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?
 A 20000 lt. B 18000 lt. C 22000 lt. D 16000 lt.
Mensuration 3D
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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?
 A 5/2π B 3/2π C 2/π D 1/2π
Mensuration 3D
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 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?
 A 2:3 B 1:5 C 2:7 D 1:8
Mensuration 3D
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Question 3 Explanation:
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
A yellow painted cube with side 1 cm is sliced into cubes of 1 cubic centimeter cubes, How many cubes will have exactly one side yellow?
 A 54 B 76 C 38 D 40
Mensuration 3D
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 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
Mensuration 3D
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Question 5 Explanation:
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
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:
 A 66.66% B 78.50% C 100% D More than 100% (i.e., some liquid will be left in the bowl)
Mensuration 3D
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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:
 A H = 2R B H = 3R C H = 2/3R D H = 3/2R
Mensuration 3D
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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:
 A 1:2 B 2:1 C 1:2^0.5 D 2^0.5:1
Mensuration 3D
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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:
 A 12 cm B 14 cm C 15 cm D 18 cm
Mensuration 3D
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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:
 A 2 cm B 3 cm C 4 cm D 6 cm
Mensuration 3D
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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.
 Question 11
A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by:
 A 2/9 cm B 4/9 cm C 9/4 cm D 9/2 cm
Mensuration 3D
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Question 11 Explanation:
Volume of the sphere = 4/3 * π * r³ = 4/3 * π * 3³ Volume of water displaced = π * r² * h = π * 4² * h Given, Volume of water displaced = Volume of sphere π * 4² * h = 4/3 * π * 3³ h = 4/3 * 3³ / 4² = 9/4
 Question 12
A solid piece of iron of dimensions 49 * 33 * 24 cm³ is molded into a sphere. The radius of the sphere is:
 A 21 cm B 28 cm C 35 cm D None of these
Mensuration 3D
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Question 12 Explanation:
Volume of a Sphere = 4/3 * π * r³ = 49 * 33 * 24 r³ = 49 * 33 * 24 * 7/22 * 3/4 = 7³ * 11 * 3 * 2³ * 3 * 1/22 * 3/4 = 7³ * 3³ = 21³ cm³ Therefore, r = 21 cm.
 Question 13
If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere, the diameter of the new sphere will be:
 A 12 cm B 24 cm C 30 cm D 36 cm
Mensuration 3D
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Question 13 Explanation:
Total volume of the sphere = 4/3 * π * (a³ + b³ + c³) = 4/3 * π * (6³ + 8³ + 10³) = 4/3 * π * (1728) = 4/3 * π * 12³ Therefore, diameter of the new sphere = 24 cm.
 Question 14
The volumes of two sphere are in the ratio of 64:27. The ratio of their surface areas is:
 A 1:2 B 2:3 C 9:16 D 16:9
Mensuration 3D
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Question 14 Explanation:
We have R³:r³ = 4³:3³. Therefore, R²:r² = 4²:3² = 16:9.
 Question 15
The radii of two cones are in the ratio 2:1, their volumes are equal. Find the ratio of their heights.
 A 1:8 B 1:4 C 2:1 D 4:1
Mensuration 3D
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Question 15 Explanation:
Let the radius and height of the two cones be R, H and r, h respectively such that R:r = 2:1. Then, Volume of the cone C = 1/3 * π * R² * H Volume of the cone c = 1/3 * π * r² * h Now, 1/3 * π * R² * H = 1/3 * π * r² * h H:h = r²:R² = 1²:2² = 1:4.
There are 15 questions to complete.

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