- Surface Areas and Volumes - Exercise 13.1
- Surface Areas and Volumes - Exercise 13.2
- Surface Areas and Volumes - Exercise 13.3
- Surface Areas and Volumes - Exercise 13.4
- Surface Areas and Volumes - Exercise 13.5
- Surface Areas and Volumes - Exercise 13.6
- Surface Areas and Volumes - Exercise 13.7
- Surface Areas and Volumes - Exercise 13.8
- Surface Areas and Volumes - Exercise 13.9
NCERT Solutions Class 9 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.2 in Chapter 13 - Surface Areas and Volumes
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A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter
being 4.4cm. (see fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area
(iii) total surface area
(Assume Ο=22/7)
Let π1 and π2 Inner and outer radii of a cylindrical pipe
π1 = 4/2 cm = 2cm
π2 = 4.4/2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
(i)the curved surface area of the outer surface of pipe = 2Ο π1 h
\begin{array}{l} =2 \times 22 / 7 \times 2 \times 77 \mathrm{cm}^{2} \\ =968 \mathrm{cm}^{2} \end{array}
(ii) curved surface area of outer surface of pipe = 2Ο π2 h
= 2 Γ 22/7 Γ 2.2 Γ 77 \mathrm{cm}^{2}
\begin{array}{l} =(22 \times 22 \times 2.2) \mathrm{cm}^{2} \\ =1064.8 \mathrm{cm}^{2} \end{array}
(iii) Total surface area of pipe = inner curved surface area + outer curved surface area + Area of
both circular ends of the pipe.
\begin{array}{l} =2 \pi r_{1} h+2 \pi r_{2} h+\left(r_{2}^{2}-r_{1}^{2}\right) \\ =9668+1064.8+2 \pi\left((2.2)^{2}-2^{2}\right) \\ =2031.8+5.28 \end{array}
2038.08 \mathrm{cm}^{2}
Therefore, the total surface area of the cylindrical pipe is 2038.08 \mathrm{cm}^{2}
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