Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 $c m^{2}$ and 2 $c m^{2}$ respectively. The rate of flow of water through the tube is 500 $c m^{3}$ $s^{- 1}$ . Find the difference of mercury levels in the U-tube.

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Solution

Given Rate of flow = 500 $c m^{3} / s e c$

$\Rightarrow v_{A} = (\frac{500}{5}) = 100 c m / s e c$

$V_{A} a_{A} = V_{B} a_{B}$

$\Rightarrow \frac{v_{A}}{v_{B}} = \frac{a_{B}}{a_{A}} = \frac{2}{5}$

$\Rightarrow 5 v_{A} = 2 v_{B}$

$\Rightarrow v_{B} = (\frac{5}{2}) v_{A}$ ...(i)

$\Rightarrow \frac{1}{2} ρ v_{A}^{2} + ρ g h_{A} + p_{A} = \frac{1}{2} ρ v_{B}^{2} + ρ g h_{B} + p_{B}$

$\Rightarrow p_{A} - p_{B} = \frac{1}{2} [p (v_{B}^{2} - v_{A}^{2})]$

$\Rightarrow h \times 13.6 \times 980 \frac{1}{2} \times 1 \times \frac{21}{4} (100)^{2}$

[Using (i)]

[because $p_{A} - p_{B}$ = ?]

$\Rightarrow h = \frac{21 \times (100)^{2}}{2 \times 13.6 \times 980 \times 4}$

= 1.969 cm

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Q. Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³ s⁻¹. Find the difference of mercury levels in the U-tube.

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respectively. The rate of flow of water through the tube is

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Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³ s^-1. Find the difference of mercury levels in the U-tube. Assume this tube lying horizontally on a table.

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5 {c m}^{2}

2 {c m}^{2}

. The rate of flow is 500

{c m}^{3} / s e c

Find difference mercury level of U-tube:

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Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm2 and 2 cm2 respectively. The rate of flow of water through the tube is 500 cm3 s−1. Find the difference of mercury levels in the U-tube.

Given Rate of flow = 500 cm3/sec ⇒vA=(5005)=100cm/sec VAaA=VBaB ⇒vAvB=aBaA=25 ⇒5vA=2vB ⇒vB=(52)vA ...(i) ⇒12ρv2A+ρghA+pA=12ρv2B+ρghB+pB ⇒pA−pB=12[p(v2B−v2A)] ⇒h×13.6×98012×1×214(100)2 [Using (i)] [because pA−pB = ?] ⇒h=21×(100)22×13.6×980×4 = 1.969 cm

Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 $c m^{2}$ and 2 $c m^{2}$ respectively. The rate of flow of water through the tube is 500 $c m^{3}$ $s^{- 1}$ . Find the difference of mercury levels in the U-tube.