Chapter 12 · Class 10 CBSE · MCQ Test
Surface Areas & Volumes MCQ Test — Class 10 CBSE
Practice 10 multiple-choice questions with instant answer reveal and explanations.
Surface Areas & Volumes — MCQ Questions
1A toy is in the shape of a cone mounted on a hemisphere of the same base radius. Which expression correctly represents the total surface area of the toy?
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Answer: CSA of cone + CSA of hemisphere
Hint: When two solids are joined, only the exposed surfaces contribute to the total surface area. The common base area is internal.
Solution:
The total surface area of a combined solid is the sum of the curved (or exposed) surface areas of its individual components.
When a cone is mounted on a hemisphere, the circular base of the cone and the circular top of the hemisphere are joined together, becoming internal surfaces.
Therefore, the exposed surface area consists only of the curved surface area of the cone and the curved surface area of the hemisphere.
2A solid is formed by placing a cylinder on top of another larger cylinder. To find the total volume of this combined solid, what approach should be used?
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Answer: Add the Volumes of both cylinders.
Hint: Volume measures the space occupied by a solid. When combining solids, their individual volumes simply add up.
Solution:
Volume is a measure of the three-dimensional space occupied by an object.
When two or more solids are combined to form a new solid, the total volume of the resulting solid is the sum of the volumes of the individual solids, assuming there are no hollow spaces or overlaps beyond their joining surfaces.
Surface area, on the other hand, measures the area of the exterior surfaces and is not directly additive in the same way.
3Ravi is calculating the total surface area of a cubical block surmounted by a hemisphere. He uses the formula: Total Surface Area = (Total Surface Area of Cube) + (Curved Surface Area of Hemisphere). What mistake, if any, did Ravi make?
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Answer: He should have subtracted the area of the base of the hemisphere from the surface area of the cube.
Hint: When the hemisphere is placed on the cubical block, the area of the top face of the cube covered by the hemisphere is no longer exposed.
Solution:
The cubical block has 6 faces. When a hemisphere is placed on top of one face, a circular area on that face is covered by the base of the hemisphere.
This covered area is no longer part of the exposed surface and must be excluded from the cube's total surface area.
Therefore, the correct approach is: (TSA of Cube) - (Area of circular base of hemisphere) + (CSA of Hemisphere).
4A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 6 cm and the diameter of the base is 4 cm. What is the volume of the toy? (Use π = 22/7)
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Answer: 41.90 cm³
Hint: Calculate the radius from the diameter. Then, find the volume of the cone and the hemisphere separately and add them.
Solution:
Given diameter = 4 cm, so radius (r) = 4/2 = 2 cm. Height of cone (h) = 6 cm.
Volume of cone = (1/3) × π × r² × h = (1/3) × π × (2)² × 6 = (1/3) × π × 4 × 6 = 8π cm³.
Volume of hemisphere = (2/3) × π × r³ = (2/3) × π × (2)³ = (2/3) × π × 8 = (16/3)π cm³.
Total Volume = Volume of cone + Volume of hemisphere = 8π + (16/3)π = (24/3)π + (16/3)π = (40/3)π cm³ ≈ (40/3) × (22/7) = 880/21 ≈ 41.90 cm³.
5A storage tank is in the shape of a cylinder with two hemispherical ends. The total length of the tank is 13.5 m and the diameter is 7 m. Find the total surface area of the tank. (Use π = 22/7)
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Answer: 297 m²
Hint: The total length includes the radius of the hemispheres at both ends. Calculate the length of the cylindrical part first.
Solution:
Diameter = 7 m, so radius (r) = 7/2 m.
Total length of the tank = 13.5 m. The height of the cylindrical part (h) = Total length - 2 × r = 13.5 - 2 × (7/2) = 13.5 - 7 = 6.5 m.
Total Surface Area = CSA of cylinder + 2 × CSA of hemisphere.
TSA = 2πrh + 2 × (2πr²) = 2πr(h + 2r) = 2 × (22/7) × (7/2) × (6.5 + 2 × 7/2) = 22 × (6.5 + 7) = 22 × 13.5 = 297 m².
6A solid metallic sphere of radius 10 cm is melted and recast into a number of smaller solid cones, each of radius 5 cm and height 2 cm. Which principle must be used to find the number of cones formed?
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Answer: The volume of the sphere is equal to the sum of the volumes of all the cones.
Hint: Melting and recasting changes the shape but not the amount of material.
Solution:
When a solid is melted and recast into other solids, the total amount of material remains constant.
The amount of material is measured by its volume.
Therefore, the volume of the original solid (the sphere) must be equal to the sum of the volumes of all the new solids (the cones).
7A cylindrical hole is drilled through the center of a solid cubical block. Which of the following statements is true regarding the resulting solid?
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Answer: Its volume will decrease, but its total surface area will increase.
Hint: Drilling a hole removes material, affecting volume. Consider the new surfaces created by the hole for surface area.
Solution:
When a cylindrical hole is drilled, material is removed from the cubical block, so its volume will definitely decrease.
For surface area, the two circular areas where the cylinder enters and exits the cube are removed from the cube's original surface. However, the curved surface area of the cylindrical hole is *added* to the total exposed surface area.
In most cases, the newly exposed curved surface area of the cylinder is greater than the two circular areas removed, leading to an overall *increase* in the total surface area.
8Consider a solid formed by a cone placed on top of a cylinder. Both have the same radius 'r'. The height of the cone is 'h₁' and the height of the cylinder is 'h₂'. Which of the following expressions correctly represents the total surface area of this combined solid?
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Answer: πr(l + 2h₂ + r) where 'l' is the slant height of the cone.
Hint: Identify all exposed surfaces: CSA of cone, CSA of cylinder, and the base of the cylinder.
Solution:
The exposed surfaces are: Curved Surface Area (CSA) of the cone, Curved Surface Area (CSA) of the cylinder, and the base area of the cylinder.
CSA of cone = πrl
CSA of cylinder = 2πrh₂
Base area of cylinder = πr²
Total Surface Area = πrl + 2πrh₂ + πr² = πr(l + 2h₂ + r).
9A metallic sphere of radius 6 cm is melted and drawn into a wire of uniform diameter 0.2 cm. What is the length of the wire?
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Answer: 28800 cm
Hint: The wire is a cylinder. Use the principle of volume conservation and convert all units to be consistent.
Solution:
Volume of sphere = (4/3)πR³, where R = 6 cm. Volume = (4/3)π(6)³ = (4/3)π(216) = 288π cm³.
The wire is a cylinder. Diameter = 0.2 cm, so radius (r) = 0.1 cm. Let the length of the wire be 'h'.
Volume of cylinder = πr²h = π(0.1)²h = 0.01πh cm³.
By volume conservation: Volume of sphere = Volume of wire. Thus, 288π = 0.01πh. Dividing by π, 288 = 0.01h. So, h = 288 / 0.01 = 28800 cm.
10A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the total length of the capsule is 14 mm and the diameter is 5 mm, to find its total surface area, one would sum the curved surface area of the cylinder and the curved surface areas of ________.
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Answer: two hemispheres
Hint: Visualize the capsule. Which parts of the hemispheres are exposed?
Solution:
The capsule consists of a cylindrical middle part and two hemispherical ends.
When calculating the total surface area, we need to consider all the exposed outer surfaces.
The exposed surfaces are the curved surface area of the cylinder and the curved surface areas of *both* hemispheres. The flat circular bases of the hemispheres are joined to the cylinder and are not exposed.
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Tips for Surface Areas & Volumes MCQs
- 1Read each question carefully and identify what is being asked before looking at the options.
- 2Try to solve the problem mentally or on paper first, then match your answer with the options.
- 3Use elimination — rule out clearly wrong options to improve your chances even when unsure.
- 4Check units, signs, and edge cases — these are common traps in Surface Areas & Volumes MCQs.
- 5Review your mistakes after completing the test to build lasting understanding.
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