NCERT Solutions for Class 9 Maths Exercise 11.2 Chapter 11

NCERT Solutions for Class 9 Maths Chapter 11 – Surface Areas and Volumes (Exercise 11) deal with the surface area of a sphere and a hemisphere. NCERT stands for National Council of Educational Research and Training. The chapter teaches how to find the area of the outer surface of round solids. This exercise adds just two formulas: the surface area of a sphere is 4πr², and the curved surface area of a hemisphere is 2πr². Here r is the radius of the solid.

These nine questions ask you to find the surface area from a radius, find it from a diameter, find the radius when the area is given, compare two surface areas as a ratio, and work out a cost from an area. Every answer is solved step by step in easy English, so students can revise quickly, parents can help at home, and teachers can use them in class. A free printable PDF of this exercise is also available for offline practice.

NCERT Solutions for Class 9 Maths Exercise 11.2 - Questions and Answers

Q1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Answer:
Surface area of a sphere = 4πr². Take π = 22/7.

(i) r = 10.5 cm
Surface area = 4 × (22/7) × 10.5 × 10.5
= 4 × 22 × 1.5 × 10.5 (because 10.5 ÷ 7 = 1.5)
= 1386 cm²

(ii) r = 5.6 cm
Surface area = 4 × (22/7) × 5.6 × 5.6
= 4 × 22 × 0.8 × 5.6 (because 5.6 ÷ 7 = 0.8)
= 394.24 cm²

(iii) r = 14 cm
Surface area = 4 × (22/7) × 14 × 14
= 4 × 22 × 2 × 14 (because 14 ÷ 7 = 2)
= 2464 cm²

Check: 14 cm is 2.5 times 5.6 cm, and 2464 ÷ 394.24 = 6.25, which is 2.5 × 2.5. Correct.

Final answer: (i) 1386 cm² (ii) 394.24 cm² (iii) 2464 cm²

Q2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m

Answer:
Radius = diameter ÷ 2. Surface area of a sphere = 4πr². Take π = 22/7.

(i) Diameter = 14 cm, so r = 7 cm
Surface area = 4 × (22/7) × 7 × 7
= 4 × 22 × 7
= 616 cm²

(ii) Diameter = 21 cm, so r = 10.5 cm
Surface area = 4 × (22/7) × 10.5 × 10.5
= 4 × 22 × 1.5 × 10.5
= 1386 cm²

(iii) Diameter = 3.5 m, so r = 1.75 m
Surface area = 4 × (22/7) × 1.75 × 1.75
= 4 × 22 × 0.25 × 1.75 (because 1.75 ÷ 7 = 0.25)
= 38.5 m²

Check: in part (ii) the radius is 10.5 cm, the same as Q1 (i), and the answer 1386 cm² matches.

Final answer: (i) 616 cm² (ii) 1386 cm² (iii) 38.5 m²

Q3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Answer:
A hemisphere has a curved part and a flat circular base.
Curved surface area = 2πr²
Area of the circular base = πr²
Total surface area = 2πr² + πr² = 3πr²
Here r = 10 cm and π = 3.14.
Total surface area = 3 × 3.14 × 10 × 10
= 3 × 3.14 × 100
= 942 cm²

Check: curved part = 2 × 3.14 × 100 = 628 cm², base = 3.14 × 100 = 314 cm², and 628 + 314 = 942 cm².

Final answer: 942 cm²

Q4. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:
Let the radius of the earth be R.
Then the diameter of the earth = 2R.
Diameter of the moon = (1/4) × 2R = R/2.
So the radius of the moon = (1/2) × (R/2) = R/4.
Surface area of the earth = 4πR²
Surface area of the moon = 4π(R/4)² = 4πR²/16
Ratio = (4πR²/16) : 4πR²
= 1 : 16

Check: the radii are in the ratio 1 : 4, and surface area depends on the square of the radius, so the areas are in the ratio 1² : 4² = 1 : 16.

Final answer: 1 : 16

Q5. 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 Rs 16 per 100 cm².

Answer:
Inner diameter = 10.5 cm, so inner radius r = 10.5 ÷ 2 = 5.25 cm.
Only the inside curved part is plated.
Inner curved surface area = 2πr²
= 2 × (22/7) × 5.25 × 5.25
= 2 × 22 × 0.75 × 5.25 (because 5.25 ÷ 7 = 0.75)
= 173.25 cm²
Cost of 100 cm² = Rs 16, so cost of 1 cm² = Rs 16/100.
Total cost = 173.25 × 16/100
= 2772/100
= Rs 27.72

Check: 100 cm² costs Rs 16, so about 1.73 times that area should cost about 1.73 × 16 = Rs 27.7. Correct.

Final answer: Rs 27.72

Q6. Find the radius of a sphere whose surface area is 154 cm².

Answer:
Let the radius be r cm.
Surface area of a sphere = 4πr²
So, 4 × (22/7) × r² = 154
88r²/7 = 154
r² = 154 × 7/88
r² = 1078/88
r² = 12.25
r = √12.25
r = 3.5 cm

Check: 4 × (22/7) × 3.5 × 3.5 = 4 × 22 × 0.5 × 3.5 = 154 cm². Correct.

Final answer: 3.5 cm

Q7. 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:
Inner radius = 5 cm
Thickness = 0.25 cm
Outer radius = inner radius + thickness
= 5 + 0.25
= 5.25 cm
Outer curved surface area = 2πr²
= 2 × (22/7) × 5.25 × 5.25
= 2 × 22 × 0.75 × 5.25
= 173.25 cm²

Check: this is the same radius as the bowl in Q5, so the same area of 173.25 cm² is expected.

Final answer: 173.25 cm²

Q8. A right circular cylinder just encloses a sphere of radius r. 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:
The sphere fits exactly inside the cylinder.
So the radius of the cylinder R = r, and the height of the cylinder h = diameter of the sphere = 2r.

(i) Surface area of the sphere = 4πr²

(ii) Curved surface area of the cylinder = 2πRh
= 2π × r × 2r
= 4πr²

(iii) Ratio = 4πr² : 4πr²
= 1 : 1

Check: both areas came out as 4πr², so the ratio has to be 1 : 1.

Final answer: (i) 4πr² (ii) 4πr² (iii) 1 : 1

Q9. 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:
Surface area of a sphere = 4πr².
First radius r₁ = 7 cm, second radius r₂ = 14 cm.
Ratio = 4πr₁² : 4πr₂²
= r₁² : r₂²
= 7 × 7 : 14 × 14
= 49 : 196
= 1 : 4

Check: with π = 22/7, the two areas are 616 cm² and 2464 cm², and 2464 ÷ 616 = 4. Correct.

Final answer: 1 : 4

NCERT Solutions for Class 9 Maths Chapter 11 – All Exercises

• NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1
• NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.2 (you are here)
• NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.3
• NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.4