Sphere is an important chapter in WBBSE Class 10 Mathematics.
This chapter mainly deals with the surface area and volume of spheres and hemispheres,
along with problems based on radius, diameter, ratios and changes in dimensions.
The practice set below covers the important question formats from the supplied
Madhyamik Mathematics Chapter 12 source, including MCQs,
True or False, Fill in the Blanks, Short Answer and Long Answer questions.
Sphere – Chapter 12
Core Idea: A sphere is a three-dimensional solid in which every
point on its surface is at the same distance from its centre. That distance is
called the radius.
Important Sphere Formulas
| Quantity |
Formula |
| Diameter |
d = 2r |
| Surface Area of Sphere |
4πr² |
| Volume of Sphere |
4/3 πr³ |
| Curved Surface Area of Hemisphere |
2πr² |
| Total Surface Area of Hemisphere |
3πr² |
| Volume of Hemisphere |
2/3 πr³ |
Multiple Choice Questions – MCQ
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1. The ratio of the curved surface areas of two solid spheres is
16 : 9. What is the ratio of their volumes?
(a) 3 : 4
(b) 27 : 64
(c) 64 : 27
Answer: (b) 27 : 64
2. The volumes of two solid spheres are in the ratio
24 : 8. Find the ratio of their curved surface areas.
(a) 1 : 8
(b) 9 : 4
(c) 8 : 9
(d) 3 : 8
Answer: (c) 8 : 9
3. If the numerical values of the volume and surface area of
a sphere are equal, what is the numerical value of its radius?
(a) 2 units
(b) 3 units
(c) 4 units
(d) 6 units
Answer: (b) 3 units
4. If the curved surface area of a solid sphere is numerically
equal to three times its volume, find its radius.
(a) 4 units
(b) 3 units
(c) 2 units
(d) 1 unit
Answer: (d) 1 unit
5. The ratio of the total surface areas of two hemispherical
solids is
1 : 4. Find the ratio of their volumes.
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) None of these
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True or False
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1. If a semicircle is rotated about its diameter as the axis,
a sphere is produced.
Answer: True
2. The total surface area of a solid sphere of radius r is
2πr square units.
Answer: False
3. If the diameter of a sphere is 2 units, its volume is
8/3π cubic units.
Answer: True
4. If the radius of a solid sphere is doubled, its volume
becomes twice the original volume.
Answer: False
5. If the radius of a solid sphere is doubled, the volume
becomes four times the original volume.
Answer: False
6. If the ratio of the volumes of two spheres is
64 : 27, the ratio of their radii is 4 : 3.
Answer: True
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Fill in the Blanks
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1. A hollow hemisphere has ______ surface.
Answer: One
2. A solid hemispherical object has ______ surfaces.
Answer: Two
3. The volume of a solid sphere of diameter d units is
______ cubic units.
Answer: πd³/6
4. A solid hemisphere has ______ surfaces.
Answer: Two
5. The volume of a solid hemisphere whose total surface area
is 108π square cm is ______ cubic cm.
Answer: 144π
6. The intersection of a sphere and a plane is a ______.
Answer: Circle
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Short Answer Questions
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1. A football has a diameter of 28 cm. How many square
centimetres of leather are required to make the football?
Required concept: Since the football is treated as a sphere,
use the surface-area formula 4πr².
2. The numerical values of the volume and surface area of
a sphere are equal. Find the numerical value of the radius of the sphere.
Answer: 3 units
3. The diameter of a sphere is 21 cm. Find its volume.
Volume = 4/3 πr³
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Long Answer Questions
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Question 1 – Melting a Sphere to Form a Cuboid
A solid iron sphere of diameter 42 cm is melted and used
to make a cuboidal brick of length 49 cm and breadth
36 cm. Find the thickness of the brick.
Volume of iron sphere = Volume of cuboidal brick
Radius of the sphere:
r = 42/2 = 21 cm
Therefore,
Volume of sphere = 4/3 π × 21³.
If the thickness of the cuboidal brick is h, then:
49 × 36 × h = 4/3 π × 21³
Using the appropriate value of π and simplifying gives the required
thickness of the brick.
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Question 2 – Volume of a Hemispherical Vessel
The curved surface area of a hemispherical vessel is
2772 cm². Find the volume of the vessel.
Curved Surface Area of Hemisphere = 2πr²
From the given curved surface area, first determine the radius of the
hemisphere and then use:
Volume of Hemisphere = 2/3 πr³
Exam Focus: In hemisphere problems, carefully identify
whether the question gives curved surface area or total surface area.
The two formulas are different.
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Question 3 – Increase in Diameter of a Hemisphere
If the diameter of a hemisphere is increased by 250%,
at what rate does its surface area increase?
Surface area varies directly as the square of the radius. Since diameter
and radius change in the same ratio, the change in surface area can be
determined by using the square relationship.
Surface Area ∝ r²
The new diameter becomes 350% of the original diameter,
so the new radius is also 3.5 times the original radius.
New Surface Area = 3.5² × Original Surface Area = 12.25 × Original Surface Area
Therefore, the increase in surface area is:
11.25 times the original surface area, or
1125%.
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Question 4 – Two Spheres Melted into a Hollow Sphere
Two solid spheres having diameters 2 cm and
12 cm are melted and converted into a hollow sphere
of thickness 0.1 dm. Find the curved surface area
of the outer surface of the new sphere.
Volume of both solid spheres = Volume of material in the hollow sphere
The first step is to calculate the total volume of the two original spheres.
Then convert the given thickness into centimetres and use the difference
between the outer and inner radii of the hollow sphere.
Important: In melting-and-recasting problems, the volume
of material remains unchanged.
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Sphere and Hemisphere – Quick Revision Chart
| Concept |
Formula / Key Point |
| Diameter |
d = 2r |
| Sphere Surface Area |
4πr² |
| Sphere Volume |
4/3 πr³ |
| Hemisphere Curved Surface Area |
2πr² |
| Hemisphere Total Surface Area |
3πr² |
| Hemisphere Volume |
2/3 πr³ |
| Volume Ratio of Spheres |
V₁ : V₂ = r₁³ : r₂³ |
| Surface Area Ratio |
A₁ : A₂ = r₁² : r₂² |
Important Points for Chapter 12
- Always convert diameter into radius before applying the sphere formula.
- Remember the difference between the curved surface area and total surface area of a hemisphere.
- For two spheres, surface-area ratios depend on the square of their radii.
- Volume ratios depend on the cube of their radii.
- In melting and recasting problems, the volume of material remains constant.
- If the radius changes, remember that surface area changes as r² while volume changes as r³.
Madhyamik Mathematics Resources
For broader WBBSE Class 10 preparation, students can also use the
Madhyamik Suggestion 2027 – All Subjects
resource on Cademy.
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The Cademy resource includes subject-wise revision support for
Mathematics along with the other Madhyamik subjects.
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Revision Tip: Practise ratio-based sphere problems and
melting-and-recasting questions carefully. These problems become much easier
when the relationship between radius, surface area and volume is clear.
