Question

A horizontally oriented tube AB of length 'l' rotates with a constant angular velocity ${}^{\prime }{\omega }^{\prime }$ about a stationary vertical axis $O_1{O}_2$ passing through the end A as shown in the figure. The tube is filled with an ideal fluid. The end A of the tube is open. [Consider $P_0=P_{atm}$ ]

A very small hole is now drilled at point 'B'. The velocity of the fluid from the hole as a function of the length of liquid column 'h' is [consider area of hole to be much smaller than area of cross-section of tube]

• A. $\omega \sqrt{2hl-h^2}$
• B. $\omega \sqrt{2h^2-hl}$
• C. $\left(\frac{\omega }{\sqrt{2}}\right)\sqrt{2hl-h^2}$
  
• D. $\omega \sqrt{4hl-h^2}$

Answer 1

The correct option is A $\omega \sqrt{2hl-h^2}$

$(P_B{)}_{inside}=P_0+\frac{\rho {\omega }^2}{2}[l^2-(l-h{)}^2]$
Applying Bernoulli’s theorem at 1 and 2, we get
$\frac{1}{2}\rho v_1^2+P_1=\frac{1}{2}\rho v_2^2+P_2...\left(i\right)$

$P_1=(P_B{)}_{inside}and{P}_2=P_0$
Also, $a{v}_2=A{v}_1...\left(ii\right)$
From (i) and (ii), we get
$\frac{1}{2}\rho (v_2^2-v_1^2)=P_1-P_2$
or, $\frac{1}{2}\rho v_2^2(1-\frac{a^2}{A^2})=\frac{\rho {\omega }^2}{2}(2hl-h^2)$
or, $\frac{1}{2}\rho v_2^2=\frac{\rho {\omega }^2}{2}(2hl-h^2)[(1-\frac{a^2}{A^2})\approx 1]$
$\Rightarrow v_2=\omega \sqrt{2hl-h^2}$