Water (density $ρ$ ) is flowing through the uniform tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)

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$\sqrt{3} ρ A v^{2}$

$2 ρ A v^{2}$

$\sqrt{2} ρ A v^{2}$

$\frac{ρ A v^{2}}{\sqrt{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} |{\vec{Δ P}}_{x}| = m v \sin 60^{\circ} = \frac{\sqrt{3}}{2} m v \\ |{\vec{Δ P}}_{y}| = \frac{m v}{2} + m v = \frac{3}{2} m v \\ \Rightarrow |{\vec{Δ P}}_{net}| = \sqrt{Δ P_{x}^{2} + Δ P_{y}^{2}} = \sqrt{(\frac{9}{4} + \frac{3}{4})} m v \\ \begin{array}{l} |{\vec{Δ P}}_{n e t}| = \sqrt{3} m v \\ \Rightarrow |{\vec{Δ F}}_{net}| = \sqrt{3} (\frac{d m}{d t}) \cdot v = \sqrt{3} ρ A v^{2} \\ (\begin{array}{l} Since, d m = A (v d t) ρ \\ \Rightarrow \frac{d m}{d t} = A ρ v \end{array}) \end{array} \end{array}$