Question

Given $h_1=h_2=3.0m$, the gauge pressure of water in the highest level CD of the tube will be:

• A. $3.0\times {10}^{4}N/m^2$
• B. $5.9\times {10}^{4}N/m^2$
• C. $7.9\times {10}^{4}N/m^2$
• D. $1.50\times {10}^{4}N/m^2$

Answer 1

The correct option is B $5.9\times {10}^{4}N/m^2$

Velocity will be same in siphon at every point.

Pressure at the exit of the siphon is $P_o$.

Applying Bernoulli's equation between the top of vessel and and exit of siphon,

$P_o=P_o-\rho g{h}_2+\frac{\rho v^2}{2}$

$⟹v=\sqrt{2g{h}_2}$ ...(1)

On applying Bernoulli's equation between A and C,

$P_o=P+\rho g{h}_1+\frac{\rho v^2}{2}$

Putting 1 in above equation,

Also gauge pressure $P_1=P_o-P$

$P_o=P_o-P_1+\rho g{h}_1+\frac{\rho \left(2g{h}_2\right)}{2}$

$P_1=\rho g(h_1+h_2)$

$P_1=1000\left(9.8\right)(3+3)$

We get $P_1=5.9\times {10}^{4}N/m^2$