Water from a tap emerges vertically downwards with an initial speed of `1.0m//s`. The cross-section area of tap is `10^(-4)m^(2)`.Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of steam `0.15 m` below the tap is

Options

A. ` 5 xx 10 ^(-4) m ^ 2 `
B. ` 1 xx 10 ^(-5 ) m ^ 2 `
C. `2 xx 10 ^(-5 ) m^2 `
D. ` 5 xx 10 ^(-5 ) m ^ 2 `

Answer

D

Explanation

`P_(0)+rhogh+(1)/(2)rhov_(1)^(2)=P_(0)+(1)/(2)rhov_(2)^(2)`
`RightarrowV_(2)^(2)=V_(1)^(2)+2gh`
`V_(2)=2m//s`
`A_(1)V_(1)=A_(2)V_(2)RightarrowA_(2)=(10^(-4))/(2)=5xx10^(-5)m^(2)`