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

2.86 m s^{- 1}

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3.0 m s^{- 1}

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3.82 m s^{- 1}

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5.7 m s^{- 1}

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Solution

The correct option is A $2.86 m s^{- 1}$
From the principle of continuity
$A_{1} v_{1} = A_{2} v_{2}$
$v_{2} = \frac{A_{1} v_{1}}{A_{2}} = \frac{4.20 \times 5.18}{7.60} = 2.86 m s^{- 1}$ .

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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 $c m^{2}$ . 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

Q. Water is moving with a speed of

5.18 m s^{- 1}

through a pipe with a cross-sectional area of

4.20 c m^{2}

. The water level 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 :

Q. Water flows through a horizontal pipe of varying cross-section at the rate of

0.5 m^{2} s^{- 1}

. The speed of the water at a point where the area of cross section is

0.05 m^{2}

Q. How many litres of water flows out of a pipe having an area of cross section of 5 cm² in 1 minute, if the speed of water in the pipe is 30 cm/sec?

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

The correct option is A 2.86 ms−1From the principle of continuityA1v1=A2v2v2=A1v1A2=4.20×5.187.60=2.86 ms−1.

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

The correct option is A $2.86 m s^{- 1}$
From the principle of continuity
$A_{1} v_{1} = A_{2} v_{2}$
$v_{2} = \frac{A_{1} v_{1}}{A_{2}} = \frac{4.20 \times 5.18}{7.60} = 2.86 m s^{- 1}$ .