The reading of a barometer fitted with a closed pipe is `3.5 xx 10^(5) Nm^(-3)`. When the valve ef the pipe is opened, the pressure read by barometer reduces to `3.0 xx 10^(5)Nm^(-2)`. Calculate the velocity of flow of water in the pipe.

Explanation

`P_(1)=3.5 xx 10^(5)Nm^(-2), P_(2)=3.0 xx 10^(5)Nm^(-2)`
We know that `(P_(1)-P_(2))=(1/2) rho(v_(2)^(2)-v_(1)^(2))`
If the value is closed , `v_(1)=0`
`(3.5 -3.0)10^(5)=1.2 xx 10^(3) (v_(2)^(2)-0)`
i.e, `0.5 xx 105=500v_(2)^(2)`
`therefore v_(2)=sqrt((5 x 10^(4))/(500)) =10ms^(-1)`.