Mensuration 3D

Question 1

Given: A water tank is being used to supply drinking water in a city due to a shortage of water. If the dimensions of the tank are 7m in length, 11m in breadth, and 3.5m in height. Find how many liters of water can be filled in the tank?

• A. 128,700 lt.
• B. 225,535 lt.
• C. 269,500 lt.
• D. 324,600 lt.

Answer: (B)

7m * 11m * 3.5m = 269,500 litres

Question 2

If Jimmy rolls a sheet of paper initially square-shaped along its length to make it a cylinder. Find what will be the ratio of the base radius of the cylinder to the side of the square?

• A. 1/2π
• B. 7/2π
• C. 9/π
• D. 11/2π

Answer: (A)

Perimeter of base circle = side of square 

2πr = a

r/a = 1 : 2π

Question 3

Given: A 5 cm cube which is cut into as many smaller cubes of 1 cm as possible. Find what will be the ratio of the surface area of the original cube to that of the sum of the surface areas of all the smaller cubes formed after cuttting? 

• A. 5:4
• B. 5:1
• C. 6:7
• D. 1:5

Answer: (D)

The volume of the original cube = 5³ = 125 cm³.
The volume of a smaller cubes = 1³ = 1 cm³. 
we will be getting total cubes = 125
The surface area of the larger cube = 6*a² = 6(5²) = 6 * 25 = 150
The surface area of each of the smaller cubes = 6 (1²) = 6.
Therefore, surface area of all of the 125, 1 cm³ cubes = 125 * 6 = 750.
Therefore, the required ratio = 150 : 750 = 1 : 5

Question 4

Given: A cubical wooden structure with sides of 5 cm is painted on all its faces. If the cubical structure is sliced into 1 cm cubes, find how many 1 cm cubes will have exactly one of their faces painted?

• A. 33
• B. 54
• C. 76
• D. 140

Answer: (B)

Given: The cube side = 5 cm

The side of cube 5cm is cut into 5 equal parts, in which each of 1 cm. Therefore, the total number of cubes of side 1 cm = 25 + 25 + 25 + 25 + 25 = 125.

In one face of cube, there are total of 9 small cubes painted.

We know that, there are 6 faces in cube.

Thus, total of 9 x 6 faces will have one face painted. (i.e.) 54

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:

• A. 40/7
• B. 80/7
• C. 70/9
• D. 13/8

Answer: (B)

Volume of cylinder= πr²h
=>22/7∗35/2∗35/2∗h=11000 {11 lt = 11000 mlt}
=>h=(11000∗7∗4)/(22∗35∗35)cm=80/7cm

Question 6

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

• A. 33.33%
• B. 66.66%
• C. 50%
• D. 100%

Answer: (D)

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 water.

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:

• A. H = 2R
• B. H = 3R
• C. H = 2/3R
• D. H = 3/2R

Answer: (A)

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:

• A. 1:2
• B. 2:1
• C. 1:2^0.5
• D. 2^0.5:1

Answer: (D)

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:

• A. 12 cm
• B. 14 cm
• C. 15 cm
• D. 18 cm

Answer: (B)

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:

• A. 2 cm
• B. 3 cm
• C. 4 cm
• D. 6 cm

Answer: (B)

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.