Question

A non viscous liquid of density $\rho$ is filled in a tube with $A$ as the area of cross section, as shown in the figure. If the liquid is slightly depressed in one of the arms, the liquid column oscillates with a frequency

• A. $\frac{1}{2\pi}\sqrt{\frac{\rho g A sin {\left(\frac{\theta_1 +\theta_2}{2}\right)}}{m}}$
• B. $\frac{1}{2\pi}\sqrt{\frac{\rho g A \left(sin \theta_{1} - sin \theta_{2}\right) }{m}}$
• C. $\frac{1}{2\pi}\sqrt{\frac{\rho g A \left(sin \theta_{1} + sin \theta_{2}\right) }{m}}$
• D. $\frac{1}{2\pi}\sqrt{\frac{\rho g A sin {\left(\frac{\theta_1 - \theta_2}{2}\right)}}{m}}$

Answer 1

Answer: (C)

Initially the level of liquid in the two limbs will be at the same height. If the liquid is depressed by $y$ in one limb, it will rise by $y$ along the length of the tube in the other limb. The force that is responsible for restoring the liquid levels in the two arms of the tube is
$F = -\Delta PA = -\left(h_{1} +h_{2}\right)\rho g A$
where $\Delta P$ is the pressure difference and $A$ is the area of cross section of the tube, $h_1$ and $h_2$ being the rise and fall of liquid levels in the two arms in vertical direction respectively.
$F = -\left(y sin \theta_{1} +y sin \theta_{2}\right)\rho gA $
$ = -\rho gA \left(sin \theta_{1} +sin \theta_{2}\right)y \quad...\left(i\right) $
$ ma = -\rho g A \left(sin \theta_{1} +sin \theta_{2}\right)y $
or $ a= -\frac{\rho gA \left(sin \theta_{1}+sin\theta_{2}\right)y}{m} $
For $SHM$
$ a = -\omega^{2}y \quad...\left(ii\right) $
Comparing $\left(i\right)$ and $\left(ii\right)$, we get
$ \omega^{2} = \rho g \frac{A\left(sin \theta_{1} +sin \theta_{1}\right)}{m} $
$\omega = \sqrt{\frac{\rho g A\left(sin \theta_{1}+sin\theta_{2}\right)}{m}}$
Frequency, $\upsilon = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{\rho gA\left(sin \theta_{1} +sin \theta_{2}\right)}{m}}$