A flat closed tube of rectangular cross-section has inner radius ‘a’ and outer radius ‘b’. It is filled completely with water and is kept flat on a horizontal surface. Now, entire water in the tube is rotated with a uniform and steady angular velocity $' ω'$ about the vertical axis of the tube. Viscous forces between the tube and water is neglected and density of water is $ρ$ . 1 and 2 are two points in water in the same horizontal plane as shown in figure. Let P be the pressure and E be energy of fluid particle per unit volume. Then

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$E_{1} - E_{2} = ρ (b^{2} - a^{2}) ω^{2}$

$P_{2} - P_{1} = \frac{1}{2} ρ (b^{2} - a^{2}) ω^{2}$

$P_{1} - P_{2} = \frac{1}{2} ρ (b^{2} - a^{2}) ω^{2}$

$E_{1} = E_{2}$

answer is A, D.

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Detailed Solution

flow is rotational (solid body rotation)
$∴ P_{1} - P_{2} = \frac{1}{2} υ w^{2} (b^{2} - a^{2}) E = P + \frac{1}{2} υ V^{2} = P + \frac{1}{2} υ r^{2} w^{2} ∴ E_{1} - E_{2} = (P_{1} - P_{2}) + \frac{1}{2} υ w^{2} (b^{2} - a^{2}) = \frac{1}{2} υ W^{2} (b^{2} - a^{2}) + \frac{1}{2} υ W^{2} (b^{2} - a^{2}) = υ W^{2} (b^{2} - a^{2})$

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