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A side wall of a wide open tank is provided with a narrowing tube through which water flows out. The cross sectional area of the tube decreases from $S = 3.0 c m^{2} t o s = 1.0 c m^{2} .$ The water level in the tank is h = 4.6 m higher than in the tube. Neglecting the viscosity of water, find the horizontal component of the force tending to pull the tube out of the tank. (Communicate the nearest integer).

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answer is 6.

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Detailed Solution

$P . O . C : π R^{2} V_{1} = π r^{2} v_{2} π y^{2} v$

$v_{2} = \sqrt{2 g h}$

Bernoulli’s : $P_{x} + \frac{1}{2} ρ v^{2} = ρ_{0} + \frac{1}{2} ρ (2 g h)$

$F = \int (ρ_{x} - ρ_{0}) \sin θ (2 π y d z)$

$d z = d y \sec θ; \tan θ = \frac{R - r}{l_{2} - l_{1}}$

$∴ F = ρ g h \frac{π (R^{2} - r^{2})}{R^{2}}$

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