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## 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

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

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)

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.

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.

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.

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.

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.

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).

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