New mercury positions after pressure rise

Assumptions from the figure: the left limb (water side) is vertical, the right limb is inclined at $15^\circ$, the bore is uniform, and the pressure at B (air, large bulb) stays at 100 kPa. Initially the mercury is in balance at $p_A=p_B=100$ kPa. $\gamma_{Hg}=13.6\times9.81=133.4$ kN/m³, $\gamma_w=9.81$ kN/m³.

  A (water)                B (air)
     |                      /
     |  drop dh            /  dL along 15 deg
     |_____ mercury ______/

Geometry. The tube has uniform bore, so the mercury length moved in each limb is equal: the vertical drop on the water side is $\Delta h$, so the movement along the incline is $\Delta L=\Delta h$. The vertical rise on the inclined side is $\Delta L\sin15^\circ$.

Pressure balance. At the water-mercury interface the water column above has lengthened by $\Delta h$:

$$p_A+\gamma_w\Delta h=p_B+\gamma_{Hg}\left(\Delta h+\Delta L\sin15^\circ\right)$$ $$130-100=\Delta h\left[133.4(1+\sin15^\circ)-9.81\right]$$ $$30=\Delta h\,(167.95-9.81)=158.14\,\Delta h$$ $$\Delta h=0.1897\ \text{m}$$ $$\Delta L=\Delta h=0.1897\ \text{m},\qquad \text{vertical rise on right}=0.1897\sin15^\circ=0.0491\ \text{m}$$

Answer: mercury on the water side drops about 19.0 cm; on the inclined side it moves about 19.0 cm along the tube (a rise of 4.9 cm).