Welcome to our comprehensive guide for MP Board Class 9 Mathematics! In this post, we provide detailed, step-by-step solutions for Exercise 11.2 (Chapter 11: Surface Areas and Volumes). This exercise focuses on finding the surface area of spheres and hemispheres.

Exercise 11.2 Step-by-Step Solutions

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

(i) 10.5 cm
Solution:
Radius (r) = 10.5 cm
Surface Area = 4πr²
= 4 × (22/7) × (10.5)²
= 4 × (22/7) × 110.25
= 4 × 22 × 15.75 = 1386 cm²

(ii) 5.6 cm
Solution:
Radius (r) = 5.6 cm
Surface Area = 4πr²
= 4 × (22/7) × 5.6 × 5.6
= 4 × 22 × 0.8 × 5.6 = 394.24 cm²

(iii) 14 cm
Solution:
Radius (r) = 14 cm
Surface Area = 4πr²
= 4 × (22/7) × 14 × 14
= 4 × 22 × 2 × 14 = 2464 cm²

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

(i) 14 cm
Solution:
Diameter = 14 cm → Radius (r) = 7 cm
Surface Area = 4 × (22/7) × 7 × 7 = 616 cm²

(ii) 21 cm
Solution:
Diameter = 21 cm → Radius (r) = 21/2 cm = 10.5 cm
Surface Area = 4 × (22/7) × 10.5 × 10.5 = 1386 cm²

(iii) 3.5 m
Solution:
Diameter = 3.5 m → Radius (r) = 3.5/2 m = 1.75 m
Surface Area = 4 × (22/7) × 1.75 × 1.75 = 38.5 m²

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

Solution:
Radius (r) = 10 cm
Total Surface Area of Hemisphere = 3πr²
= 3 × 3.14 × (10)²
= 3 × 3.14 × 100 = 942 cm²

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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.

Solution:
Initial radius (r₁) = 7 cm
Final radius (r₂) = 14 cm
Ratio of Surface Areas = (4πr₁²) / (4πr₂²) = (r₁/r₂)²
= (7 / 14)² = (1 / 2)² = 1/4
Therefore, the required ratio is 1 : 4.

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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.

Solution:
Inner diameter = 10.5 cm → Inner radius (r) = 5.25 cm
Inner Curved Surface Area = 2πr²
= 2 × (22/7) × 5.25 × 5.25 = 173.25 cm²
Cost of tin-plating per 100 cm² = ₹16
Total Cost = (173.25 / 100) × 16 = ₹27.72

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

Solution:
Surface Area = 4πr² = 154
4 × (22/7) × r² = 154
r² = (154 × 7) / (4 × 22) = 12.25
r = √12.25 = 3.5 cm

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

Solution:
Let the diameter of the Earth be 'd'. Radius of Earth (r_e) = d/2.
Diameter of Moon = d/4. Radius of Moon (r_m) = d/8.
Ratio of Surface Areas = (4πr_m²) / (4πr_e²) = (r_m / r_e)²
= [(d/8) / (d/2)]² = (2/8)² = (1/4)² = 1/16
Therefore, the required ratio is 1 : 16.

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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.

Solution:
Inner radius (r) = 5 cm
Thickness = 0.25 cm
Outer radius (R) = 5 + 0.25 = 5.25 cm
Outer CSA = 2πR²
= 2 × (22/7) × 5.25 × 5.25 = 173.25 cm²

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Question 9: A right circular cylinder just encloses a sphere of radius r. Find:

(i) surface area of the sphere,
Solution: Surface Area of sphere = 4πr²

(ii) curved surface area of the cylinder,
Solution: Height of cylinder (h) = 2r, Radius = r.
CSA of cylinder = 2πrh = 2πr(2r) = 4πr²

(iii) ratio of the areas obtained in (i) and (ii).
Solution: Ratio = 4πr² / 4πr² = 1/1 = 1 : 1

Master MP Board Class 9 Maths Chapter 11 Exercise 11.2 with step-by-step solutions on surface area of spheres and hemispheres.