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Question 57
A hollow iron pipe is 28 cm long and its external diameter is 10 cm.If the thickness of the pipe is 2 cm and iron weighs $$5\ g/cm^3$$ , then the weight of the pipe is:
Solution
Given hollow cylinder and Volume of it is given by $$\pi(R^{2}-r^{2})h$$
given 2R=10 cm
R=5 cm
Thickness=t=2 cm
r=R-t
r=5-2
r=3 cm
Volume=$$\pi(R^{2}-r^{2})h$$
=(22/7)*(25-9)*28
=22*16*4
=1408 cubic cm
Mass=density*volume
=5*1408
=7040 grams
=7.040 kg
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