Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1. Find the surface area of a sphere of radius:

(i) 10.5cm (ii) 5.6cm (iii) 14cm

Sol.

(i) Given Radius = 10.5 cm

Surface area =  4πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(10.5)2

= 1386cm2

 

(ii) Given radius = 5.6cm

Surface area =  4πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(5.6)2

= 394.24cm2

 

(iii) Given radius = 14cm

Surface area =  4πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(14)2

= 2464cm2

 

Q2. Find the surface area of a sphere of diameter:

(i) 14cm (ii) 21cm (iii) 3.5cm

Sol.

(i) Given Diameter = 14 cm

Radius = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics= 7cm

Surface area =  4πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(7)2

= 616cm2

 

(ii) Given Diameter = 21cm

Radius = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 10.5cm

Surface area =  4πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(10.5)2

= 1386cm2

 

(iii) Given diameter = 3.5cm

Radius = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 1.75cm

Surface area =  4πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(1.75)2

= 38.5cm2

 

Q3. Find the total surface area of a hemisphere and a solid hemisphere each of radius 10cm (∏ = 3.14)

Sol.

The surface area of the hemisphere =  2πr2

=  2×3.14×(10)2

= 628cm2

The surface area of solid hemisphere =   2πr2

=  3×3.14×(10)2

= 942cm2

 

Q4. The surface area of a sphere is 5544cm2, find its diameter.

Sol.

Surface area of a sphere is 5544cm2

4πr2 = 5544

4×3.14×(r)2 = 5544

r2 Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r = 21cm

Diameter = 2(radius)

= 2(21)  = 42cm

 

Q5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin plating it on the inside at the rate of Rs.4 per 100cm2.

Sol.

Inner diameter of hemispherical bowl = 10.5cm

Radius =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 5.25cm

Surface area of hemispherical bowl =  2πr2

= 2×3.14×(5.25)2

= 173.25cm2

Cost of tin plating 100cm2 area = Rs.4

Cost of tin plating 173.25cm2 area = Rs. Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  = Rs. 6.93

Thus, the cost of tin plating the inner side of hemispherical bowl is Rs.6.93

 

Q6. The dome of a building is in the form of a hemisphere. Its radius is 63dm. Find the cost of painting it at the rate of Rs.2 per sq m.

Sol.

Dome radius = 63dm = 6.3m

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Inner surface area of dome =  2πr2

= 2×3.14×(6.3)2

= 249.48m2

Now, cost of 1m2 = Rs.2

Therefore cost of 249.48m2 = Rs. (249.48×2)  = Rs.498.96

 

Q7. Assuming the earth to be a sphere of radius 6370km, how many square kilometers is the area of the land if three-fourths of the earth’s surface is covered by water?

Sol.

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics of earth surface is covered by water

Therefore Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics of earth surface is covered by land

Therefore Surface area covered by land =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×4πr2

 Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 127527400km2

 

Q8. A cylinder of same height and radius is placed on top of a hemisphere. Find the curved surface area of the shape if the length of the shape is 7cm.

Sol.

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Given length of the shape = 7cm

But length = r+r

2r = 7cm

r = 3.5cm

Also; h = r

Total surface area of shape =  2πrh+2πr2

= 2πrr+2πr2

=  2πr2+2πr2

= 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(3.5)2  = 154cm2

 

Q9. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16cm and its height is 15cm. Find the cost of painting the toy at Rs.7 per 100cm2

Sol.

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Diameter of cone = 16cm

Radius of cone = 8cm

Height of cone = 15cm

Slant height of cone = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 17cm

Therefore Total curved surface area of toy

=  πrl+2πr2

= Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×8×17+2×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×82

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm2

Now, cost of 100cm2 = Rs.7

1cm2 = Rs.Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Hence cost ofEx-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= Rs. 58.08

 

Q10. a storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4m and its length is 8m, find the cost of painting it on the outside at the rate of Rs.10 per m2.

Sol.

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Diameter of a cylinder = 1.4m

Therefore radius of cylinder =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 0.7m

Height of cylinder = 8m

Therefore surface area of tank =  2πrh+2πr2

= 2×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×0.7×8+2×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(0.7)2

= Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  = 38.28m2

Now cost of 1m2 = Rs.10

Therefore cost of 38.28m2 = Rs. 382.80

 

Q11. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Sol.

Let the diameter of the earth be d

Then,

Diameter of moon will beEx-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Radius of earth =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Radius of moon =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Surface area of moon =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Surface area of earth =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Required ratio =  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

  Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Thus the required ratio of the surface areas is Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q12. A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6cm, find the cost of painting it, given the cost of painting is Rs.5 per 100cm2

Sol.

Given that only the rounded surface of the dome to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done.

Now, circumference of the dome = 17.6cm

Therefore 2πr = 17.6

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×r = 17.6m

So, the radius of the dome =  2πr2

=  2×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×2.8×2.8

= 49.28m2

Cost of painting 100cm2 is Rs.5

So, the cost of painting 1m2 = Rs.500

Therefore the cost of painting the whole dome = Rs. 500×49.28

= Rs. 24640

 

Q13. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2.

Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Sol.

Wooden sphere radius = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 10.5cm

Surface area of a wooden sphere =  4πr2 = 4×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(10.5)2 = 1386cm2

Radius r of cylindrical support = 1.5cm

Height h of cylindrical support = 7cm

Curved surface area of cylindrical support =  2πrh = 2×Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×1.5×7 = 66cm2

Area of circular end of cylindrical support =  πr2 = Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(1.5)2 = 7.07cm2

Area to be painted silver =  8(1386−7.07)cm2 = 8(1378.93)cm2 = 11031.44cm2

Cost occurred in painting silver colour =  (11034.44×0.25) = Rs.2757.86

Area to be painted black =  (8×66)cm2 = 528cm2

Cost occurred in painting black colour = (528×5.05) = Rs.26.40

Therefore total cost in painting = Rs.2757.86 + Rs.26.40 = Rs.2784.26

The document Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-21.1, Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. How do you calculate the surface area of a sphere?
Ans. The surface area of a sphere can be calculated using the formula A = 4πr², where A represents the surface area and r is the radius of the sphere.
2. How do you find the volume of a sphere?
Ans. The volume of a sphere can be calculated using the formula V = (4/3)πr³, where V represents the volume and r is the radius of the sphere.
3. Can you provide an example of calculating the surface area of a sphere?
Ans. Sure! Let's say the radius of a sphere is 5 cm. To calculate its surface area, we use the formula A = 4πr². Plugging in the value, we get A = 4π(5)² = 4π(25) = 100π cm².
4. How can I find the radius if I know the volume of a sphere?
Ans. To find the radius of a sphere if you know its volume, you can rearrange the volume formula V = (4/3)πr³ to solve for r. Taking the cube root of both sides, you get r = (3V/4π)^(1/3).
5. Is there a connection between the surface area and volume of a sphere?
Ans. Yes, there is a connection between the surface area and volume of a sphere. The surface area represents the total area covered by the sphere, while the volume represents the amount of space enclosed by the sphere. As the radius of a sphere increases, both the surface area and volume increase.
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