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. `5.0xx10^(-4) m^(2)`
B. `1.0xx10^(-5)m^(2)`
C. `5.0xx10^(-5) m^(2)`
D. `2.0xx10^(-5) m^(2)`

Answer

C

Explanation

Decrease in potential energy = increase in kinetic energy
`therefore" "rhogh=1/2rho(v_(f)^(2)-v_(i)^(2))`
`or" "2(10)(0.15)-v_(f)^(2)-(10)^(2)`
`or" "v_(f)= 2 ms^(-1)`
Now, from continuity equation,
`A_(1)v_(1)=A_(2)v_(2)" or "Aprop1/v`
Velocity has become two times, Hence, area of cross-section will remain half.