Volume of a Sphere - Class 9 Maths MCQ

This set of Class 9 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Volume of a Sphere”.

1. Volume of a sphere of radius r is equal to __________
a) \(\frac{4}{3} πr^3\)
b) πr²h
c) \(\frac{1}{3}\) πr²h
d) \(\frac{2}{3} πr^3\)
View Answer

Answer: a
Explanation: Volume of a sphere is \(\frac{4}{3}\) π times the cube of a radius.
Therefore, volume of a sphere is equal to \(\frac{4}{3} πr^3\).

2. Volume of a hemisphere of radius r is equal to __________
a) \(\frac{4}{3} πr^3\)
b) πr²h
c) \(\frac{1}{3}\) πr²h
d) \(\frac{2}{3} πr^3\)
View Answer

Answer: d
Explanation: Volume of a hemisphere is half the volume of a sphere.
We know that volume of a sphere of radius r is equal to \(\frac{4}{3} πr^3\).
Therefore, volume of a hemisphere is equal to \(\frac{2}{3} πr^3\).

3. What is the radius of a sphere having volume of 4851cm³? (Take π = \(\frac{22}{7}\))
a) 9 cm
b) 11 cm
c) 10.5 cm
d) 11.3 cm
View Answer

Answer: c
Explanation: We know that volume of a sphere of radius r is equal to \(\frac{4}{3} πr^3\).
\(\frac{4}{3} πr^3\) = 4851
\(\frac{4}{3} * \frac{22}{7} * r^3\) = 4851
\(r^3 = \frac{9261}{8}\)
Therefore, r = \(\frac{21}{2}\) = 10.5 cm.

4. Curved surface area of a hemisphere is equal to 2772 cm², what is its volume? (Take π = \(\frac{22}{7}\))
a) 19464 cm³
b) 19404 cm³
c) 19427 cm³
d) 19425 cm³
View Answer

Answer: b
Explanation: We know that curved surface area of a hemisphere is equal to 2πr².
2πr² = 2772
r² = \(\frac{2772*7}{2*22}\) = 441
Therefore, r = 21cm
Now, we know that volume of a hemisphere is equal to \(\frac{2}{3}\) πr³.
Hence, V = \(\frac{2}{3}\) πr³
= \(\frac{2*22*21*21*21}{3*7}\)
= 19404 cm³.

5. A hollow sphere of outer radius of 4 cm and thickness of 3 cm is to be made from metal. What is the total amount of metal required (in cm³) to make the sphere? (Take π = \(\frac{22}{7}\))
a) 281
b) 264
c) 225
d) 227
View Answer

Answer: b
Explanation: Outer radius r_o = 4cm
Inner radius r_i = outer radius – thickness
= 4 – 3
r_i = 1cm
This means that the sphere is hollow and volume of metal required = \(\frac{4}{3} πr_o^3 – \frac{4}{3} πr_i^3\)
= \(\frac{4}{3} π (4^3 – 1^3)\)
= \(\frac{4*22*63}{3*7}\)
= 264 cm³.

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6. Cost of whitewashing a sphere is 1232₹ at the rate of 2₹/cm². What is the volume of a sphere? (Take π = \(\frac{22}{7}\))
a) \(\frac{4324}{3}\)
b) \(\frac{4317}{5}\)
c) \(\frac{4312}{3}\)
d) \(\frac{4312}{7}\)
View Answer

Answer: c
Explanation: Area of a sphere = \(\frac{Total \,cost \,of \,whitewashing}{Rate \,of \,whitewashing/cm^2}\)
= \(\frac{1232}{2}\)
= 616 cm²
We know that total surface area of a sphere = 4πr²
4πr² = 616
r² = \(\frac{616*7}{22*4}\)
= 49
Therefore, r = 7cm
Now, volume of a sphere = \(\frac{4}{3} πr^3\)
= \(\frac{4}{3} * \frac{22}{7}\) * 7³
= \(\frac{4312}{3}\).

Sanfoundry Global Education & Learning Series – Mathematics – Class 9.

To practice all chapters and topics of class 9 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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