The period of oscillation of mercury of mass  $m$ and destiny $\rho$  poured into a bent tube of cross sectional area $s$  whose right arm forms an angle ${60}^{°}$  with the vertical is

If the two limbs of the U-tube make angle $0_1$ and $0_2$ respectively with the vertical, then the time period of oscillation of the liquid is given by

$T-2\pi \sqrt{\frac{m}{S\rho (\cos {\theta }_1+\cos {\theta }_2)g}}$

In the given case, we have  ${\theta }_1=0^{\circ }$and  ${\theta }_2={60}^{\circ }$, so we get

$T-2\pi \sqrt{\frac{m}{S\rho (\cos 0^{\circ }+\cos {60}^{\circ })g}}-2\pi \sqrt{\frac{2m}{3\rho gS}}$