Question

The figure shows a $U-$ tube of uniform cross-sectional area $A$ having a liquid of mass $M$ and density $\rho$. If the liquid in the column is slightly displaced, then it oscillates with a time period of :
[$g=$ acceleration due to gravity]

• A. $T=2\pi \sqrt{\frac{M}{g}}$
• B. $T=2\pi \sqrt{\frac{MA}{g\rho }}$
• C. $T=2\pi \sqrt{\frac{M}{A\rho g}}$
• D. $T=2\pi \sqrt{\frac{M}{2A\rho g}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{M}{2A\rho g}}$

If the level of liquid is depressed by $y$ in column $Q$, then the level of liquid in column $P$ will be $2y$ higher than the liquid level in column $Q$ as shown in the figure.


Let $m$ be the mass of increased height of liquid column in $P$.
The weight of extra liquid in the column $P$,
$W=mg=\rho Vg=A\left(2y\right)\rho g$
This will act as the restoring force on the entire liquid column.
i.e $F_r=-2Ay\rho g$
$\therefore$ Acceleration $\left(a\right)=\frac{-2Ay\rho g}{M}$
From $a=-{\omega }^{2}y$,
$\omega =\sqrt{\frac{2A\rho g}{M}}$
Thus, the liquid executes SHM.
Time period of oscillation of liquid
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{M}{2A\rho g}}$
Hence, option (d) is the correct answer.