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NCERT · Class 9 · Maths

Chapter 11: Surface Areas and Volumes

Practice each question, then open the answer or worked solution.

Exercise 13.4

9 questions
Q1Written answer

Find the surface area of a sphere of radius

(i) 10.5cm (ii) 5.6cm (iii) 14cm

(Assume π=22/7)

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Worked solution

Formula: Surface area of a sphere (SA) = 4πr2

(i) Radius of a sphere, r = 10.5 cm

SA = 4×(22/7)×10.52 = 1386

Surface area of a sphere is 1386 cm2

(ii) Radius of a sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.62 = 394.24

Surface area of a sphere is 394.24 cm2

(iii) Radius of a sphere, r = 14cm

SA = 4πr2

= 4×(22/7)×(14)2

= 2464

Surface area of a sphere is 2464 cm2

Q2Written answer

Find the surface area of a sphere of diameter

(i) 14cm (ii) 21cm (iii) 3.5cm

(Assume π = 22/7)

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Worked solution

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for the surface area of sphere = 4πr2

= 4×(22/7)×72 = 616

Surface area of a sphere is 616 cm2

(ii) Radius (r) of sphere = 21/2 = 10.5 cm

Surface area of a sphere = 4πr2

= 4×(22/7)×10.52 = 1386

Surface area of a sphere is 1386 cm2

Therefore, the surface area of a sphere having a diameter 21 cm is 1386 cm2

(iii) Radius(r) of a sphere = 3.5/2 = 1.75 cm

Surface area of a sphere = 4πr2

= 4×(22/7)×1.752 = 38.5

Surface area of a sphere is 38.5 cm2

Q3Written answer

Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]

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Worked solution

Radius of the hemisphere, r = 10cm

Formula: Total surface area of the hemisphere = 3πr2

= 3×3.14×102 = 942

The total surface area of the given hemisphere is 942 cm2.

Q4Written answer

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

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Worked solution

Let r1 and r2 be the radii of the spherical balloon and spherical balloon when air is pumped into it, respectively. So,

r1 = 7cm

r2 = 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4πr12/4πr22

= (r1/r2)2

= (7/14)2 = (1/2)2 = ¼

Therefore, the ratio between the surface areas is 1:4.

Q5Written answer

A hemispherical bowl made of brass has an inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7)

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Worked solution

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for the surface area of hemispherical bowl = 2πr2

= 2×(22/7)×(5.25)2 = 173.25

Surface area of the hemispherical bowl is 173.25 cm2

Cost of tin-plating 100 cm2 area = Rs 16

Cost of tin-plating 1 cm2 area = Rs 16 /100

Cost of tin-plating 173.25 cm2 area = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

Q6Written answer

Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

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Worked solution

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr2 = 154

r2 = (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

Q7Written answer

The diameter of the moon is approximately one-fourth of the diameter of the earth.

Find the ratio of their surface areas.

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Worked solution

If the diameter of the earth is said d, then the diameter of the moon will be d/4 (as per the given statement).

Radius of earth = d/2

Radius of moon = ½×d/4 = d/8

Surface area of moon = 4π(d/8)2

Surface area of earth = 4π(d/2)2

Ncert solutions class 9 chapter 13-6

The ratio between their surface areas is 1:16.

Q8Written answer

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)

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Worked solution

Given:

Inner radius of the hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of the hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of the hemispherical bowl = 2πr2, where r is the radius of the hemisphere.

= 2×(22/7)×(5.25)2 = 173.25 cm2

Therefore, the outer curved surface area of the bowl is 173.25 cm2.

Q9Written answer

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in(i) and (ii).

Ncert solutions class 9 chapter 13-7

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Worked solution

(i) Surface area of the sphere = 4πr2, where r is the radius of sphere

(ii) Height of the cylinder, h = r+r =2r

The radius of the cylinder = r

CSA of the cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr2

(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)

= 4πr2/4πr2 = 1/1

The ratio of the areas obtained in (i) and (ii) is 1:1.