Question

A hollow cylindrical tube , open at both ends is made of 2 cm thick iron. if the external diameter is 50 cm and the length of the tube is 140 cm, find the volume

Answer 1

Thickness =2cm
Ex.diameter=50cm
Therefore,Ex.radius=25cm
Now,In.radius=(25-2)cm
=23cm
Length=140cm
Therefore,Volume=

$\pi ({ex.r}^{2} - {in.r}^{2} ) \times h$
=22/7(625-529)×140
=22/7×96×140
=42240cu.cm

Answer 2

Solution :-

outer diameter = 50 cm

Wall needs to be 2 cm thick

That means the inner diameter must be 46 cm.

Cross sectional area = π× outer dia^2 / 4 - π× inner_dia^2/4

Volume = Cross sectional area × length

Volume = π./4×(outer^2 - inner^2) × length

Volume of external surface cylinder = 22/7× 50×2× 140 = 44000 cm^3

Volume of internal surface cylinder = 22/7×48×2×140 = 42240 cm^3

Plug in your numbers

Volume = 42200 cm^3

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@GauravSaxena01