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

Answer

D

Explanation

The equation of continuity
`A_1v_1 = A_2v_2`
`rArr A xx 4 = 2/3 A xx v_2`
`rArr v_2= 6ms^(-1)`
From bernoulli's theorem,
`p+rhogh_1 + 1/2 rhov_1^2 = p + rho gh_2 + 1/2 rho v_2^2`
or `g(h_1-h_2)=1/2(v_2^2-v_1^2)`
or `gxxh=1/2[(6)^2-(4)^2] " " (therefore h_1 - h_2=h)`
or `10xxh=1/2[36-16]`
or `h=20/30=1m`