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by sukhman » Sat Oct 05, 2013 6:45 pm
The External radius of a steel pipe is 1.6 cm and its thickness is 1 cm. If 1 cm*3 of steel pipe weighs 20 gms, then find the weight of the steel pipe of length 28 cm.
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by vinay1983 » Sat Oct 05, 2013 6:57 pm
sukhman wrote:The External radius of a steel pipe is 1.6 cm and its thickness is 1 cm. If 1 cm*3 of steel pipe weighs 20 gms, then find the weight of the steel pipe of length 28 cm.
No options?Strange!

It is cylinder, so r=1.6 or it can be average radius i.e r=1.6+0.6=2.2/2=1.1

V= pie r^2 h
=22/7*28*r^2
=88* 1.6^2 or 88* 1.1^2
= 225.28 cubic cm 106.48 cubic cm

So if 1 cubic cm weighs 20gm then

225.28 * 20gms 106.48 * 20gms

4505.6 gms 2129.6 gms

Actually I don't think we need to take the average radius here. So i will go with the former value.

hope i am correct
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by theCodeToGMAT » Sat Oct 05, 2013 9:10 pm
R1 = 1.6cm
Thickness means cross sectional thickness..

Volume of the Pipe = pi (1)^2 * 28 = 87.92

Total Weight = 1758.4


Definitely not a GMAT Question.. Is it another Arun Sharma's Question?
Last edited by theCodeToGMAT on Sun Oct 06, 2013 12:20 am, edited 1 time in total.
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by ganeshrkamath » Sun Oct 06, 2013 12:02 am
sukhman wrote:The External radius of a steel pipe is 1.6 cm and its thickness is 1 cm. If 1 cm*3 of steel pipe weighs 20 gms, then find the weight of the steel pipe of length 28 cm.
Volume of the pipe of length 28 cm = pi * (1.6^2 - 0.6^2) * 28
= 22/7 * 2.2 * 28
= 22 * 2.2 * 4
= 193.6 cm^3

Weight = 20 * 193.6 = 3872 gms

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Last edited by ganeshrkamath on Sun Oct 06, 2013 10:13 pm, edited 1 time in total.
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by vipulgoyal » Sun Oct 06, 2013 9:39 pm
V = pi * H (R^2 - r^2)
V = 22/7 * 28 * (1.6^2 - .6^2)
V = 193.6 * 20 = 3872
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