Answer
The mercury in the left arm rises $~~4.12~mm~~$ above its initial height.
Work Step by Step
We can find the height of mercury with the same weight as $11.2~cm$ of water:
$V_m~\rho_m = V_w~\rho_w$
$h_m~A~\rho_m = h_w~A~\rho_w$
$h_m = \frac{h_w~\rho_w}{\rho_m}$
$h_m = \frac{(11.2~cm)~(1000~kg/m^3)}{13,600~kg/m^3}$
$h_m = 0.824~cm$
$h_m = 8.24~mm$
Since the liquid in the right arm falls as the liquid in the left arm rises, the mercury in the left arm would rise half of this height.
The mercury in the left arm rises $~~4.12~mm~~$ above its initial height.
