A horizontally oriented tube $A B$ of length $l$ rotates with a constant angular velocity $ω$ about a stationary vertical axis $O O'$ passing through the end $A$ . The tube is filled with an ideal fluid. The end $A$ of the tube is open, the closed end $B$ has a very small orifice. Find the velocity of the fluid relative to the tube as a function of the column height $h$ .

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$ω h \sqrt{\frac{4 l}{h} - 1}$

$ω h \sqrt{\frac{4 h}{l} - 1}$

$ω h \sqrt{\frac{2 l}{h} - 1}$

$ω l \sqrt{\frac{2 h}{l} - 1}$

answer is A.

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Detailed Solution

Take an element of fluid of thickness $d x$ . By Newton's second law

$m a = m ω^{2} x o r v \frac{d v}{d x} = ω^{2} x o r \int_{0}^{v} v d v = \int_{l - h}^{h} ω^{2} x d x ∴ v = ω h \sqrt{\frac{2 l}{h} - 1}$

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