## Chapter 13 Surface Areas and Volumes Exercise 13.4

Question 1: Find the surface area of a sphere of radius:

i) 10.5 cm

ii) 5.6 cm

iii) 14 cm

Answer:

i) Given

Radius of the sphere = 10.5 cm

Surface area of sphere = ?

TSA of sphere = 4πr²

= 4 × 22/7 × 10.5 × 10.5

= 4 × 22 × 1.5 × 10.5

= 1386 cm²

ii) Given

Radius of the sphere = 5.6 cm

Surface area of sphere = ?

TSA of sphere = 4πr²

= 4 × 22/7 × 5.6 × 5.6

= 4 × 22 × 0.8 × 5.6

= 394.24 cm²

iii) Given

Radius of the sphere = 14 cm

Surface area of sphere = ?

TSA of sphere = 4πr²

= 4 × 22/7 × 14 × 14

= 4 × 22 × 2 × 14

= 2464 cm²

Question 2: Find the surface area of a sphere of diameter:

i) 14 cm

ii) 21 cm

iii) 3.5 m

Answer:

i) Given

Diameter of the sphere = 14 cm

Surface area of sphere = ?

Radius = Diameter/2 = 14/2 = 7 cm

TSA of sphere = 4πr²

= 4 × 22/7 × 7 × 7

= 4 × 22 × 7

= 616 cm²

ii) Given

Diameter of the sphere = 21 cm

Surface area of sphere = ?

Radius = Diameter/2 = 21/2 cm

TSA of sphere = 4πr²

= 4 × 22/7 × 21/2 × 21/2

= 22 × 3 × 21

= 1386 cm²

iii) Given

Diameter of the sphere = 3.5 m

Surface area of sphere = ?

Radius = Diameter/2 = 3.5/2 m

TSA of sphere = 4πr²

= 4 × 22/7 × 3.5/2 × 3.5/2

= 22 × 0.5 × 3.5

= 38.5 m²

Question 3: Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer:

Given

Radius of hemisphere = 10 cm

Total Surface Area of hemisphere = ?

TSA of hemisphere = 3πr²

= 3 × 3.14 × 10 × 10

= 942 cm²

Therefore, the TSA of the hemisphere is 942 cm².

Question 4: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer:

Given

Radius of balloon before air r1 = 7 cm

Radius of balloon after air r2 = 14 cm

Ratio of TSA = ?

= TSA of balloon before air/TSA of balloon after air

= 4π(7)²/4π(14²)

= (1)²/(2)²

= 1/4

= 1:4

Therefore, the ratio of surface areas of the balloon in the two cases is 1:4.

Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Answer:

Given

Inner diameter of hemispherical bowl = 10.5 cm

Inner radius of hemispherical bowl = 10.5/2 cm

Inner CSA of hemisphere = 2πr²

= 2 × 22/7 × 10.5/2 × 10.5/2

= 11 × 1.5 × 10.5

= 173.25 cm²

Cost of tin-plating per 100 cm² = ₹16

Cost of tin-plating 173.25 cm² = ?

= 173.25 × 16/100

= 2772/100

= ₹27.72

Therefore, the cost of tin-plating the bowl is ₹27.72.

Question 6: Find the radius of a sphere whose surface area is 154 cm².

Answer:

Given

TSA of sphere = 154 cm²

Radius of the sphere = ?

TSA of sphere = 4πr²

154 = 4 × 22/7 × r²

154 = 88r²/7

154 × 7/88 = r²

7 × 7/4 = r²

r² = 49/4

r = √49/4

r = 7/2 = 3.5 cm

Therefore, the radius of the sphere is 3.5 cm.

Question 7: The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

Given

Diameter of the moon is 1/4 of the diameter of the earth.

Ratio of their surface areas = ?

Let ‘d’ be the diameter of the moon and ‘D’ be the diameter of earth.

d = 1/4 D

2r = 1/4 × 2R

4r = R

The ratio of their surface area = ?

= 4πr²/4πR²

= r²/R²

= r²/(4r)²

= r²/16r²

= 1/16

= 1:16

Therefore, the ratio of their surface areas is 1:16.

Question 8: A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer:

Given

Thickness of the bowl = 0.25 cm

Inner radius of the bowl = 5 cm

Outer curved surface area of the bowl = ?

Outer radius = Inner radius + Thickness = 5 + 0.25 = 5.25 cm

CSA of hemisphere = 2πr²

= 2 × 22/7 × 5.25 × 5.25

= 2 × 22 × 0.75 × 5.25

= 173.25 cm²

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

Question 9: 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).

Answer:

Given

Diameter of sphere = height of cylinder

2r = h

i) Surface area of the sphere = 4πr²

ii) CSA of cylinder = 2πrh

= 2πr(2r)

= 4πr²

iii) Ratio of their area = 4πr²/4πr² = 1:1

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