# R.D. Sharma Solutions Class 9th: Ch 21 Surface Area and Volume of a Sphere Exercise 21.1

#### Chapter 21 Surface Area and Volume of a Sphere R.D. Sharma Solutions for Class 9th Exercise 21.1

**Exercise 21.1**

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

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

(i) Given Radius= 10.5 cm

Surface area= 4πr

= 4×22/7×(10.5)

= 1386cm

(ii) Given radius= 5.6cm

Surface area= 4πr

= 4×22/7×(5.6)

= 394.24cm

(iii) Given radius= 14cm

Surface area= 4πr

= 4×22/7×(14)

= 2464cm

2. Find the surface area of a sphere of diameter.

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

(i) Given Diameter = 14 cm

Radius = Diameter/2

= 14/2 = 7cm

Surface area = 4πr

= 4×22/7×(7)

= 616cm

(ii) Given Diameter = 21cm

Radius = Diameter/2

= 2/12 = 10.5cm

Surface area = 4πr

= 4×22/7×(10.5)2

= 1386cm

(iii) Given diameter = 3.5cm

Radius = Diameter/2

= 3.5/2 = 1.75cm

Surface area = 4πr

= 4×22/7×(1.75)2

= 38.5cm

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

The surface area of the hemisphere= 2πr

= 2×3.14×(10)

= 628cm

The surface area of solid hemisphere= 2πr

= 3×3.14×(10)

= 942cm

4. The surface area of a sphere is 5544 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

Inner diameter of hemispherical bowl= 10.5cm

Radius= 10.5/2 = 5.25cm

Surface area of hemispherical bowl= 2πr

=2×3.14×(5.25)

=173.25cm(ii) 5.6 cm

(iii) 14 cm

**Solution**(i) Given Radius= 10.5 cm

Surface area= 4πr

^{2}= 4×22/7×(10.5)

^{2}= 1386cm

^{2}(ii) Given radius= 5.6cm

Surface area= 4πr

^{2}= 4×22/7×(5.6)

^{2}= 394.24cm

^{2}(iii) Given radius= 14cm

Surface area= 4πr

^{2}= 4×22/7×(14)

^{2}= 2464cm

^{2}2. Find the surface area of a sphere of diameter.

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

**Solution**(i) Given Diameter = 14 cm

Radius = Diameter/2

= 14/2 = 7cm

Surface area = 4πr

^{2}= 4×22/7×(7)

^{2}= 616cm

^{2}(ii) Given Diameter = 21cm

Radius = Diameter/2

= 2/12 = 10.5cm

Surface area = 4πr

^{2}= 4×22/7×(10.5)2

= 1386cm

^{2}(iii) Given diameter = 3.5cm

Radius = Diameter/2

= 3.5/2 = 1.75cm

Surface area = 4πr

^{2}= 4×22/7×(1.75)2

= 38.5cm

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

**Solution**The surface area of the hemisphere= 2πr

^{2}= 2×3.14×(10)

^{2}= 628cm

^{2}The surface area of solid hemisphere= 2πr

^{2}= 3×3.14×(10)

^{2}= 942cm

^{2}4. The surface area of a sphere is 5544 cm

^{2}, find its diameter .**Solution**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**Inner diameter of hemispherical bowl= 10.5cm

Radius= 10.5/2 = 5.25cm

^{2}Surface area of hemispherical bowl= 2πr

^{2}=2×3.14×(5.25)

^{2}^{2}

Cost of tin plating 100cm

^{2}area= Rs.4

Cost of tin plating 173.25cm

^{2}area= Rs. 4×173.25/100 = Rs. 6.93

Thus, the cost of tin plating the inner side of hemispherical 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= 63dm= 6.3m

Inner surface area of dome= 2πr

^{2}

= 2×3.14×(6.3)

^{2}

= 249.48m

^{2}

Now, cost of 1m

^{2}= Rs.2

Therefore cost of 249.48m

^{2}= Rs. (249.48×2) = Rs.498.96 .

7. Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth's surface is covered by water?

**Solution**

34th of earth surface is covered by water

Therefore 1/4th of earth surface is covered by land

Therefore Surface area covered by land= 1/4×4πr

^{2}

= 1/4×4×22/7×(6370)

^{2}

=127527400km

^{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**

But length = r+r

2r = 7cm

r = 3.5cm

Also; h = r

Total surface area of shape = 2πrh+2πr

^{2}

= 2πrr+2πr

^{2}

= 2πr

^{2}+2πr

^{2}

= 4×2/27×(3.5)

^{2}= 154cm

^{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**

10. A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder by 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**

Diameter of a cylinder= 1.4m

Therefore radius of cylinder= 1.4/2 = 0.7m

Height of cylinder = 8m

Therefore surface area of tank= 2πrh+2πr

^{2}

= (2 × 2/27 × 0.7 × 8) + [2 × 22/7 × (0.7)

^{2}]

=176/5+77/25 = 38.28m

^{2}

Now cost of 1m

^{2}= Rs.10

Therefore, cost of 38.28m

^{2}= Rs. 382.80