Question

A fluid while flowing through a tube as shown passes through points A and B. Find the velocity of the fluid at point B.

• A. $\sqrt{\frac{4}{15}}\text{m/s}$
• B. $\sqrt{\frac{1}{15}}\text{m/s}$
• C. $\sqrt{\frac{2}{15}}\text{m/s}$
• D. $\sqrt{\frac{8}{15}}\text{m/s}$

Answer 1

The correct option is D $\sqrt{\frac{8}{15}}\text{m/s}$

According to question

On applying equation of continuity at point $1$ and $2$,
$A_1{v}_1=A_2{v}_2$
$\Rightarrow v_1=\frac{A_2{v}_2}{A_1}=\frac{2\times v_2}{4}$
$\Rightarrow v_1=\frac{v_2}{2}$.........(1)
On applying bernoulli's equation at point $1$ and $2$,
$P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2$
Since height is same at both the points.
$\Rightarrow P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)$
$P_1-P_2=\frac{1}{2}\rho (v_2^2-\frac{v_2^2}{4})$ $[$ from (1) $]$
$P_1-P_2=\frac{3}{8}\rho v_2^2$............(2)
Also, $P_1-P_2=\rho g{h}_o$
$[$ Pressure difference in column $]$
$P_1-P_2=\rho g\times 0.02$ $[h_o=2\text{cm}=0.02\text{m}]$
$P_1-P_2=\frac{1}{5}\rho$..........(3)
From (2) and (3),
$\frac{3}{8}\rho v_2^2=\frac{1}{5}\rho$
$\Rightarrow v_2=\sqrt{\frac{8}{15}}\text{m/s}$
Speed of water at point $B$ is $\sqrt{\frac{8}{15}}\text{m/s}$