Physics - Fluid Question with Solution | TestHub

Using the equation of continuity:

$A_1{v}_1=A_2{v}_2$

$(4\times 10^{-2})\times 2=(2\times 10^{-2})\times v_2$

$v_2=4$ m/s

Using Bernoulli's equation for a horizontal pipe:

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

$4\times 10^4+\frac{1}{2}\times 1000\times (2{)}^2=P_2+\frac{1}{2}\times 1000\times (4{)}^2$

$40000+2000=P_2+8000$

$P_2=42000-8000=34000$ Pa

$P_2=3.4\times 10^4$ Pa

Answer: $3.4\times 10^4$ Pa