At first, the mercury will attain the same height in both the limbs. Now a column of water 13.4 cm high is poured into the left limb and, due to this, the mercury level falls through a depth of x cm from its initial position. Since the cross-sectional areas of both the limbs are the same, the mercury level will rise through a height of x cm from its original position in the other limb.
Now, pressure at point C = pressure at point E
or, `13.4xx1xxg=2x xx13.4xxgor,x=1/2=0.5cm`
The mercury level in the other limb will rise by 0.5 cm from its original position.