Find the displacement of mercury in a vertical tube So we have a 1m tube sealed at both ends. It is initially horizontally placed and the middle 0.1m contains mercury and the two ends contain air. Now the tube is turned vertical, we have to find the amount of mercury displaced.

Let the pressure in the upper end be $P_1$ and the pressure in the bottom end be $P_2$. The area of cross-section is $A$ and the lengths in the upper and bottom end are $(0.45-x)$ and $(0.45+x)$,respectively.Let $x$ be the mercury displacement.

By Boyle's Law
$$P_1V_1=P_2V_2$$
$$P_1A(0.45-x)=P_2A(0.45+x)$$
A single equation is insufficient to solve this, what should I do.
I assume the Hg is a competent barrier between the parcels of air and the parcels both have an initial pressure $P_0$ of 1 bar (760mm of Hg). Each air parcel has an identical initial energy content of $$ P_0V_0 = P_0A(0.45\mathrm{m}) $$

With hydrostatic equilibrium, $\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}}$ would all be due to the weight of the Hg, IOW, in mm of Hg terms, would be 100mm (0.1m) of Hg. -- This is your second equation. $$\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}} = 100\, \mathrm{\scriptstyle mmHg} $$From Boyle's Law we know that $$ P_{\mathrm{top}} = P_0 \frac{V_0}{V_{\mathrm{top}}} \quad\quad\text{and}\quad\quad P_{\mathrm{bot}} = P_0 \frac{V_0}{V_{\mathrm{bot}}} $$ Substituting, $$ \Delta P = P_0(\frac{V_0}{V_{\mathrm{bot}}} - \frac{V_0}{V_{\mathrm{top}}}) $$ or $$ (\Delta P) V_{\mathrm{top}} V_{\mathrm{bot}} = P_0V_0(V_{\mathrm{top}} - V_{\mathrm{bot}}) $$ Substituting again, $$ (\Delta P) A^2 ( 0.45\mathrm{m} - x )( 0.45\mathrm{m} + x ) = P_0A^2(0.45\mathrm{m}) ( (0.45\mathrm{m} - x) - (0.45\mathrm{m} + x) ) $$ eliminating $A^2$ and simplifying $$ \Delta P ( (0.45\mathrm{m})^2 - x^2) = -2P_0 x(0.45\mathrm{m}) $$ or in quadratic form $$ x^2 - 2\frac{P_0}{\Delta P}(0.45\mathrm{m})x - (0.45\mathrm{m})^2 = 0 $$ which would be $$ x^2 - (6.84\mathrm{m})x - 0.2025\mathrm{m}^2 = 0 $$ You can take it from there.