Difference between revisions of "1950 AHSME Problems/Problem 33"
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− | Knowing that <math>d^2 | + | Knowing that <math>d^2 \alpha A A</math>, where <math>d</math> is the pipe diameter and <math>A</math> is the cross-sectional area we simply get <math>6^2 = 36 = \boxed{\textbf{(D)}\ 36}</math>. |
This works because the diameter of one of the other pipes is <math>1</math>, which is not affected by powers. | This works because the diameter of one of the other pipes is <math>1</math>, which is not affected by powers. |
Revision as of 00:55, 26 August 2024
Problem
The number of circular pipes with an inside diameter of inch which will carry the same amount of water as a pipe with an inside diameter of
inches is:
Solution
It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter ,
Cross sectional area of pipe with diameter
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
So the answer is .
Solution 2 (Solution 1 but easier)
Knowing that , where
is the pipe diameter and
is the cross-sectional area we simply get
.
This works because the diameter of one of the other pipes is , which is not affected by powers.
~PeterDoesPhysics
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
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All AHSME Problems and Solutions |
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