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))`

Options

A. 2 m
B. 1 m
C. 0.5 m
D. 4 m

Answer

B

Explanation

The equation of continuity
`A_(1)v_(1)=A_(2)v_(2) Rightarrow Axx4=(2)/(3)Axxv_(2) Rightarrow v_(2)=6ms^(-1)`
From Bernoulli's theoram,
`p+pgh_(1)+(1)/(2)pv_(1)^(2)=p+pgh_(2)+(1)/(2)pv_(2)^(2)`
`or g(h_(1)-h_(2))=(1)/(2)(v_(2)^(2)-v_(1)^(2)) or g xx h=(1)/(2)[(6)^(2)-(4)^(2)](therefore h_(1)-h_(2)=h)`
`or 10xxh=(1)/(2)[36-16] or h=(20)/(20)=1m`