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(D) When the liquid column is displaced by a small distance $y$ from its equilibrium position,a restoring force is generated due to the difference in

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$2 \pi \sqrt{\frac{L}{2 g}}$

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(D) When the liquid column is displaced by a small distance $y$ from its equilibrium position,a restoring force is generated due to the difference in the heights of the liquid levels in the two arms.The difference in height between the two arms becomes $2y$.The restoring force $F$ is equal to the weight of the excess liquid column of height $2y$: $F = -(A \cdot 2y \cdot \rho \cdot g)$,where $A$ is the cross-sectional area and $\rho$ is the density.The mass of the liquid column is $m = A \cdot L \cdot \rho$.Using Newton's second law,$F = ma$:$A \cdot L \cdot \rho \cdot \frac{d^2 y}{dt^2} = -2y \cdot A \cdot \rho \cdot g$$\frac{d^2 y}{dt^2} = -(\frac{2g}{L})y$Comparing this with the standard $SHM$ equation $\frac{d^2 y}{dt^2} = -\omega^2 y$,we get $\omega^2 = \frac{2g}{L}$.Therefore,the time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{2g}}$.

$A$ $U$-tube of a uniform bore has its arms vertical. The total length of the liquid in the two arms of the $U$-tube is $L$. The time period $T$ of the oscillation of the liquid column when it is displaced by $y$ is ($g$ = acceleration due to gravity).

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