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The horizontal bottom of a wide vessel with an ideal fluid has a round orifice of radius $R_{1}$ over which a round closed cylinder is mounted, whose radius $R_{2} > R_{1}$ . The clearance between the cylinder and the bottom of the vessel is very small, the fluid density is $ρ$ . Find the static pressure of the fluid in the clearance as a function of the distance $r$ from the axis of the orifice (and the cylinder), if the height of the fluid is equal to $h$ .

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$P = P_{0} + ρ g h (1 - \frac{{R^{3}}_{1}}{r^{3}})$

$P = P_{0} + ρ g h (1 - \frac{{R^{2}}_{1}}{r^{2}})$

$P = P_{0} + ρ g h (1 - \frac{R_{1}}{r})$

$P = P_{0} + ρ g h (1 - \sqrt{\frac{R_{1}}{r}})$

answer is C.

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Detailed Solution

Using equation of continuity between sections $A$ and $B$ , we have

$(2 πrd) v_{B} = (2 π R_{1} d) v_{A} o r r v_{B} = R_{1} v_{A} \dots \dots (i)$

Now using Bernoulli's equation between $A$ and $C$ and between $A$ and $B$ , we have

$P_{a} + \frac{1}{2} ρ v_{A}^{2} = P_{a} + \frac{1}{2} ρ g h \dots \dots (i i) a n d P_{a} + \frac{1}{2} ρ v_{A}^{2} = P_{a} + \frac{1}{2} ρ v_{B}^{2} \dots \dots (i i i)$

After simplifying, we get

$P = P_{0} + ρ g h (1 - \frac{R_{1}^{2}}{r^{2}})$

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