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Standard XII

Physics

Terminal Velocity

Water is movi...

Question

Water is moving with a speed of 5.18 ms $^{-} 1$ through a pipe with a cross-sectional area of 4.20 $c m^{2}$ . The water gradually descends 9.66 m as the pipe increase in area to 7.60 $c m^{2}$ . The speed of flow at the lower level is

3.0 $m s^{- 1}$

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5.7 $m s^{- 1}$

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3.82 $m s^{- 1}$

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2.86 $m s^{- 1}$

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Solution

The correct option is D
2.86 $m s^{- 1}$

$a_{1} v_{1} = a_{2} v_{2}$

$\Rightarrow 4.20 \times 5.18 = 7.60 \times v_{2} \Rightarrow v_{2} = 2.86 m / s$

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Water is moving with a speed of 5.18 ms−1 through a pipe with a cross-sectional area of 4.20 cm2. The water gradually descends 9.66 m as the pipe increase in area to 7.60 cm2. The speed of flow at the lower level is

The correct option is D 2.86 ms−1 a1v1=a2v2 ⇒4.20×5.18=7.60×v2⇒v2=2.86m/s

Water is moving with a speed of 5.18 ms $^{-} 1$ through a pipe with a cross-sectional area of 4.20 $c m^{2}$ . The water gradually descends 9.66 m as the pipe increase in area to 7.60 $c m^{2}$ . The speed of flow at the lower level is

The correct option is D
2.86 $m s^{- 1}$

$a_{1} v_{1} = a_{2} v_{2}$

$\Rightarrow 4.20 \times 5.18 = 7.60 \times v_{2} \Rightarrow v_{2} = 2.86 m / s$