A siphon tube is discharging a liquid of density
`900(kg)/(m^(3))` as shown in figure. `(P_(0)=1.01xx10^(5) N//m^(2))`

Pressure at point B is

Options

A. `4.25x10^(4) N//m^(2)`
B. `6.25xx10^(4)N//m^(2)`
C. `2.50xx10^(4)N//m^(3)`
D. `2.0xx10^(5)N//m^(2)`

Answer

A

Explanation

Applying Bernoull's theorem at A and B, we
have `P_(0)+0+0=P_(B)+(1)/(2)pv^(2)+pg(1.5)`
or `P_(B)=(1.01xx10^(5))-(1)/(2)xx900xx(10)^(2)-900`
`xx10xx1.5=4.25xx10^(4) N//m^(2)uparrow`