Figure shows a liquid flowing through a tube at the rate of 0-1 m 3 - s- The tube is branched into two semi-circular tubes of cross-sectional area A -3 and 2 A -3 of sections P and Q respectively- 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 A- 30 m - sB- 35 m - sC- 15 m - sD- 5 m - s

The correct option is D 5 m/sFrom the equation of continuity, Rate =r=A/3V_P+(A A/3)V_Q ∴ V_P+2V_Q=3r/A=3× 0.1/10^ 2=30 m/s As, V_P=20 m/s, so V_Q= 5 m/s. ...

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Standard XII

Physics

Stopping Potential

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 semi-circular tubes of cross-sectional area $\frac{A}{3}$ and $\frac{2 A}{3}$ of sections P and Q respectively. 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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15 m / s

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Solution

The correct option is A $5 m / s$
From the equation of continuity,
Rate $= r = \frac{A}{3} V_{P} + (A - \frac{A}{3}) V_{Q}$

$∴ V_{P} + 2 V_{Q} = \frac{3 r}{A} = \frac{3 \times 0.1}{10^{- 2}} = 30 m / s$

As, $V_{P} = 20 m / s$ , so $V_{Q} = 5 m / s$ .

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In the figure shown a liquid is flowing through a tube at the rate of 0.1 m³/sec. The tube is branched into two semi circular tubes of cross sectional area A/3 and 2A/3. The velocity of liquid at Q is (the cross-section of the main tube (A) = 10^-2 m² and V_P = 20 m/sec.):

In the figure shown, a liquid is flowing through a tube at the rate of 0.1 $m^{3}$ /sec. The tube is randed into two semicircular tubes of cross sectional area $\frac{A}{3}$ and 2 $\frac{A}{3}$ . The velocity of liquid at Q is $V_{Q}$ and Velocity at P is $V_{P}$ (The cross-sectional area of the main is A)

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Figure shows a liquid flowing through a tube at the rate of 0.1 m3/s. The tube is branched into two semi-circular tubes of cross-sectional area A3 and 2A3 of sections P and Q respectively. The velocity of liquid at Q is (the cross section of the main tube is A=10−2 m2 and VP=20 m/s)

The correct option is A 5 m/sFrom the equation of continuity, Rate =r=A3VP+(A−A3)VQ ∴VP+2VQ=3rA=3×0.110−2=30 m/s As, VP=20m/s, so VQ=5 m/s.

The correct option is A $5 m / s$
From the equation of continuity,
Rate $= r = \frac{A}{3} V_{P} + (A - \frac{A}{3}) V_{Q}$

$∴ V_{P} + 2 V_{Q} = \frac{3 r}{A} = \frac{3 \times 0.1}{10^{- 2}} = 30 m / s$

As, $V_{P} = 20 m / s$ , so $V_{Q} = 5 m / s$ .