The U tube of uniform bore of cross-sectional area A has been set...

Let the liquid be displaced by a small distance $x$ in one arm of the U-tube.

The liquid level in the other arm rises by $x$, creating a level difference of $2x$ between the two arms.

The restoring force acting on the liquid is due to the weight of this excess liquid column.

$F = -(2x) A d g$

The acceleration of the liquid column of mass $M$ is $a = \dfrac{F}{M}$.

$a = -\left(\dfrac{2 A d g}{M}\right) x$

Comparing this with the standard equation of simple harmonic motion $a = -\omega^2 x$, we get:

$\omega^2 = \dfrac{2 A d g}{M} \Rightarrow \omega = \sqrt{\dfrac{2 A d g}{M}}$

The time period of oscillation is $T = \dfrac{2\pi}{\omega}$.

$T = 2\pi \sqrt{\dfrac{M}{2 A g d}}$

Answer: $2\pi\sqrt{\dfrac{M}{2Agd}}$