Question

A $U$ tube of uniform bore of cross-sectional area $A$ has been set up vertically with open ends facing up. Now $mgm$ of a liquid of density $d$ is poured into it. The column of liquid in this tube will oscillate with a period $T$ such that

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

Answer 1

Answer: (D)

If the level of liquid is depressed by $ycm$ on one side, then the level of liquid in column $P$ is $2ycm$ higher than $B$ as shown.

The weight of extra liquid on the side $P=2Aydg$.
This becomes the restoring force on mass $M$.
$\therefore$ Restoring acceleration $=\frac{-2Aydg}{M}$
This relation satisfies the condition of SHM i.e., $a\propto -y.$
Hence time period
$T=2\pi \sqrt{\frac{\text{ Displacement }}{\mid \text{ Acceleration }\mid }}$
$=2\pi \sqrt{\frac{\frac{y}{2Aydg}}{M}}$
$\Rightarrow T=2\pi \sqrt{\frac{M}{2Adg}}$