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) πr^{2}h

c) \(\frac{1}{3}\) πr^{2}h

d) \(\frac{2}{3} πr^3\)

View Answer

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) πr^{2}h

c) \(\frac{1}{3}\) πr^{2}h

d) \(\frac{2}{3} πr^3\)

View Answer

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 4851cm^{3}? (Take π = \(\frac{22}{7}\))

a) 9 cm

b) 11 cm

c) 10.5 cm

d) 11.3 cm

View Answer

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 cm^{2}, what is its volume? (Take π = \(\frac{22}{7}\))

a) 19464 cm^{3}

b) 19404 cm^{3}

c) 19427 cm^{3}

d) 19425 cm^{3}

View Answer

Explanation: We know that curved surface area of a hemisphere is equal to 2πr

^{2}.

2πr

^{2}= 2772

r

^{2}= \(\frac{2772*7}{2*22}\) = 441

Therefore, r = 21cm

Now, we know that volume of a hemisphere is equal to \(\frac{2}{3}\) πr

^{3}.

Hence, V = \(\frac{2}{3}\) πr

^{3}

= \(\frac{2*22*21*21*21}{3*7}\)

= 19404 cm

^{3}.

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 cm^{3}) to make the sphere? (Take π = \(\frac{22}{7}\))

a) 281

b) 264

c) 225

d) 227

View Answer

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 cm

^{3}.

6. Cost of whitewashing a sphere is 1232₹ at the rate of 2₹/cm^{2}. 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

Explanation: Area of a sphere = \(\frac{Total \,cost \,of \,whitewashing}{Rate \,of \,whitewashing/cm^2}\)

= \(\frac{1232}{2}\)

= 616 cm

^{2}

We know that total surface area of a sphere = 4πr

^{2}

4πr

^{2}= 616

r

^{2}= \(\frac{616*7}{22*4}\)

= 49

Therefore, r = 7cm

Now, volume of a sphere = \(\frac{4}{3} πr^3\)

= \(\frac{4}{3} * \frac{22}{7}\) * 7

^{3}

= \(\frac{4312}{3}\).

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

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