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
$=1100+308$
$=1408 \mathrm{~cm}^{2} .$
