Chapter 21 Surface Area and Volumes of a Sphere RD Sharma Solutions Exercise 21.1 Class 9 Maths
Chapter Name  RD Sharma Chapter 21 Surface Area and Volumes of a Sphere Exercise 21.1 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 21.1 Solutions
1. Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution
(i) Given radius = 10.5 cm
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Solution
(i) Diameter = 14 cm
Radius = Diameter/2 = 14/2 = 7 cm
Radius = Diameter/2 = 14/2 = 7 cm
3. Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (Use Ï€ = 3.14)
Solution
The surface area of the hemisphere = 2Ï€r^{2} .
= 2 × 3.14 × (10)^{2} = 628 cm^{2}
The surface area of solid hemisphere = 3Ï€r^{2}
= 3 × 3.14 × (10)^{2} = 942 cm^{2}
= 2 × 3.14 × (10)^{2} = 628 cm^{2}
The surface area of solid hemisphere = 3Ï€r^{2}
= 3 × 3.14 × (10)^{2} = 942 cm^{2}
4. The surface area of a sphere is 5544 cm^{2} , find its diameter.
Solution
⇒ r = 21 cm.
Diameter = 2 (radius)
= 2(21 cm)
= 42 cm.
= 2(21 cm)
= 42 cm.
5. 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. 4 per 100 cm^{2}.
Solution
Given
Inner diameter of hemisphere bowl = 10.5 cm
Radius = 10.5/2 cm = 5.25 cm
Surface area of hemispherical bowl = 2Ï€r
= 2[22/7] × (5.25)^{2} cm^{2}
= 173.25 cm^{2}
Surface area of hemispherical bowl = 2Ï€r
= 2[22/7] × (5.25)^{2} cm^{2}
= 173.25 cm^{2}
Cost of tin planning 100 cm^{2} area = Rs. 4
Cost of tin planning 173.25 cm^{2} area = Rs.[(4 × 173.25)/100] = Rs. 6.93
Cost of tin planning 173.25 cm^{2} area = Rs.[(4 × 173.25)/100] = Rs. 6.93
Thus, The cost of tin plating the inner side of hemisphere bowl is Rs. 6.93
6. The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs. 2 per sq. m.
Solution
Dome Radius = 63 d m = 6.3 m
Inner S.A of dome = 2Ï€r^{2} = 2 × 22/7 × (6.3)^{2} = 249.48 m^{2}
Now, cost of 1 m^{2} = Rs. 2
∴ Cost of 249. 48 m^{2} = Rs. [2×249.48]
= Rs. 498.96
Now, cost of 1 m^{2} = Rs. 2
∴ Cost of 249. 48 m^{2} = Rs. [2×249.48]
= Rs. 498.96
7. Assuming the earth to be a sphere of radius 6370 km, how many square kilo metres is area of the land, if three  fourth of the earth's surface is covered by water ?
Solution
3/4 th of earth surface is covered by water
∴ 1/4 th earth surface is covered by land
∴ Surface area covered by land = 1/4 × 4Ï€r^{2}
= 1/4 × 4 × 22/7 × (6370)^{2}
= 127527.4 km^{2}
= 1/4 × 4 × 22/7 × (6370)^{2}
= 127527.4 km^{2}
8. A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.
Solution
Given length of the shape = 7 cm
But length = r + r
⇒ 2r = 7 cm
⇒ 2r = 7 cm
⇒ r = 7/2 cm
⇒ r = 3.5 cm
Also; h = r
Total S.A of shape = 2Ï€rh + 2Ï€r^{2} = 2Ï€r ×r + 2Ï€r^{2}
= 2Ï€r^{2} + 2Ï€r^{2}
=4Ï€r^{2}
= 4 × 22/7 × (3.5)^{2}
= 154 cm^{2}
= 2Ï€r^{2} + 2Ï€r^{2}
=4Ï€r^{2}
= 4 × 22/7 × (3.5)^{2}
= 154 cm^{2}
9. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at Rs. 7 per 100 cm^{2} .
Solution
Diameter of cone = 16 cm
∴ Radius of cone 8 cm
Height of cone = 15 cm
10. A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m^{2} .
Solution
11. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the
ratio of their surface areas.
ratio of their surface areas.
Solution
Let the diameter of the earth is d then, diameter of moon will be d/4
Radius of earth = d/2
Radius of moon = (d/2)/4 = d/8
Radius of earth = d/2
Radius of moon = (d/2)/4 = d/8
12. A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 cm, find the cost of painting it, given the cost of painting is Rs. 5 per l00 cm^{2} .
Solution
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 = 17.6 m.
Therefore, 17.6 = 2Ï€r
2 ×22/7 × r = 17.6 m.
Therefore, 17.6 = 2Ï€r
2 ×22/7 × r = 17.6 m.
So, the radius of the dome = 17.6 × 7/(2×22) = 2.8m
The curved surface area of the dome = 2Ï€r^{2}
= 2 × 22/7 × 2.8 × 2.8 cm^{2} =49.28 m^{2}
Now, cost of painting 100 cm^{2} is Rs. 5.
So, cost of painting 1m^{2} = Rs. 500
Therefore, cost of painting the whole dome
The curved surface area of the dome = 2Ï€r^{2}
= 2 × 22/7 × 2.8 × 2.8 cm^{2} =49.28 m^{2}
Now, cost of painting 100 cm^{2} is Rs. 5.
So, cost of painting 1m^{2} = Rs. 500
Therefore, cost of painting the whole dome
= Rs. 500 × 49.28 = Rs. 24640
13. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed in small supports as shown in Fig. below. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1 .5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm^{2} and black paint costs 5 paise per cm^{2} .
Solution
Wooden sphere radius= (21/2) cm = 10.5 cm
Surface area of a wooden sphere
= 66 cm^{2} Area of circular end of cylindrical support = Ï€r^{2} = [22/7(1.5)^{2} ] = 7.07 cm^{2}
Area to be painted silver = [8 × (1386  7.07)] cm^{2}
= 8(1378.93) cm^{2}
= 8(1378.93) cm^{2}
= 11031.44 cm^{2}
Cost occurred in painting silver color
= Rs. (11031.44 × 0.25) = Rs. 2757.86
Area to painted black = (8 × 66) cm^{2} = 528 cm^{2}
Area to painted black = (8 × 66) cm^{2} = 528 cm^{2}
Cost occurred in painting black color = Rs. (528 × 0.05) = Rs. 26.40
∴ Total cost occurred in painting = Rs. (2757.86 + 26.40) = Rs. 2784.26
∴ Total cost occurred in painting = Rs. (2757.86 + 26.40) = Rs. 2784.26