I'm trying to solve the question b) in this linkhttp://i36.tinypic.com/2h3wmtx.jpg(adsbygoogle = window.adsbygoogle || []).push({});

My reasoning is

[tex]

dK = dm v^2

[/tex]

[tex]

dm = \rho dV = \rho A(x) dx

[/tex]

where A(x) is a function of how the cross sectional area varies along the tube from left to right.

[tex]

A(x) = \pi r(x)^2

[/tex]

where r(x) is how the radius of the cross sectional area varies and is given by

[tex]

r(x) = r(1+x/l)

[/tex]

this gives

[tex]

dm = \rho \pi r^2 (1+x/l)^2

[/tex]

the answer given gives the (1+x/l)^2 term in the denominator but the method i have used will give it in the numerator, i don't see the flaw in what i have done though unfortunately.

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