Question

U-tube moves with a constant speed $v$ parallel to the surface of a stationary liquid. The cross-section area of the lower part of the tube lowered into the liquid, is equal to $S_{1}$ and that of the top part located over the liquid is $S_{2}$ . Friction and formation of waves should be neglected. The density of the fluid is $\rho $ . Neglect difference in heights at both the openings of the tube.



The velocity of the liquid coming out of the top part as seen by an observer on the ground will be

• A. $v\left(\frac{S_{1}}{S_{2}}\right)$
• B. $v\left(1 + \frac{S_{1}}{S_{2}}\right)$
• C. zero
• D. none of these

Answer 1

Answer: (B)

Using the equation of continuity, we get
$S_{1}v_{1}=S_{2}v_{2}$
$v_{1}=v$ (liquid is entering in pipe with a speed $v$ with respect to the pipe)
$\Rightarrow v_{2}=\frac{S_{1}}{S_{2}}v$
$v_{2}:$ velocity of liquid at $S_{2}$ with respect to pipe
$\therefore \, \, $ With respect to ground, the speed of liquid will be,
$v+\frac{S_{1}}{S_{2}}v=v\left(1 + \frac{S_{1}}{S_{2}}\right)$