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  • RD Sharma Solution for Class 9 Chapter 21 - Download PDF
    • RD Sharma Solution for Class 9 Chapter 21 - Question with Answers
  • Surface Area and Volume of a Sphere FAQs
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RD Sharma Solutions Class 9 Maths Chapter 21 - Surface Area and Volume of a Sphere
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RD Sharma Solutions Class 9 Maths Chapter 21 - Surface Area and Volume of a Sphere

By Swati Singh

|

Updated on 5 May 2025, 11:08 IST

RD Sharma Class 9 Chapter 21 Solutions – Surface Area and Volume of a Sphere offers detailed answers to all textbook questions, curated by experienced mathematics educators at Infinity Learn. You can download the free PDF of these solutions to strengthen your exam preparation and boost your performance in both board and competitive exams.

RD Sharma Solution for Class 9 Chapter 21 - Download PDF

Here are the RD Sharma Solutions Class 9 Maths Chapter 25 Probability Solutions, designed to help students prepare effectively for their exams. By referring to these solutions and practicing the problems, students can boost their confidence and improve their scores.

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RD Sharma Solution for Class 9 Chapter 21 - Question with Answers

1. Find the surface area of a sphere with radius 7 cm.

Answer: Surface Area = 4πr² = 4 × 3.14 × 7² = 615.75 cm²

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2. Calculate the volume of a sphere with radius 3 cm.

Answer: Volume = (4/3)πr³ = (4/3) × 3.14 × 27 = 113.04 cm³

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3. A sphere has surface area 804.25 cm². Find its radius.

Answer: r = √(Surface Area / 4π) = √(804.25 / 12.56) ≈ 8 cm

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4. A sphere has a volume of 4188.79 cm³. Find the radius.

Answer: r = ∛(3V / 4π) = ∛(3×4188.79 / 12.56) ≈ 10 cm

5. Find the surface area of a sphere of diameter 14 cm.

Answer: r = 7 cm → Surface Area = 4πr² = 615.75 cm²

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6. Find the volume of a sphere of diameter 10 cm.

Answer: r = 5 cm → Volume = (4/3)πr³ = 523.33 cm³

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7. If the radius of a sphere is doubled, by what factor does the surface area increase?

Answer: Increases by a factor of 4

8. If the radius of a sphere is tripled, how many times does the volume increase?

Answer: Increases by a factor of 27

9. What is the radius of a sphere whose surface area is 201.06 cm²?

Answer: r = √(201.06 / 12.56) ≈ 4 cm

10. What is the volume of a sphere with radius 6 cm?

Answer: Volume = (4/3) × π × 216 = 904.32 cm³

11. Find the surface area of a sphere whose radius is 12 cm.

Answer: Surface Area = 4πr² = 4 × 3.14 × 144 = 1809.6 cm²

12. A sphere has a diameter of 18 cm. What is its volume?

Answer: r = 9 cm → Volume = (4/3)πr³ = (4/3) × 3.14 × 729 = 3053.62 cm³

13. What is the surface area of a sphere with radius 5.5 cm?

Answer: Surface Area = 4 × 3.14 × 5.5² = 380.15 cm²

14. The volume of a sphere is 113.1 cm³. Find its radius.

Answer: r = ∛[(3 × 113.1) / (4 × 3.14)] ≈ 3 cm

15. A solid metallic sphere has radius 10 cm. What is its surface area?

Answer: Surface Area = 4 × π × 100 = 1256 cm²

16. If the volume of a sphere is 268.08 cm³, find its radius.

Answer: r = ∛[(3 × 268.08) / (4 × 3.14)] ≈ 4 cm

17. A ball has a radius of 2.5 cm. Calculate its volume.

Answer: Volume = (4/3) × π × 15.625 ≈ 65.45 cm³

18. Find the surface area of a hemisphere with radius 8 cm.

Answer: Surface Area = 3πr² = 3 × 3.14 × 64 = 602.88 cm²

19. A hemisphere has a volume of 904.32 cm³. Find its radius.

Answer: Volume of sphere = 1808.64 → r = ∛[(3 × 1808.64) / (4 × 3.14)] ≈ 6 cm

20. What is the curved surface area of a hemisphere with radius 10 cm?

Answer: CSA = 2πr² = 2 × 3.14 × 100 = 628 cm²

21. A hollow sphere has an internal radius of 6 cm and external radius of 8 cm. Find the volume of material used.

Answer: V = (4/3)π(R³ – r³) = (4/3) × 3.14 × (512 – 216) = 1230.4 cm³

22. Find the total surface area of a hemisphere with radius 7 cm.

Answer: TSA = 3πr² = 3 × 3.14 × 49 = 461.58 cm²

23. A spherical toy has a surface area of 2010.6 cm². Find its radius.

Answer: r = √(2010.6 / 12.56) ≈ 12.6 cm

24. Find the volume of a metal sphere with diameter 16 cm.

Answer: r = 8 → V = (4/3) × 3.14 × 512 = 2144.66 cm³

25. The surface area of a sphere is 1256 cm². What is its radius?

Answer: r = √(1256 / 12.56) = 10 cm

26. A spherical water tank has a radius of 1.5 m. Find its volume.

Answer: V = (4/3)πr³ = (4/3) × 3.14 × 3.375 = 14.13 m³

27. Find the volume of a sphere whose surface area is 154 cm².

Answer: r = √(154 / 12.56) ≈ 3.5 cm → V = (4/3) × π × 42.875 = 179.5 cm³

28. A lead sphere has radius 2 cm. How much lead is used in making it?

Answer: V = (4/3)πr³ = (4/3) × 3.14 × 8 = 33.49 cm³

29. A sphere has a radius 6.5 cm. Calculate its surface area.

Answer: Surface Area = 4πr² = 4 × 3.14 × 42.25 = 530.26 cm²

30. A football is spherical with radius 11 cm. Find its volume.

Answer: V = (4/3)πr³ = (4/3) × 3.14 × 1331 = 5575.57 cm³
 

Surface Area and Volume of a Sphere FAQs

What is the formula to calculate the surface area of a sphere?

The surface area of a sphere is given by the formula:
Surface Area = 4πr²,
where r is the radius of the sphere and π = 3.14 (or 22/7).

What is the difference between a sphere and a hemisphere?

A sphere is a perfectly round 3D shape, like a ball. A hemisphere is half of a sphere, obtained by cutting a sphere through its center.

How is the surface area of a hemisphere calculated?

There are two types:

  • Curved Surface Area (CSA) = 2πr²

  • Total Surface Area (TSA) = 3πr² (includes the base circle)

If the radius of a sphere is doubled, how does the surface area change?

If the radius is doubled, the surface area increases by 4 times, because surface area ∝ r².

If the radius of a sphere is tripled, how does the volume change?

The volume increases by 27 times, because volume ∝ r³.

How do you find the radius of a sphere from its surface area?

Use the formula:
r = √(Surface Area / 4π)

How do you find the radius of a sphere from its volume?

Use the formula:
r = ∛[(3 × Volume) / (4π)]

Why is π (pi) used in the formulas for spheres?

π (pi) is a mathematical constant used in calculations involving circles and spheres, as it represents the ratio of the circumference to the diameter of a circle.

Are RD Sharma solutions useful for competitive exams like NTSE or Olympiads?

Yes. RD Sharma’s solutions build strong conceptual clarity, which is very helpful for competitive exams like NTSE, Olympiads, and entrance-level math contests.

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