Water from a tap emerges vertically downward with an initial speed of `1.0ms^(-1)`. The cross-sectional area of the tap is `10^(-4) m^(2)`. Assume that the flow is steady. What is the cross-sectional area of the stream 0.15 m below the tap? Use `g=10 ms^(-2)`.

Explanation

Here, `upsilon_(1) = 1.0 ms^(-1) , a_(1) = 10^(4) m^(2)`,
`h=0.15 m, a_(2) =?`
As `P_(1) = rho gh_(1) +1/2 rho upsilon_(1)^(2) = P_(2) +rho gh_(2) +1/2 rho upsilon_(2)^(2)`
Since `P_(1) = P_(2)`, so
`rho gh_(1) +1/2 rho upsilon_(1)^(2) = rho gh_(2) +1/2 rho upsilon_(2)^(2)`
`or upsilon_(2)^(2) = upsilon_(1)^(2) + 2 g(h_(1)-h_(2)) = upsilon_(1)^(2) + 2gh`
`=(1.0)^(2) + 2 xx 10 xx 0.15 = 4`
or, `upsilon_(2) = 2 ms^(-1)`
Now, `a_(1) upsilon_(1) =a_(2) upsilon_(2)`
or `a_(2) = (a_(1)upsilon_(1))/(upsilon_2) = (10^(-4)xx 1.0)/(2.0) = 5 xx 10^(-5)m^(2)`.