NCERT Solutions for Class 9 Maths - Surface Areas and Volumes (Exercise 11.2)

Q1. Find the surface area of a sphere of radius:
(i) 10.5cm
(ii) 5.6cm
(iii) 14cm
(Assume π = 22 / 7)
Ans: Formula: Surface area of sphere (SA) = 4πr²
(i) Radius of sphere, r = 10.5 cm
SA = 4 × (22 / 7) × 10.52 = 1386
Surface area of sphere is 1386 cm²
(ii) Radius of sphere, r = 5.6cm
Using formula, SA = 4 × (22 / 7) × 5.62 = 394.24
Surface area of sphere is 394.24 cm²
(iii) Radius of sphere, r = 14cm
SA = 4πr²
= 4 × (22 / 7) × (14)²
= 2464
Surface area of sphere is 2464 cm².

Q2. Find the surface area of a sphere of diameter:
(i) 14cm
(ii) 21cm
(iii) 3.5cm
(Assume π = 22 / 7)
Ans: (i) Radius of sphere, r = diameter / 2 = 14 / 2 cm = 7 cm
Formula for 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 sphere = 4πr²
= 4 × (22 / 7) × 10.52 = 1386
Surface area of a sphere is 1386 cm²
Therefore, the surface area of a sphere having diameter 21cm is 1386 cm²
(iii) Radius(r) of sphere = 3.5 / 2 = 1.75 cm Surface area of sphere = 4πr2
= 4 × (22 / 7) × 1.752 = 38.5
Surface area of a sphere is 38.5 cm².

Q3. Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Ans: Radius of hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3πr²
= 3 × 3.14 × 10² = 942
The total surface area of given hemisphere is 942 cm².

Q4. 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.
Ans: Let r₁ and r₂ be the radii of 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)
= 4r₁² / 4r₂²
= (r₁ / r₂)²
= (7 / 14)² = (1 / 2)² = 1 / 4
Therefore, the ratio between the surface areas is 1 : 4.

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. (Assume π = 22 / 7)
Ans: Inner radius of hemispherical bowl, say r = diameter / 2 = (10.5) / 2 cm = 5.25 cm
Formula for Surface area of hemispherical bowl = 2πr²
= 2 × (22 / 7) × (5.25)² = 173.25
Surface area of 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.

Q6. Find the radius of a sphere whose surface area is 154 cm². (Assume π = 22/7)
Ans: 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.

Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Ans: If diameter of earth is said d, then the diameter of moon will be d / 4 (as per given statement)
Radius of earth = d / 2
Radius of moon = 1 / 2 × d / 4 = d / 8
Surface area of moon = 4π(d / 8)²
Surface area of earth = 4π(d / 2)²
Ratio of their Surace areas = 4π(d / 8)² / 4π (d / 2)² = 4 / 64 = 1 / 16
The ratio between their surface areas is 1 : 16.

Q8. 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)
Ans: Given:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Formula for outer CSA of hemispherical bowl = 2πr², where r is radius of hemisphere
= 2 × (22 / 7) × (5.25)² = 173.25
Therefore, the outer curved surface area of the bowl is 173.25 cm².

Q9. A right circular cylinder just encloses a sphere of radius r (see Fig). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in(i) and (ii).
Ans: (i) Surface area of sphere = 4πr², where r is the radius of sphere
(ii) Height of cylinder, h = r + r = 2r
Radius of cylinder = r
CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr²
(iii) Ratio between areas = (Surface area of sphere) / CSA of Cylinder)
= 4r² / 4r²
= 1 / 1
Ratio of the areas obtained in (i) and (ii) is 1 : 1.