Mathematics Questions and Answers – Volume of a Sphere

«
»

This set of Mathematics 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 \(\frac{4}{3} πr^3\).
advertisement

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.
advertisement
advertisement

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 ro = 4cm
Inner radius ri = outer radius – thickness
= 4 – 3
ri = 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.
advertisement

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.

advertisement

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter