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Standard XII

Physics

Equation of Continuity

Figure shows ...

Question

Figure shows a liquid flowing through a tube at the rate of $0.1 m^{3} / s$ . The tube is branched into two semicircular tubes of cross-sectional area A/3 (upper) and 2A/3 (lower). The velocity of liquid at Q is (the cross section of the main tube is $A = 10^{- 2} m^{2}$ and $v_{P} = 20 m / s)$

5 m/s

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30 m/s

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35 m/s

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none of these

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Solution

The correct option is B 5 m/s
From the equation of continuity we get-
$R a t e = 0.1 = \frac{A}{3} v_{p} + (2 \frac{A}{3}) v_{q}$
or
$0.1 = \frac{10^{- 2}}{3} \times 20 + \frac{2}{3} (10^{- 2}) v_{q}$
or
$0.1 = \frac{10^{- 2}}{3} (20 + 2 v_{q})$
or
$v_{q} = \frac{30 - 20}{2} = 5 m / s$

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Similar questions

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0.1 m^{3} / s

. The tube is branched into semicircular tubes of cross-sectional area

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A = 10^{- 2} m^{2}

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Figure shows a liquid flowing through a tube at the rate of 0.1m3/s. The tube is branched into two semicircular tubes of cross-sectional area A/3 (upper) and 2A/3 (lower). The velocity of liquid at Q is (the cross section of the main tube is A=10−2m2 and vP=20m/s)

The correct option is B 5 m/sFrom the equation of continuity we get-Rate=0.1=A3vp+(2A3)vqor0.1=10−23×20+23(10−2)vqor0.1=10−23(20+2vq)orvq=30−202=5m/s

Figure shows a liquid flowing through a tube at the rate of $0.1 m^{3} / s$ . The tube is branched into two semicircular tubes of cross-sectional area A/3 (upper) and 2A/3 (lower). The velocity of liquid at Q is (the cross section of the main tube is $A = 10^{- 2} m^{2}$ and $v_{P} = 20 m / s)$

The correct option is B 5 m/s
From the equation of continuity we get-
$R a t e = 0.1 = \frac{A}{3} v_{p} + (2 \frac{A}{3}) v_{q}$
or
$0.1 = \frac{10^{- 2}}{3} \times 20 + \frac{2}{3} (10^{- 2}) v_{q}$
or
$0.1 = \frac{10^{- 2}}{3} (20 + 2 v_{q})$
or
$v_{q} = \frac{30 - 20}{2} = 5 m / s$