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Water from a tap emerges vertically downwards with initial velocity 4 ms^-1. The cross-sectional area of the tap is A. The flow is steady and pressure is constant throughout the stream of water. The distance h vertically below the tap, where the cross - sectional area of the stream becomes (2/3)A, is (g = 10 ms²)

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0.5 m

1 m

1.5 m

2.2 m

answer is B.

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Detailed Solution

By equation of continuity

$A_{1} v_{1} = A_{2} v_{2} \Rightarrow A \times 4 = \frac{2}{3} A \times v_{2} \Rightarrow v_{2} = 6 {ms}^{- 1}$

now Bernoulli's theorem,

$p + {ρgh}_{1} + \frac{1}{2} ρ {v_{1}}^{2} = p + {ρgh}_{2} + \frac{1}{2} ρ {v_{2}}^{2}$

$g (h_{1} - h_{2}) = \frac{1}{2} ({v_{2}}^{2} - {v_{1}}^{2})$

$g \times h = \frac{1}{2} [6^{2} - 4^{2}]$

$10 \times h = \frac{1}{2} [36 - 16] \Rightarrow h = \frac{20}{20} = 1 m$

Hence the correct answer is 1m.

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