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

1. Surface area of a sphere of radius “r” is given by __________

a) 2πr^{2}

b) 4πr^{2}

c) 3πr^{2}

d) πr^{2}

View Answer

Explanation: If a sphere has radius equal to ”r”, then its surface area is given by 4πr

^{2}.

2. Curved surface area of hemisphere of radius “r” is equal to __________

a) 2πr^{2}

b) 4πr^{2}

c) 3πr^{2}

d) πr^{2}

View Answer

Explanation: Hemisphere is created when a sphere is divided in two equal parts.

Hence, the curved surface are of a hemisphere = \(\frac{Surface \,area \,of \,a \,sphere}{2}\)

= \(\frac{4πr^2}{2}\)

= 2πr

^{2}.

3. Total surface are of a hemisphere of radius “r” is equal to __________

a) 2πr^{2}

b) 4πr^{2}

c) 3πr^{2}

d) πr^{2}

View Answer

Explanation: Total surface area of a hemisphere = Curved surface area + Base area

= 2πr

^{2}+ πr

^{2}

= 3πr

^{2}.

4. Total surface area of a hemisphere is 4158cm^{2}, the diameter of the hemisphere is equal to __________ cm. (Take π = \(\frac{22}{7}\))

a) 40

b) 20

c) 21

d) 42

View Answer

Explanation: We know that total surface area of a hemisphere = 3πr

^{2}

3πr

^{2}= 4158

3 * \(\frac{22}{7}\) * r

^{2}= 4158

r

^{2}= 441

Hence, r = 21cm

Therefore, diameter of hemisphere = 2r = 2 * 21 = 42cm.

5. If surface area of a sphere of radius “R” is equal to curved surface area of a hemisphere of radius “r”, what is the ratio of R/r?

a) 1/2

b) 1/√2

c) 2

d) √2

View Answer

Explanation: We know that surface area of a sphere of radius “R” is given by 4πR

^{2}and

curved surface area of hemisphere of radius “r” is equal to 2πr

^{2}.

It is given that surface area of a sphere of radius “R” is equal to curved surface area of a hemisphere of radius “r”.

Hence, 4πR

^{2}= 2πr

^{2}

2R

^{2}= r

^{2}

\(\frac{R^2}{r^2} = \frac{1}{2}\)

\(\frac{R}{r} = \frac{1}{\sqrt{2}}\).

6. A sphere of radius “r” is fitted in a cylinder of height “h” such that the top of the sphere reaches only half the height of cylinder as shown in the figure. What is the ratio of curved surface area of a cylinder to surface area of a sphere?

a) \(\sqrt{2}\):1

b) 4:1

c) 2:1

d) 1:2

View Answer

Explanation: It can be seen that h = 4r … (1)

We know that curved surface area of a cylinder = 2πrh

And surface area of a sphere = 4πr

^{2}

(We know that curved surface area of a cylinder)/(And surface area of a sphere) = 2πrh/(4πr

^{2})

= \(\frac{2πr(4r)}{4πr^2}\) (From result (1))

= \(\frac{2πrh}{4πr^2}\)

= 2/1.

7. A hemispheric dome of radius 3.5m is to be painted at a rate of ₹600/m^{2}. What is the cost of painting it? (Take π = \(\frac{22}{7}\))

a) ₹46200

b) ₹45000

c) ₹47260

d) ₹48375

View Answer

Explanation: We know that curved surface are of a hemisphere = 2πr

^{2}

= 2 * \(\frac{22}{7}\) * (3.5)

^{2}

= 77m

^{2}

Now, the cost of painting 1m

^{2}of area = ₹600

Then, the cost of painting 77m

^{2}of area = 77 * 600

= ₹46200.

8. A balloon’s radius increases 3 times when the air is pumped in it. If the surface area of the balloon after pumping the air is A_{2} and its surface area before pumping the air is A_{1}, then what is the value of \(\frac{A_1}{A_2}\) ?

a) \(\frac{9}{1}\)

b) \(\frac{1}{3}\)

c) \(\frac{3}{1}\)

d) \(\frac{1}{9}\)

View Answer

Explanation: Let the radius of the balloon before pumping the air be “r”, then

A

_{1}= 4πr

^{2}

The radius of the balloon after pumping the air = 3r (given)

Hence, A

_{2}= 4π(3r)

^{2}

Therefore, \(\frac{A_1}{A_2} = \frac{4πr^2}{4π(3r)^2} = \frac{1}{9}\).

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

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