2:1.

Question 9 |

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is:

12 cm | |

14 cm | |

15 cm | |

18 cm |

**Mensuration 3D**

**Discuss it**

Question 9 Explanation:

Volume of the hollow sphere = 4/3 * π * (R³-r³) = 4/3 * π * (4³-2³) = 4/3 * π * 56 cm³
Let the height of the cone be h cm. Then,
1/3 * π * 4 * 4 * h = 4/3 * π * 56
h = (4 * 56 / 4 * 4) = 14 cm.

Question 10 |

A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is:

2 cm | |

3 cm | |

4 cm | |

6 cm |

**Mensuration 3D**

**Discuss it**

Question 10 Explanation:

Volume of Sphere = 4/3 * π * r³ = 4/3 * π * 3³
Volume of Cone = 1/3 * π * r² * h = 1/3 * π * 6² * h
Given,
Volume of Sphere = Volume of Cone
4/3 * π * 3³ = 1/3 * π * 6² * h
h = 4 * 3³ / 6² = 3 cm.

There are 16 questions to complete.