Water flows through a horizontal pipe of varying cross-seetions. The pressure of water equals 0.1 m of mercury at a place where the velocity of flow is `0.4 ms^(-1).` What will be the pressure at another place, where the velocity of flow is `0.5ms^(-1)`.

Explanation

Given `P_(1)=0.1"M of Hg"=0.1 xx 13600 xx 9.8`
`v_(1)=0.4ms^(-1) =1332Nm^(-2) P_(2)=? V_(2)=0.5 ms^(-1)`
Applying `(P_(1)-P_(2))=1/2 rho(v_(2)^(2)-v_(1)^(2))` we get , `=1/2 xx 10^(3) (0.5^(2)-0.4^(2))`
`13328-P_(2)=500(0.9) (0.1)`
`therefore P_(2)=13328-45=13283 Nm^(-2)`
`or P_(2)=(13283)/(13600 xx 9.8)=0.0996"m of mercury"`