S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 A

 Exercise 15 A

Question 1

Ans: (i) Given

$\begin{aligned}&h=12 \mathrm{~cm}, \\&r=7 \mathrm{~cm} .\end{aligned}$

$\therefore$ Curved surface arca $=2 \pi r h$'

$\begin{aligned} &=2 \times \frac{22}{7} \times 7 \times 12 \\ &=44 \times 12 \\ &=528 \mathrm{~cm}^{2}\end{aligned}$

(ii) Given,

$\begin{aligned}&h=10 \mathrm{~cm}, \\&r=7 \mathrm{~cm} .\end{aligned}$

Curved Surface area = $2 \pi r h$

$=440 \mathrm{~cm}^{2}$

Total surface area $=2 πrh (h+r)$

$\begin{aligned}&=2 \times \frac{22}{7} \times 7(10+7) \\&=44 \times 17 \\&=748 \mathrm{~cm}^{2}\end{aligned}$

(iii) Given

$\begin{aligned}&h=5 \mathrm{~cm}, \\&r=21 \mathrm{~cm}\end{aligned}$

∴ Curved Surface area = 2πr h

$=2 \times \frac{22}{7} \times 21 \times 5$

$=44 \times 15$

$=660 \mathrm{~cm}^{2}$

Total Surface area = $2 \pi r(h+r)$

$=2 \times \frac{22}{7} \times 21(5+21)$

$=44 \times 3 \times 26$

$=132 \times 26$

$=3432 \mathrm{~cm}^{2}$

(iv) Given

$\begin{aligned}&h=20 \mathrm{~cm} \\&r=14 \mathrm{~cm}\end{aligned}$

$\therefore$ Curved Surface area $=2πrh 

 $\begin{aligned} &=2 \times \frac{22}{7} \times 14 \times 20 \\=& 44 \times 40 \end{aligned}$

$=1760 \mathrm{~cm}^{2}$

Total surface area= $2 \pi r(h+r)$

$=2 \times \frac{22}{7} \times 14(20+14)$

$=44 \times 2 \times 34$

$=88 \times 34$

$=2992 \mathrm{~cm}^{2}$

(v) Given h= 16m

r= 10.5m

 Curved surface arca=2πrh 

$=44 \times 24$

$=105.5 m^{2}$

Total surface area $=2πr (h+r)$

$=66 \times 26.5$

$=1749 \mathrm{~m}^{2}$

Question 2

Ans: Given,

Radius (r) = $\frac{7}{2} m$

Depth (h) = 4m

$\therefore$ Total area of the-wet surface.

$=2 \pi r h+\pi r^{2}$

$=\pi r(2 h+r)$

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

$\begin{aligned} &=11\left(8+\frac{7}{2}\right) \\ &=11\left(\frac{16+7}{2}\right) \\ &=11 \times \frac{23}{2} \\ &=\frac{253}{2} \\ &=126.5 \mathrm{~m}^{2} . \end{aligned}$

Question 3

Ans: Given, 

External Radius, (r) $\frac{14}{2}=7 \mathrm{~cm} .$

Thickness = 2cm

So, inner radius (r) = 7-2=5cm

Height (h) = 20cm

Total surface area = $2 \pi R h+2 \pi r h+2 \pi R^{2}-2 \pi r^{2} .$

$\begin{aligned} &=2 \pi\left(R h+r h+R^{2}-r^{2}\right) . \\ &=2 \times \frac{22}{7}\left(7 \times 20+5 \times 20+(7)^{2}-(5)^{2}\right) \\ &=\frac{44}{7}(140+100+49-25) \\ &=\frac{44}{7}(240+24) . \\ &=\frac{44}{7} \times 264 . \\ &=\frac{11616}{7} \\=& 1659.43 \mathrm{~cm}^{2} \end{aligned}$

Question 4

Ans: Given, 

Curved Surface Area = $110 \mathrm{~cm}^{2}$

Heights $(h)=5 \mathrm{~cm}$.

$\therefore$ Curved surface area $=2 \pi r h$

$110=2 \times \frac{22}{7} \times r \times 5 .$

$110=\frac{44 \times 5}{7} \times r$

$\frac{110\times 7}{44\times 5}=r$

$\frac{7}{2}=r$

$3 \cdot 5=r$

Hence the radius of the cylinder is 3.5cm

Question 5

Ans: Given, 

Curved Surface area = $13 \cdot 2 \mathrm{~cm}^{2}$

Radius $=6 \mathrm{~cm}$.

$\therefore \quad$ Curved surface arca= $2 \pi r h$

$13.2=2 \times \frac{22}{7} \times 6 \times \mathrm{h}$

$\frac{13.2}{10}=\frac{44 \times 6}{7} \times h$

$\frac{132 \times 7}{44 \times 6 \times 10}=h$

$\frac{132 \times 7}{44 \times 60}$=h .

$\begin{aligned} \frac{7}{20} &=h \\ 0.35 &=h \\ \therefore h &=0.35 \mathrm{~cm}\end{aligned}$

Hence the height of the cylinder is 0.35cm

Question 6

Ans: Given,

Radius of garden roller (r) =$\frac{75}{2}cm$

width $(\mathrm{h})=105 \mathrm{~cm}$

 

Curved surface area = $2 \pi \mathrm{rh}$

$=2 \times \frac{22}{7} \times \frac{75}{2} \times 105$

$=22 \times 15 \times 75$

$=330 \times 75$

$=24750 \mathrm{~cm}^{2} .$

Area covered in 14 revolutions 

$=24750 \times 14$

$=346500 \mathrm{~cm}^{2} .$

Question 7

Ans: Given, 

Radius of each cylindrical pillar = $=\frac{50}{2} \mathrm{~m}=25 \mathrm{~m}=\frac{25}{100} \mathrm{~m}$

Height = 4m

$\therefore$ Curved surface area $=2πrh

$=6.28 \mathrm{~m}^{2}$

Area of 10 cylinder = $=6.28 \times 10$

$=62.8 \mathrm{~m}^{2}$

Cost of painting = 50 paise per $m^{2}$

Total cost = $=\frac{628}{10} \times \frac{50}{100}$

$=\frac{314}{10}$

$=₹ 31.4$

Question 8

Ans: Given, 

Radius of a roller = $\frac{84}{2}=42 \mathrm{~cm}$

Height $(h)=120 \mathrm{~cm}$

Curved surface area = $2 \pi r h$

$=2 \times \frac{22}{7} \times 42 \times 120 .$

$=44 \times 420$

$=31680 \mathrm{~cm}^{2} .$

$=\frac{31680}{100 \times 100}\mathrm{m}^{2}$

$=\frac{3168}{1000} \mathrm{~m}^{2}$

Area of leveling playground by 500 revolution 

$=\frac{3168}{1000} \times 500$

$=\frac{15840}{10}$

$=1584 \mathrm{~m}^{2}$

Cost of leveling the ground at the rate of 30 paisa per meter square

$=1584 \times \frac{30}{100}$

$=\frac{4752}{10}$

$=₹ 475.2$

Question 9

Ans: Given, 

Curved surface area of cylinder = $1000 \mathrm{~cm}^{2}$

Diameter of wire =5mm = 1cm

$=0.5 \mathrm{~cm}$

$2 \pi r h=1000 \mathrm{~cm}^{2}$

$\therefore$ Number of coils $=\frac{h}{\text { Diameter wire }}$

$=\frac{h}{\frac{0.5}{10}}$

$=\frac{10h}{5}$=2h

So total length of wire = $2 \pi R \times 2 h$

$=2 \times 2 \pi \mathrm{Rh} .$

$=2 \times 1000 .$

$=2000 \mathrm{~cm}$

$=\frac{2000}{100} \mathrm{m}$

=20m

Question 10

Ans: Given,

Height = 20cm

Exterior radius = $\frac{25}{2} \mathrm{~cm}=12.5 \mathrm{~cm}$

Thickness of pipe = 1cm

$\begin{aligned} \therefore \quad \text { Interior radius }(x) &=12.5-1 \\ &=11.5 \mathrm{~cm} \end{aligned}$

∴ Total surface area = $2 \pi R h+2 \pi r h+2 r R^{2}-2 \pi r^{2}$

$\begin{aligned} &=2 \pi \cdot\left(R h+r h+R^{2}-r^{2}\right) \\ &=2 \times \frac{22}{7}\left(12-5 \times 20+11.5 \times 20+(12-5)^{2}-(11-5)^{2}\right) \\ &=\frac{44}{7}(250+230+156.25-132-25) . \\ &=\frac{44}{7}(480+24) \\ &=\frac{44}{7} \times 504 \\ &=\frac{22176}{7} \\ &=3168 \mathrm{~cm}^{2}\end{aligned}$

Question 11

Ans: Given,

No. of circular plates = 50 

Radius of each plates = 7cm

Thickness = 0.5cm

∴ Height thickness of 50 plates = $0.5 \times 50$

$25 \mathrm{~cm}$

∴ Curved surface area = 2πrh

$=2 \times \frac{22}{7} \times 7 \times 25$

$=44 \times 25$

$=1100 \mathrm{~cm}^{2} .$

Area of top and bottom = $2 \pi r^{2}$ 

$=2 \times \frac{22}{7} \times-7-x 7$

$=44 \times 7$

$=308 \mathrm{~cm}^{2} .$

∴ Total surface area = Curved surface area + Area of top and bottom

$=1100+308$

$=1408 \mathrm{~cm}^{2} .$