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 .

The diameter of one end of a tube is 2 cm and that of another end is 3

1.	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 .
Answer» SOLUTION :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)xx1xx10^(3)[(2)^(2)-(0.89)^(2)]+1xx10^(3)xx9.8xx2.5` `thereforeP_(2)=1.76xx10^(5)N//m^(2)`