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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 4.4 cm. Find its
i) inner curved surface area
ii) outer curved surface area
iii) total surface area.
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Answer :

Given :
Inner radius (r) of cylindrical pipe = \(\frac{4}{2}\) cm = 2 cm
outer radius (R) of cylindrical pipe = \(\frac{4.4}{2}\) cm = 2.2 cm
Height (h) Of cylindrical pipe = Length Of cylindrical pipe = 77 cm

i) CSA of inner surface of pipe
= \(2{\pi}rh\)
= \(2 × \frac{22}{7} × 2 × 77\) \({cm}^2\)
= \(2 × 22 × 2 × 11\) \({cm}^2\)
= 968 \({cm}^2\)

ii) CSA of inner surface of pipe
= \(2{\pi}{R}h\)
= \(2 × \frac{22}{7} × 2.2 × 77\)
= \(2 × 22 × 2.2\) \({cm}^2\)
= 1064.8 \({cm}^2\)

iii) Total surface area of pipe
= CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
= \(2{\pi}{r}h\) + \(2{\pi}{R}h\) + \(2{\pi} [(R)^2 - (r)^2]\)
= \([968 + 1064.8 + 2{\pi}{(2.2)^2 - (2)^2}] {cm}^2\)
= \(2032.8 + 2 × \frac{22}{7} × 0.84\) \({cm}^2\)
= (2032.8 + 5.28) \({cm}^2\)
= 2038.08 \({cm}^2\)

Therefore, the total surface area of the cylindrical pipe is 2038.08 \({cm}^2\).

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