Rbse Solutions for Class 9 Maths Chapter 11 Exercise 11.2 | Surface Area of a Sphere

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.