The diameter of one end of a tube is 2 cm and that of another end is 3 cm . Velocity and pressure of water at narrow end are `2ms^(-1)and1.5xx10^(5)Nm^(9-2)` respectively . If the height differnce between narrow and broad ends is 2.5 m , find the velocity and pressure of water at the broad end .(Density of water is `1xx10^(3)kgm^(-3))`. The narrow end is higher .

Explanation

The narrow end of the flow tube :
`d_(1)=2cm`
`thereforer_(1)=1cm=1xx10^(-2)m`
`v_(1)=2ms^(-1)`
`P_(1)=1.5xx10^(5)Nm^(-2)`
The broad end of the flow tube :
`d_(2)=3cm`
`thereforer_(2)=1.5cm=1.5xx10^(-2)m`
`v_(2)=?,P_(2)=?`
`A_(1)v_(1)=A_(2)v_(2)`
`thereforev_(2)=(A_(1))/(A_(2))v_(1)=(pir_(1)^(2))/(pir_(2)^(2))*v_(1)`
`thereforev_(2)=(r_(1)^(2))/(r_(2)^(2))*v_(1)=((1xx10^(-2))^(2))/((1.5xx10^(-2))^(2))xx2`
`thereforev_(2)=089m//s`
According to Bernoulli.s equation,
`P_(1)+(1)/(2)rhov_(1)^(2)+rhogy_(1)=P_(2)+(1)/(2)rhov_(1)^(2)+rhogy_(2)`
`thereforeP_(2)=P_(1)+(1)/(2)rho(v_(1)^(2)-v_(2)^(2))+rhog(y_(1)-y_(2))`
`=(1.5xx10^(5))+(1)/(2)xx1 xx10^(3)[(2)^(2)-(0.89)^(2)]+1xx10^(3)xx9.8xx2.5`
`thereforeP_(2)=1.76xx10^(5)N//m^(2)`