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. 0.5m
B. 1m
C. 1.5m
D. 2.2m

Answer

B

Explanation

Contiuity equation,
`A_(1)V_(1)=A_(2)V_(2)`
`implies A_(1)xx4= 2/3 A_(1)xxV_(2)`
`therefore V_(2)=6m/s`
Bernoullis equation,
`P+rho gh_(1) + 1/2 rhoV_(1)^(2) = P + rhogh_(2) + 1/2rhoV_(2)^(2)`
i.e.,`g(h_(1)-h_(2))=1/2(V_(2)^(2)-V_(1)^(2))`
`therefore h=1/2 (V_(2)^(2)-V_(1)^(2)) . 1/g [because h_(1)-h_(2)=h]`