Physics - Fluid Question with Solution | TestHub

From the equation of continuity:

$A_A{v}_A=A_B{v}_B$

Given $A_A=4{\text{ cm}}^2$ and $A_B=2{\text{ cm}}^2$:

$4v_A=2v_B\Rightarrow v_B=2v_A$

Using Bernoulli's equation for a horizontal tube:

$P_A+\frac{1}{2}\rho v_A^2=P_B+\frac{1}{2}\rho v_B^2$

$\Rightarrow P_A-P_B=\frac{1}{2}\rho (v_B^2-v_A^2)$

Substituting $v_B=2v_A$:

$P_A-P_B=\frac{1}{2}\rho ((2v_A{)}^2-v_A^2)=\frac{3}{2}\rho v_A^2$

Taking the density of water in CGS units as $\rho =1{\text{ g/cm}}^3$, the pressure difference is:

$P_A-P_B=\frac{3}{2}{v}_A^2$

Answer: $\frac{3}{2}{v}_A^2$