RD Sharma Solutions Class 9 Surface Area And Volume Of Sphere Exercise 21.1

RD Sharma Solutions Class 9 Chapter 21 Exercise 21.1

RD Sharma Class 9 Solutions Chapter 21 Ex 21.1 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= \(4\pi r^{2}\)

\(=4\times \frac{22}{7}\times \left ( 10.5 \right )^{2}\)

=1386\(cm^{2}\)

 

(ii) Given radius= 5.6cm

Surface area= \(4\pi r^{2}\)

\(=4\times \frac{22}{7}\times \left ( 5.6 \right )^{2}\)

=394.24\(cm^{2}\)

(iii) Given radius= 14cm

Surface area= \(4\pi r^{2}\)

\(=4\times \frac{22}{7}\times \left ( 14 \right )^{2}\)

=2464\(cm^{2}\)

 

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

 

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

 

Sol.

(i) Given Diameter= 14 cm

Radius=\(\frac{Diameter}{2}\)

\(=\frac{14}{2}=7cm\)

Surface area= \(4\pi r^{2}\)

\(=4\times \frac{22}{7}\times \left ( 7 \right )^{2}\)

=616\(cm^{2}\)

 

(ii) Given Diameter= 21cm

Radius=\(\frac{Diameter}{2}\)

\(=\frac{21}{2}=10.5cm\)

Surface area= \(4\pi r^{2}\)

\(=4\times \frac{22}{7}\times \left ( 10.5 \right )^{2}\)

=1386\(cm^{2}\)

 

(iii) Given diameter= 3.5cm

Radius=\(\frac{Diameter}{2}\)

\(=\frac{3.5}{2}=1.75cm\)

Surface area= \(4\pi r^{2}\)

\(=4\times \frac{22}{7}\times \left ( 1.75 \right )^{2}\)

= 38.5\(cm^{2}\)

 

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

 

Sol.

The surface area of the hemisphere= \(2\pi r^{2}\)

= \(2\times 3.14\times \left ( 10 \right )^{2}\)

= 628\(cm^{2}\)

The surface area of solid hemisphere=  \(2\pi r^{2}\)

= \(3\times 3.14\times \left ( 10 \right )^{2}\)

=942\(cm^{2}\)

 

Q4. The surface area of a sphere is 5544\(cm^{2}\), find its diameter.

 

Sol.

Surface area of a sphere is 5544\(cm^{2}\)

\(4\pi r^{2}\)= 5544

\(4\times 3.14\times \left ( r \right )^{2}\)=5544

\(r^{2}=\frac{5544\times 7}{88}\)

\(r= \sqrt{21cm\times 21cm}\)

r=21cm

Diameter=\(2\left ( radius \right )\)

=\(2\left ( 21 \right )\) = 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 100\(cm^{2}\).

 

Sol.

Inner diameter of hemispherical bowl= 10.5cm

Radius= \(\frac{10.5}{2}= 5.25cm\)

Surface area of hemispherical bowl= \(2\pi r^{2}\)

=\(2\times 3.14\times \left ( 5.25 \right )^{2}\)

=173.25\(cm^{2}\)

Cost of tin plating 100\(cm^{2}\) area= Rs.4

Cost of tin plating 173.25\(cm^{2}\) area= Rs. \(\frac{4\times 173.25}{100}\) = 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

 

1

 

Inner surface area of dome= \(2\pi r^{2}\)

=\(2\times 3.14\times \left ( 6.3 \right )^{2}\)

=249.48\(m^{2}\)

Now, cost of 1\(m^{2}\)= Rs.2

Therefore cost of 249.48\(m^{2}\)= Rs. \(\left ( 249.48\times2 \right )\) = 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.

\(\frac{3}{4}^{th}\) of earth surface is covered by water

Therefore \(\frac{1}{4}^{th}\) of earth surface is covered by land

Therefore Surface area covered by land= \(\frac{1}{4}\times 4\pi r^{2}\)

= \(\frac{1}{4}\times 4\times \frac{22}{7}\times \left ( 6370 \right )^{2}\)

=127527400\(km^{2}\)

 

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.

 

2

 

Given length of the shape= 7cm

But length= r+r

2r= 7cm

r= 3.5cm

Also; h=r

Total surface area of shape= \(2\pi rh+2\pi r^{2}\)

=\(2\pi rr+2\pi r^{2}\)

= \(2\pi r^{2}+2\pi r^{2}\)

=\(4\times\frac{22}{7}\times\left ( 3.5 \right )^{2}\) = 154\(cm^{2}\)

 

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 100\(cm^{2}\)

 

Sol.

 

3

 

Diameter of cone= 16cm

Radius of cone= 8cm

Height of cone= 15cm

Slant height of cone= \(\sqrt{8^{2}+15^{2}}\)

=\(\sqrt{64+225}\)

=\(\sqrt{289}\) = 17cm

Therefore Total curved surface area of toy

= \(\pi rl+2\pi r^{2}\)

=\(\frac{22}{7}\times8\times17+2\times\frac{22}{7}\times8^{2}\)

= \(\frac{5808}{7}cm^{2}\)

Now, cost of 100\(cm^{2}\)= Rs.7

1\(cm^{2}\)= Rs. \(\frac{7}{100}\)

Hence cost of \(\frac{5808}{7}cm^{2}\)= Rs. \(\frac{5808}{7}\times\frac{7}{100}\)

= 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 \(m^{2}\).

 

Sol.

 

4

 

Diameter of a cylinder= 1.4m

Therefore radius of cylinder= \(\frac{1.4}{2}\)=0.7m

Height of cylinder= 8m

Therefore surface area of tank= \(2\pi rh+2\pi r^{2}\)

=\(2\times\frac{22}{7}\times0.7\times8+2\times\frac{22}{7}\times\left ( 0.7 \right )^{2}\)

=\(\frac{176}{5}+\frac{77}{25}\) = 38.28\(m^{2}\)

Now cost of 1\(m^{2}\)= Rs.10

Therefore cost of 38.28\(m^{2}\)= 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 be \(\frac{d}{4}\)

Radius of earth= \(\frac{d}{2}\)

Radius of moon= \(\frac{\frac{d}{2}}{4}\)= \(\frac{d}{8}\)

Surface area of moon= \(4\pi\left ( \frac{d}{8} \right )^{2}\)

Surface area of earth= \(4\pi\left ( \frac{d}{2} \right )^{2}\)

Required ratio= \(\frac{4\pi\left ( \frac{d}{8} \right )^{2}}{4\pi\left ( \frac{d}{2} \right )^{2}}\)

= \(\frac{4}{64}\) = \(\frac{1}{16}\)

Thus the required ratio of the surface areas is \(\frac{1}{16}\)

 

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 100\(cm^{2}\)

 

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 \(2\pi r\)=17.6

\(2\times\frac{22}{7}\times r=17.6m\)

So, the radius of the dome= \(2\pi r^{2}\)

= \(2\times \frac{22}{7}\times 2.8\times 2.8\)

=49.28\(m^{2}\)

Cost of painting 100\(cm^{2}\) is Rs.5

So, the cost of painting 1\(m^{2}\)= Rs.500

Therefore the cost of painting the whole dome= Rs. \(500\times 49.28\)

=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 \(cm^{2}\).

 

5

 

Sol.

Wooden sphere radius= \(\frac{21}{2}\)=10.5cm

Surface area of a wooden sphere= \(4\pi r^{2}=4\times\frac{22}{7}\times\left(10.5 \right )^{2}=1386cm^{2}\)

Radius r of cylindrical support=1.5cm

Height h of cylindrical support=7cm

Curved surface area of cylindrical support= \(2\pi r h=2\times\frac{22}{7}\times 1.5\times7\\ =66cm^{2}\)

Area of circular end of cylindrical support= \(\pi r^{2}=\frac{22}{7}\times \left(1.5 \right )^{2}=7.07cm^{2}\)

Area to be painted silver= \(8\left(1386-7.07 \right )cm^{2}\\ =8\left(1378.93 \right )cm^{2}\\ =11031.44cm^{2}\)

Cost occurred in painting silver colour= \(\left(11034.44\times0.25 \right )=Rs.2757.86\)

Area to be painted black= \(\left(8\times66 \right )cm^{2}=528cm^{2}\)

Cost occurred in painting black colour=\(\left(528\times5.05 \right )=Rs.26.40\)

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