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Water from a tap emerges vertically down...

Water from a tap emerges vertically downwards with initial velocity `4ms^(-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=10m//s^(2))`

A

0.5 m

B

1 m

C

1.5 m

D

2.2 m

Text Solution

Verified by Experts

The correct Answer is:
B
The equation of continuity
`A_(1)v_(1)=A_(2)v_(2), Axx4=(2)/(3)Axxv_(2), " From Bernoulli.s theorem "P+rhogh_(1)+(1)/(2)rhov_(1)^(2)=P+rhogh_(2)+(1)/(2)rhov_(2)^(2) or g(h_(1)-h_(2))=(1)/(2)(v_(2)^(2)-v_(1)^(2))`
`gxxh=(1)/(2)[(6)^(2)-(4)^(2)][because h_(1)-h_(2)=h]rArr h=1m`
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NARAYNA-MECHANICAL PROPERTIES OF FLUIDS-EXERCISE - II (C.W) (EQUATION OF CONTINUITY, BERNOULLI.S THEOREM AND ITS APPLICATIONS)
  1. Water from a tap emerges vertically downwards with initial velocity 4m...

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