3D Mensuration (Volume, Surface Area) for Competitive Exams

3D Mensuration (Volume, Surface Area)

Q. A cone and a cylinder have the same base radius and height. How do their volumes compare?

A. Cone has half the volume of the cylinder
B. Cone has the same volume as the cylinder
C. Cone has double the volume of the cylinder
D. Cone has one-third the volume of the cylinder

Solution

Volume of cone = (1/3)πr²h; Volume of cylinder = πr²h. Therefore, cone's volume is one-third of the cylinder's volume.

Correct Answer: A — Cone has half the volume of the cylinder

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Q. A cone has a base radius of 2 cm and a height of 6 cm. What is the volume of the cone?

A. 8π cm³
B. 12π cm³
C. 4π cm³
D. 16π cm³

Solution

Volume of a cone = (1/3)πr²h = (1/3)π(2)²(6) = (1/3)π(4)(6) = 8π cm³.

Correct Answer: B — 12π cm³

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Q. A cone has a base radius of 2 m and a height of 6 m. What is its surface area?

A. 16π
B. 20π
C. 24π
D. 28π

Solution

The surface area of a cone is given by SA = πr(r + l), where l is the slant height. Here, l = √(r² + h²) = √(2² + 6²) = √40. Thus, SA = π(2)(2 + √40) = 20π.

Correct Answer: B — 20π

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Q. A cone has a base radius of 3 m and a height of 4 m. What is the surface area of the cone?

A. 15π
B. 18π
C. 21π
D. 24π

Solution

The surface area of a cone is given by SA = πr(r + l), where l is the slant height. Here, l = √(3² + 4²) = 5, so SA = π(3)(3 + 5) = 24π.

Correct Answer: C — 21π

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Q. A cone has a base radius of 4 cm and a height of 9 cm. What is its volume?

A. 12π
B. 36π
C. 48π
D. 72π

Solution

The volume of a cone is given by V = (1/3)πr²h. Here, r = 4 and h = 9, so V = (1/3)π(4)²(9) = 48π.

Correct Answer: B — 36π

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Q. A cone has a base radius of 5 cm and a height of 12 cm. What is its surface area?

A. 50π cm²
B. 65π cm²
C. 70π cm²
D. 80π cm²

Solution

The surface area of a cone is given by SA = πr(r + l), where l is the slant height. Here, l = √(5² + 12²) = 13, so SA = π(5)(5 + 13) = 90π cm².

Correct Answer: B — 65π cm²

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Q. A cylinder has a height of 10 cm and a volume of 100π cm³. What is the radius of the base?

A. 2 cm
B. 3 cm
C. 4 cm
D. 5 cm

Solution

Using the volume formula V = πr²h, we have 100π = πr²(10). Thus, r² = 10, so r = √10 cm.

Correct Answer: B — 3 cm

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Q. A cylinder has a height of 10 cm and a volume of 200π cm³. What is the radius of the base?

A. 4 cm
B. 5 cm
C. 6 cm
D. 7 cm

Solution

Using the volume formula V = πr²h, we have 200π = πr²(10). Thus, r² = 20, so r = √20 = 4.47 cm.

Correct Answer: B — 5 cm

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Q. A cylinder has a volume of 100π cm³ and a height of 10 cm. What is the radius of the base?

A. 2 cm
B. 3 cm
C. 4 cm
D. 5 cm

Solution

Volume = πr²h. Therefore, 100π = πr²(10) => r² = 10 => r = √10 ≈ 3.16 cm.

Correct Answer: C — 4 cm

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Q. A cylindrical tank has a radius of 3 meters and a height of 5 meters. What is the volume of the tank in cubic meters?

A. 45π
B. 30π
C. 15π
D. 60π

Solution

The volume of a cylinder is given by the formula V = πr²h. Here, r = 3 and h = 5, so V = π(3)²(5) = 45π.

Correct Answer: A — 45π

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Q. A cylindrical tank has a radius of 3 meters and a height of 5 meters. What is the total surface area of the tank (including the top and bottom)?

A. 56.52 m²
B. 94.25 m²
C. 75.40 m²
D. 37.68 m²

Solution

Total surface area = 2πr(h + r) = 2π(3)(5 + 3) = 2π(3)(8) = 48π ≈ 150.8 m².

Correct Answer: B — 94.25 m²

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Q. A hemisphere has a radius of 3 cm. What is its volume?

A. 18π cm³
B. 27π cm³
C. 36π cm³
D. 9π cm³

Solution

Volume of a hemisphere = (2/3)πr³ = (2/3)π(3)³ = (2/3)π(27) = 18π cm³.

Correct Answer: A — 18π cm³

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Q. A rectangular prism has dimensions 3 cm, 4 cm, and 5 cm. What is its volume?

A. 60 cm³
B. 12 cm³
C. 15 cm³
D. 20 cm³

Solution

Volume of a rectangular prism = length × width × height = 3 × 4 × 5 = 60 cm³.

Correct Answer: A — 60 cm³

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Q. A rectangular prism has dimensions 3 m, 4 m, and 5 m. What is its surface area?

A. 47 m²
B. 60 m²
C. 70 m²
D. 80 m²

Solution

The surface area of a rectangular prism is given by SA = 2(lw + lh + wh). Here, SA = 2(3*4 + 3*5 + 4*5) = 2(12 + 15 + 20) = 2(47) = 94 m².

Correct Answer: B — 60 m²

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Q. A rectangular prism has dimensions 4 m, 3 m, and 5 m. What is its surface area?

A. 47 m²
B. 60 m²
C. 70 m²
D. 80 m²

Solution

The surface area of a rectangular prism is given by SA = 2(lw + lh + wh). Here, SA = 2(4*3 + 4*5 + 3*5) = 2(12 + 20 + 15) = 2(47) = 94 m².

Correct Answer: B — 60 m²

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Q. A rectangular prism has dimensions 4 m, 5 m, and 6 m. What is its surface area?

A. 94 m²
B. 60 m²
C. 70 m²
D. 80 m²

Solution

The surface area of a rectangular prism is given by SA = 2(lw + lh + wh). Here, SA = 2(4*5 + 4*6 + 5*6) = 94 m².

Correct Answer: A — 94 m²

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Q. A sphere has a radius of 7 cm. What is its surface area?

A. 154 cm²
B. 196 cm²
C. 308 cm²
D. 616 cm²

Solution

Surface area of a sphere = 4πr² = 4π(7)² = 4π(49) = 196 cm².

Correct Answer: B — 196 cm²

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Q. If a cube has a volume of 64 cubic centimeters, what is the length of one side of the cube?

A. 4 cm
B. 8 cm
C. 16 cm
D. 2 cm

Solution

Volume of a cube = side³. Therefore, side = ∛64 = 4 cm.

Correct Answer: A — 4 cm

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Q. If a sphere has a volume of 288π cubic centimeters, what is its radius?

A. 6 cm
B. 8 cm
C. 9 cm
D. 10 cm

Solution

The volume of a sphere is given by V = (4/3)πr³. Setting (4/3)πr³ = 288π, we find r³ = 216, thus r = 6 cm.

Correct Answer: B — 8 cm

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Q. If the height of a cylinder is doubled while keeping the radius constant, how does the volume change?

A. It remains the same
B. It doubles
C. It triples
D. It quadruples

Solution

Volume of a cylinder = πr²h. If height is doubled, volume becomes 2πr²h, which is double the original volume.

Correct Answer: B — It doubles

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Q. If the radius of a sphere is halved, how does its volume change?

A. It remains the same
B. It doubles
C. It halves
D. It reduces to one-eighth

Solution

Volume of a sphere = (4/3)πr³. If radius is halved, volume becomes (4/3)π(1/2)³ = (4/3)π(1/8) = (1/6)π, which is one-eighth of the original volume.

Correct Answer: D — It reduces to one-eighth

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Q. If the surface area of a cube is 216 square meters, what is the length of one side of the cube?

A. 6 meters
B. 9 meters
C. 12 meters
D. 15 meters

Solution

The surface area of a cube is given by the formula SA = 6a². Setting 6a² = 216, we find a² = 36, thus a = 6 meters.

Correct Answer: B — 9 meters

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Q. If the surface area of a cylinder is 150π cm² and the height is 10 cm, what is the radius?

A. 5 cm
B. 6 cm
C. 7 cm
D. 8 cm

Solution

The surface area of a cylinder is given by SA = 2πr(h + r). Setting 150π = 2πr(10 + r), we find r = 5 cm.

Correct Answer: A — 5 cm

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Q. If the surface area of a cylinder is 150π cm² and the height is 5 cm, what is the radius?

A. 5 cm
B. 3 cm
C. 4 cm
D. 2 cm

Solution

The surface area of a cylinder is given by SA = 2πr(h + r). Setting 150π = 2πr(5 + r), we find r = 4 cm.

Correct Answer: C — 4 cm

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Q. What is the total surface area of a sphere with a radius of 7 cm?

A. 49π
B. 98π
C. 144π
D. 196π

Solution

The total surface area of a sphere is given by SA = 4πr². Here, r = 7, so SA = 4π(7)² = 196π.

Correct Answer: B — 98π

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Q. What is the volume of a cube with a side length of 4 cm?

A. 16 cm³
B. 32 cm³
C. 64 cm³
D. 80 cm³

Solution

The volume of a cube is given by V = a³. Here, V = 4³ = 64 cm³.

Correct Answer: C — 64 cm³

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3D Mensuration (Volume, Surface Area) MCQ & Objective Questions

Understanding 3D Mensuration, specifically the concepts of volume and surface area, is crucial for students preparing for school and competitive exams. Mastering these topics not only enhances your mathematical skills but also boosts your confidence in tackling objective questions. Regular practice of MCQs and important questions in this area can significantly improve your exam performance and conceptual clarity.

What You Will Practise Here

Formulas for calculating volume and surface area of common 3D shapes like cubes, cylinders, cones, and spheres.
Understanding the relationship between different dimensions and their impact on volume and surface area.
Application of the Pythagorean theorem in solving problems related to 3D shapes.
Real-life applications of 3D mensuration concepts in various fields.
Diagrams and visual representations to aid in understanding complex concepts.
Practice questions that simulate exam conditions to enhance time management skills.
Common problem-solving strategies and tips for tackling tricky MCQs.

Exam Relevance

The topic of 3D Mensuration is a significant part of the mathematics syllabus in CBSE, State Boards, and competitive exams like NEET and JEE. You can expect questions that require you to apply formulas, solve real-world problems, and interpret data from diagrams. Common question patterns include finding the volume or surface area of a given shape, comparing different shapes, and solving multi-step problems that integrate various concepts.

Common Mistakes Students Make

Confusing the formulas for volume and surface area, leading to incorrect answers.
Neglecting to include units in their final answers, which can result in loss of marks.
Misinterpreting the dimensions given in a problem, especially in word problems.
Overlooking the importance of drawing diagrams to visualize the problem.
Rushing through calculations, which often leads to simple arithmetic errors.

FAQs

Question: What are the key formulas for volume and surface area of a cylinder?
Answer: The volume of a cylinder is given by V = πr²h, and the surface area is A = 2πr(h + r).

Question: How can I improve my speed in solving 3D mensuration problems?
Answer: Regular practice with timed quizzes and understanding the underlying concepts will help improve your speed and accuracy.

Now is the time to sharpen your skills! Dive into our collection of 3D Mensuration (Volume, Surface Area) MCQ questions and practice questions to test your understanding and prepare effectively for your exams. Your success is just a practice away!

3D Mensuration (Volume, Surface Area)

3D Mensuration (Volume, Surface Area) MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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