Question

An ideal liquid is flowing in two pipes one is inclined and second is horizontal. Both the pipes are connected by two vertical tubes of length $h_1$ and $h{­}_2$ as shown in fig. The flow is streamline in both the pipes. If velocity of liquid at A, B and C are 2 m/s, 4 m/s and 4 m/s respectively, the velocity at D will be

• A. $4m/s$
• B. $\sqrt{14}m/s$
• C. $\sqrt{28}m/s$
• D. $2m/s$

Answer 1

The correct option is C $\sqrt{28}m/s$

Applying Bernoulli’s equation at A and C,
$\frac{1}{2}\rho v_A^2+\rho g{h}_1+P_A=\frac{1}{2}\rho v_C^2+\rho g{h}_2+P_C...\left(1\right)$
at B and D,
$\frac{1}{2}\rho v_B^2+P_B=\frac{1}{2}\rho v_D^2+P_D...\left(2\right)$
Also,
$P_B=P_A+\rho g{h}_1$
$P_D=P_C+\rho g{h}_2$
$\Rightarrow P_B-P_D=(P_A+\rho g{h}_1)-(P_C+\rho g{h}_2)$
$=\frac{1}{2}\rho (v_C^2-v_A^2)from\left(1\right)$
Substituting this result in (2),
$\frac{1}{2}\rho v_D^2=\frac{1}{2}\rho v_B^2+\frac{1}{2}\rho (v_C^2-v_A^2)$
$v_D^2=v_B^2+v_C^2-v_A^2$
$=(4{)}^2+(4{)}^2-(2{)}^2$
$V_D=\sqrt{28}m{s}^{-1}$