Welcome to AssamSchool. This article contains 30 Extra Questions from Surface Areas and Volumes from Class 9 Mathematics Chapter 11. These Surface Areas and Volumes Class 9 Extra Questions are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Surface Areas and Volumes Class 9 Extra Questions
1. Find the curved surface area of a cone with radius \(7 cm\) and slant height \(24 cm\).
2. True or False: The total surface area of a hemisphere is \(2\pi r^2\).
3. A cone has a radius of \(14 cm\) and a volume of \(2464 cm³\). Find its height.
4. A spherical balloon has a radius of \(10 cm\). If it is inflated to double its radius, find the ratio of its surface areas before and after inflation.
5. A hemispherical tank with an inner radius of \(2.1 m\) is made of steel \(0.5 cm\) thick. Calculate the volume of steel used.
6. A solid sphere of radius \(6 cm\) is melted and recast into small cones, each with radius \(2 cm\) and height \(3 cm\). How many cones are formed?
7. Calculate the volume of a sphere with radius \(3.5 cm\).
8. Convert the volume of a hemisphere (radius \(14 cm\)) into liters. (1 liter = \(1000 cm³\))
9. A conical tent has a base radius of \(7 m\) and a slant height of \(25 m\). Find the cost of canvas required at \(₹50\) per \(m²\).
10. The surface area of a sphere is \(616 cm²\). Find its radius.
11. A cylindrical tank of radius \(7 m\) contains water. A sphere of radius \(3.5 m\) is submerged into it. Find the water level rise.
12. What is the volume of a cone with radius \(3 cm\) and height \(4 cm\)?
13. The curved surface area of a cone is \(220 cm²\), and its slant height is \(10 cm\). Find its radius.
14. A hemispherical dome has a total surface area of \(462 cm²\). Find its volume.
15. True or False: The volume of a hemisphere is \(\frac{1}{3}\pi r^3\).
16. A joker’s cap (cone) has a base circumference of \(44 cm\) and slant height \(25 cm\). Find its curved surface area.
17. A sphere and a cone have the same radius and volume. Find the ratio of the cone’s height to the sphere’s radius.
18. Find the total surface area of a hemisphere with radius \(21 cm\).
19. A metallic sphere of radius \(8 cm\) is melted to form smaller spheres of radius \(2 cm\) each. How many small spheres are formed?
20. A cone, hemisphere, and cylinder have the same base and height. Compare their volumes.
21. A sphere has a volume of \(288\pi \, \text{cm}^3\). Find its radius.
22. A conical pit with a diameter of \(7 m\) and depth \(12 m\) is dug. Find its capacity in kiloliters.
23. A cylinder encloses a sphere of radius \((r)\). Find the ratio of their surface areas.
24. The diameter of a hemispherical bowl is \(14 cm\). Find its curved surface area.
25. A hollow sphere’s outer radius is \(12 cm\), and thickness is 1 cm. Find the volume of the metal used.
26. A right triangle with legs \(6 cm\) and \(8 cm\) is revolved about the hypotenuse. Find the volume of the resulting solid.
27. The volume of a cone is \(154 \, \text{cm}^3\) with radius \(7 cm\). Find its height.
28. A hemispherical bowl holds \(0.303\, liters\) of milk. Find its radius.
29. A solid cylinder of radius \((r)\) and height \((h)\) is melted to form \((n)\) cones of radius \((2r)\) and height \((h/3)\). Find \((n)\).
30. Find the surface area of a sphere with diameter \(21 cm\).
Answers:
- Solution:
CSA of cone = \( \pi r l = \frac{22}{7} \times 7 \times 24 = 528 \, \text{cm}^2 \) - Solution:
False. TSA of hemisphere = \(3\pi r^2\) - Solution:
Volume = \(\frac{1}{3}\pi r^2 h \Rightarrow 2464\)
\(= \frac{1}{3} \times \frac{22}{7} \times 14^2 \times h \Rightarrow h \)
\(= 12 \, \text{cm}\) - Solution:
Ratio = \(\frac{4\pi (10)^2}{4\pi (20)^2} = \frac{1}{4}\) - Solution:
Outer radius = \(2.1 + 0.005 = 2.105 \, \text{m}\)
Volume of steel = \(\frac{2}{3}\pi (2.105^3 – 2.1^3) \approx 0.139 \, \text{m}^3\) - Solution:
Volume of sphere = \(\frac{4}{3}\pi (6)^3 = 288\pi \, \text{cm}^3\)
Volume of one cone = \(\frac{1}{3}\pi (2)^2 \times 3 = 4\pi\)
Number of cones = \(288\pi / 4\pi = 72\) - Solution:
Volume = \(\frac{4}{3}\pi (3.5)^3 = 179.67 \, \text{cm}^3\) - Solution:
Volume = \(\frac{2}{3}\pi (14)^3 = 11498.67 \, \text{cm}^3 = 11.498 \, \text{liters}\) - Solution:
CSA of tent = \( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \, \text{m}^2 \)
Cost = \(550 \times 50 = ₹27,500\) - Solution:
\(4\pi r^2 = 616 \Rightarrow r^2 = 49 \Rightarrow r = 7 \, \text{cm}\) - Solution:
Volume of sphere = \(\frac{4}{3}\pi (3.5)^3 = 179.67 \, \text{m}^3\)
Let water rise = \(h\)
Then, \(\pi (7)^2 h = 179.67 \Rightarrow h = 1.16 \, \text{m}\) - Solution:
Volume = \(\frac{1}{3}\pi (3)^2 \times 4 = 12\pi \, \text{cm}^3 = 37.71 \, \text{cm}^3\) - Solution:
\( \pi r l = 220 \Rightarrow \frac{22}{7} \times r \times 10 = 220 \Rightarrow r = 7 \, \text{cm} \) - Solution:
TSA = \(3\pi r^2 = 462 \Rightarrow r = 7 \, \text{cm}\)
Volume = \(\frac{2}{3}\pi (7)^3 = 718.67 \, \text{cm}^3\) - Solution:
False. Volume of hemisphere = \(\frac{2}{3}\pi r^3\) - Solution:
Radius = \(\frac{44}{2\pi} = 7 \, \text{cm}\)
CSA = \( \pi \times 7 \times 25 = 550 \, \text{cm}^2 \) - Solution:
\(\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h \Rightarrow h = 4r\)
Ratio \(h : r = 4 : 1\) - Solution:
TSA = \(3\pi (21)^2 = 4158 \, \text{cm}^2\) - Solution:
Volume ratio = \(\frac{(8)^3}{(2)^3} = 64\)
Number of small spheres = 64 - Solution:
Volume ratio \(Cone : Hemisphere : Cylinder\) = \(1 : 2 : 3\) - Solution:
\(\frac{4}{3}\pi r^3 = 288\pi \Rightarrow r^3 = 216 \Rightarrow r = 6 \, \text{cm}\) - Solution:
Volume = \(\frac{1}{3}\pi (3.5)^2 \times 12 = 154 \, \text{m}^3 = 154 \, \text{kL}\) - Solution:
Cylinder surface area = \(2\pi r (r + 2r) = 6\pi r^2\)
Sphere surface area = \(4\pi r^2\)
Ratio = \(6\pi r^2 : 4\pi r^2 = 3 : 2\) - Solution:
CSA = \(2\pi (7)^2 = 308 \, \text{cm}^2\) - Solution:
Inner radius = \(12 – 1 = 11 \, \text{cm}\)
Volume of metal = \(\frac{4}{3}\pi (12^3 – 11^3) = 634.87 \, \text{cm}^3\) - Solution:
Hypotenuse = \(10 \, \text{cm}\)
Radius of rotation = \(\frac{6 \times 8}{10} = 4.8 \, \text{cm}\)
Volume = \(2 \times \frac{1}{3}\pi (4.8)^2 \times 10 = 486.4 \, \text{cm}^3\) - Solution:
\(154 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times h \Rightarrow h = 3 \, \text{cm}\) - Solution:
\(\frac{2}{3}\pi r^3 = 303 \Rightarrow r^3 = 144 \Rightarrow r \approx 5.24 \, \text{cm}\) - Solution:
Volume of cylinder = \(n \times \frac{1}{3}\pi (2r)^2 \times \frac{h}{3}\)
\( \pi r^2 h = n \times \frac{4}{9}\pi r^2 h \Rightarrow n = \frac{9}{4} = 2.25 \)
Not possible; check problem constraints. - Solution:
Radius = \(10.5 \, \text{cm}\)
Surface area = \(4\pi (10.5)^2 = 1386 \, \text{cm}^2\)
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Abdur Rohman is an Electrical Engineer from Charaideo, Assam, who wears multiple hats as a part-time teacher, blogger, entrepreneur, and digital marketer. Passionate about education, he founded The Assam School blog to provide free, comprehensive textbook solutions, MCQs (Multiple Choice Questions), and other academic content for students from Class V to XII.