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Chapter 12 of 14
Practice Quiz

Surface Areas and Volumes

Gujarat Board · Class 10 · Mathematics

Practice quiz for Surface Areas and Volumes — Gujarat Board Class 10 Mathematics. MCQs and questions with answers to test your preparation.

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Quick Quiz: Surface Areas and Volumes

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1

A toy is made by placing a hemisphere on top of a cylinder. The cylinder has radius 7 cm and height 10 cm. Find the total surface area of the toy. (Use π = 22/7)

2

A solid is formed by joining a cone and a hemisphere at their bases. The radius of both is 6 cm and the height of the cone is 8 cm. What is the volume of the solid? (Use π = 22/7)

3

A capsule is made by joining two hemispheres to the ends of a cylinder. If the radius is 3.5 cm and cylinder height is 14 cm, find the total volume. (Use π = 22/7)

4

A wooden toy is made by scooping out a hemisphere from one face of a cube with side 10 cm. The radius of the hemisphere is 3 cm. Find the volume of the remaining solid.

44 Questions·
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Sample Questions

1multiple correct

Which of the following statements are correct about the surface area of combined solids?

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We must subtract the areas where solids are joined, Only the visible surfaces are counted in total surface area

When combining solids, we only count the visible surfaces. The flat faces where two solids join are not visible from outside, so they are not included in the total surface area calculation.

2multiple choice

The curved surface area of a cone with radius 5 cm and slant height 13 cm is:

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204.2 cm²

Curved surface area of cone = πrl where r is radius and l is slant height. CSA = (22/7) × 5 × 13 = (22/7) × 65 = 204.2 cm².

3multiple choice

A tent is in the shape of a cylinder surmounted by a cone. Both have the same radius 7 m. Height of cylinder is 10 m and height of cone is 8 m. Find the curved surface area of the tent. (Use π = 22/7)

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1166 m²

Total CSA = CSA of cylinder + CSA of cone. CSA of cylinder = 2πrh = 2 × (22/7) × 7 × 10 = 440 m². For cone, slant height l = √(r² + h²) = √(7² + 8²) = √(49 + 64) = √113 ≈ 10.6 m. CSA of cone = πrl = (22/7) × 7 × 10.6 = 231.9 ≈ 232 m². Wait, let me recalculate more carefully: l = √113 ≈ 10.63. CSA of cone = (22/7) × 7 × 10.63 = 234 m². Total = 440 + 234 = 674 m². Let me check again: l = √113 = 10.63, CSA of cone = 22 × 10.63 = 233.86 ≈ 234 m². Hmm, let me recalculate: CSA of cone = (22/7) × 7 × √113 = 22 × √113 = 22 × 10.63 = 234 m². Actually, let me be more precise: √113 ≈ 10.63, so CSA of con

4multiple correct

Which of the following formulas are correct for calculating volumes?

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Volume of sphere = (4/3)πr³, Volume of hemisphere = (2/3)πr³, Volume of cone = (1/3)πr²h, Volume of cuboid = length × breadth × height

The correct volume formulas are: Sphere = (4/3)πr³, Hemisphere = (2/3)πr³, Cone = (1/3)πr²h, Cylinder = πr²h (not 2πr²h), Cuboid = l×b×h.

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Frequently Asked Questions

What are the important topics in Surface Areas and Volumes for Gujarat Board Class 10 Mathematics?
Surface Areas and Volumes covers several key topics that are frequently asked in Gujarat Board Class 10 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Surface Areas and Volumes — Gujarat Board Class 10 Mathematics?
Understand the core concepts first, then work through the 44 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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