A rectangular tube of uniform cross …
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A rectangular tube of uniform cross section has three liquids of
densities
1 2 ,
and
3
. Each liquid column has length l equal to
length of sides of the equilateral triangle. Find the length x of the
liquid of density
1
in the horizontal limb of the tube, if the
triangular tube is kept in the vertical plane.
Posted by Sanshray Gupta 4 years, 5 months ago
- 1 answers
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Sia ? 4 years, 5 months ago
Q : A rectangular tube of uniform cross section has three liquids of densities ρ1, ρ2 and ρ3. Each liquid column has length l equal to length of sides of the equilateral triangle. Find the length x of the liquid density ρ1 in the horizontal limb of the tube, if the triangular tube is kept in the vertical plane.
Solution:
At junction force applied by left to coloumn = force applied by light of coloumn.
(1 − x) cos30o p1 gA + x cos30o p2 gA = (1 − x) cos30o p2 gA + x cos30o p3 gA
{tex}\Rightarrow{/tex} (1 − x)p1 + xp2 = (1 − x)p2 + xp3
{tex}\Rightarrow{/tex} p11 − p1x + xp2 = 1p2 − xp2 + p3x
{tex}\Rightarrow{/tex} xp2 − xp1 + xp2 − xp3 = 1p2 − 1p1
{tex}\Rightarrow{/tex} x(2p2 − p1 − p3) = 1(p2 − p1)
x = {tex}\frac {1(p_2 − p_1)}{2p_2 − p_1 − p_3}{/tex}
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