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Standard XII

Physics

Equation of Continuity

The velocity ...

Question

The velocity head of a stream of water is 40 cm. Then, the velocity of flow of water is :

2.8

m s^{- 1}

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5.6

m s^{- 1}

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7.2

m s^{- 1}

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9.8

m s^{- 1}

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Solution

The correct option is A $2.8$ $m s^{- 1}$
Velocity head of water $= 40 c m$ $= \frac{V^{2}}{2 g}$
$\Rightarrow V^{2} = \frac{40}{100} \times 2 \times 10$
$\Rightarrow V^{2} = 8 \Rightarrow V = \sqrt{8} m / s$
$\Rightarrow V = 2.8 m / s$

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The velocity head of a stream of water is 40 cm. Then, the velocity of flow of water is :

The correct option is A 2.8 ms−1Velocity head of water =40 cm =V22g⇒V2=40100×2×10⇒V2=8⇒V=√8 m/s⇒V=2.8 m/s

The correct option is A $2.8$ $m s^{- 1}$
Velocity head of water $= 40 c m$ $= \frac{V^{2}}{2 g}$
$\Rightarrow V^{2} = \frac{40}{100} \times 2 \times 10$
$\Rightarrow V^{2} = 8 \Rightarrow V = \sqrt{8} m / s$
$\Rightarrow V = 2.8 m / s$