One arm of a communicating vessel containing mercury is closed by a piston $20 cm$ above the mercury. The other arm is open. The mercury level is the same in both arms, whose cross–sectional areas are  $2c{m}^2$. In an isothermal process the piston is pushed down by 10 cm. Find the change in height (x cm) of right arm (Nearest integer)
 

If the mercury level sinks by x cm, then the rise in the other arm is also x cm. If we calculate in the Hg cm unit of pressure, then we get a very simple equation for the requested rise in the level.
The initial pressure of the enclosed air is  $p_0=76$ Hg-cm, the final pressure is  $p_1=(76+2x)$ Hg-cm.

According to Boyle’s law  $p_0{h}_{0}A=(p_0+2x)(h_0+\Delta h+x)A$
In our case  $\Delta h=-10cm=-h_0/2$. Substituting this and simplifying by A gives  $p_0{h}_0=(p_0+2x)(\frac{h_0}{2}+x)$, numerically  $76Hgcm.20cm=(76+2x)Hgcm.(10+x)cm$
From here, (omitting the dimensions) equation
  $x^2+48x-380=0$
Is acquired, whose positive solution is   $x=\frac{-48+\sqrt{{48}^2+4.380}}{2}cm=6.9cm$