Question

If the cross-sectional area of limb $\text{I}$ is $A_1$ and that of limb $\text{II}$ is $A_2$, then the velocity of the liquid in the tube will be, (cross-sectional area of tube is very small)

• A. $\sqrt{2g(X-Y)}$
• B. $\frac{A_1}{A_2}\sqrt{2g(X-Y)}$
• C. $\frac{A_2}{A_1}\sqrt{2g(X-Y)}$
• D. $0$

Answer 1

The correct option is A $\sqrt{2g(X-Y)}$


$P_1=P_0+\rho gX$

$P_2=P_0+\rho gY$

$\therefore \Delta p=P_1-P_2=\rho g(X-Y)$

From Bernoulli's equation between $1$ and $2$,

$\text{Difference in pressure energy}=\text{difference in kinetic energy}$

$\rho g(X-Y)=\frac{1}{2}\rho V^2$

$\Rightarrow V=\sqrt{2g(X-Y)}$

Hence, option $\left(\text{A}\right)$ is correct.