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Bernoulli's Principle

A liquid flow...

Question

A liquid flows along a horizontal pipe AB of uniform cross-section. The difference between the level of the liquid in tubes P and Q is $10 c m$ . The diameter of the tubes P and Q are the same. Then:
( $g = 9.8 {m s}^{- 2}$ )

level in P is greater than that of Q and velocity of flow is

1.4 m / s

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level in Q is greater than that of P and velocity of flow is

1.4 m / s

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level in P is greater than that of Q and velocity of flow is

0.7 m / s

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level in Q is greater than that of P and velocity of flow is

0.7 m / s

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Solution

The correct option is B level in Q is greater than that of P and velocity of flow is $1.4 m / s$
Assuming laminar flow, velocity of liquid inside P is zero as it is perpendicular to the direction of flow of liquid. Hence, liquid level in P is same as the height of the pipe AB.
For pipe Q, it is inclined against the direction of water and there is extra pressure on the bottom of Q. This causes rise in level in Q.
Applying Bernoulli theorem for Q,
$P_{o} + ρ g h = P_{o} + \frac{1}{2} ρ v^{2}$
$v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 10} = 1.4 m / s$

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A liquid flows along a horizontal pipe AB of uniform cross-section. The difference between the level of the liquid in tubes P and Q is 10 cm. The diameter of the tubes P and Q are the same. Then: (g=9.8ms−2)

A liquid flows along a horizontal pipe AB of uniform cross-section. The difference between the level of the liquid in tubes P and Q is $10 c m$ . The diameter of the tubes P and Q are the same. Then:
( $g = 9.8 {m s}^{- 2}$ )