Rbse Solutions for Class 9 Maths Chapter 11 Exercise 11.2 | Surface Area of a Sphere
1. Find the surface area of a sphere of radius
(i) 10.5cm (ii) 5.6cm (iii) 14cm
(Assume π=22/7)
Solution:
Formula: Surface area of a sphere (SA) = 4πr2
(i) Radius of a sphere, r = 10.5 cm
SA = 4×(22/7)×10.52 = 1386
Surface area of a sphere is 1386 cm2
(ii) Radius of a sphere, r = 5.6cm
Using formula, SA = 4×(22/ 7)×5.62 = 394.24
Surface area of a sphere is 394.24 cm2
(iii) Radius of a sphere, r = 14cm
SA = 4πr2
= 4×(22/7)×(14)2
= 2464
Surface area of a sphere is 2464 cm2
2. Find the surface area of a sphere of diameter
(i) 14cm (ii) 21cm (iii) 3.5cm
(Assume π = 22/7)
Solution:
(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm
Formula for the 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 a sphere = 4πr2
= 4×(22/7)×10.52 = 1386
Surface area of a sphere is 1386 cm2
Therefore, the surface area of a sphere having a diameter 21 cm is 1386 cm2
(iii) Radius(r) of a sphere = 3.5/2 = 1.75 cm
Surface area of a sphere = 4πr2
= 4×(22/7)×1.752 = 38.5
Surface area of a sphere is 38.5 cm2
3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]
Solution:
Radius of the hemisphere, r = 10cm
Formula: Total surface area of the hemisphere = 3πr2
= 3×3.14×102 = 942
The total surface area of the given hemisphere is 942 cm2.
4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Let r1 and r2 be the radii of the 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)
= 4πr12/4πr22
= (r1/r2)2
= (7/14)2 = (1/2)2 = ¼
Therefore, the ratio between the surface areas is 1:4.
5. A hemispherical bowl made of brass has an inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7)
Solution:
Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm
Formula for the surface area of hemispherical bowl = 2πr2
= 2×(22/7)×(5.25)2 = 173.25
Surface area of the 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.
6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)
Solution:
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.
7. The diameter of the moon is approximately one-fourth of the diameter of the earth.
Find the ratio of their surface areas.
Solution:
If the diameter of the earth is said d, then the diameter of the moon will be d/4 (as per the given statement).
Radius of earth = d/2
Radius of moon = ½×d/4 = d/8
Surface area of moon = 4π(d/8)2
Surface area of earth = 4π(d/2)2
$$\text{Ratio of their Surface areas} = \frac{4\pi \left(\frac{d}{8}\right)^2}{4\pi \left(\frac{d}{2}\right)^2}$$
This implies:
- Sphere 1 (Numerator): Has a radius of $r_1 = \frac{d}{8}$.
- Sphere 2 (Denominator): Has a radius of $r_2 = \frac{d}{2}$.
2. Simplifying the Ratio
The common term $4\pi$ cancels out:
$$\text{Ratio} = \frac{\left(\frac{d}{8}\right)^2}{\left(\frac{d}{2}\right)^2}$$
Simplify the squares:
$$\text{Ratio} = \frac{\frac{d^2}{64}}{\frac{d^2}{4}}$$
The term $d^2$ cancels out:
$$\text{Ratio} = \frac{\frac{1}{64}}{\frac{1}{4}} = \frac{1}{64} \times \frac{4}{1} = \frac{4}{64}$$
3. Final Result
Reducing the fraction:
$$\mathbf{\text{Ratio} = \frac{1}{16}}$$
Conclusion: The calculation correctly finds the ratio of the surface area of a sphere with radius $\frac{d}{8}$ to the surface area of a sphere with radius $\frac{d}{2}$, resulting in a ratio of 1:16.
The ratio between their surface areas is 1:16.
8. 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)
Solution:
Given:
Inner radius of the hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of the hemispherical bowl = (5+0.25) cm = 5.25 cm
Formula for outer CSA of the hemispherical bowl = 2πr2, where r is the radius of the hemisphere.
= 2×(22/7)×(5.25)2 = 173.25 cm2
Therefore, the outer curved surface area of the bowl is 173.25 cm2.
9. A right circular cylinder just encloses a sphere of radius r (see fig. 11.10). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in(i) and (ii).
Solution:
(i) Surface area of the sphere = 4πr2, where r is the radius of sphere
(ii) Height of the cylinder, h = r+r =2r
The radius of the cylinder = r
CSA of the cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr2
(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)
= 4πr2/4πr2 = 1/1
The ratio of the areas obtained in (i) and (ii) is 1:1.
