Water from a tap emerges vertically downwards with an initial velocity `V_(0)`. Assume pressure is constant throughout the stream of water and the flow is steady. Find the distance form the tap at which cross-sectional area of stream is half of the cross-sectional area of stream at the tap.

Options

A. `V_(0)^(2)//2g`
B. `3V_(0)^(2)//2g`
C. `2V_(0)^(2)//g`
D. `5V_(0)^(2)//2g`

Answer

B

Explanation

`V_(2)^(2)=V_(0)^(2)+2gh` and `A_(1)v_(0)=A_(2)V_(2)`…….i
solving we get `(A_(2))/(A_(1))=(V_(0))/(sqrt(V_(0)^(2)+2gh))`
`(A_(2))/(A_(1)=1/2=(V_(0))/(sqrt(V_(0)^(2)=2gh))`
`4V_(0)^(2)=V_(0)^(2)+2gh`
`h=(3V_(0)^(2))/(2g)`