Inverted manometer differential level

The figure dimensions are not fully legible, so this reading is used and stated. Specific weight of water $\gamma_w=9.81$ kN/m³, $g=9.81$.

Assumed reading: the pipe centre lies 0.5 m below the free surface in the tank, so the pipe is under a static head of 0.5 m of water. Left leg of the inverted U (water) rises 0.25 m from the pipe to the lighter fluid ($S=0.83$); the manometer reading $R=0.15$ m (left interface is higher than the right one); the right leg water goes down 0.20 m to a mercury U-tube whose other limb is open. We find the differential mercury level $h$.

        ______ oil S=0.83 ______
       |                        |
 water |                  R     | water 0.20
 0.25  |                        |
 pipe  P                        Hg  |--h--| open

Pressure at pipe (gauge):

$$p_P=\gamma_w(0.5)=4.905\ \text{kPa}$$

At the left oil-water interface (rise of 0.25 m in water):

$$p_{L}=4.905-9.81(0.25)=2.453\ \text{kPa}$$

Across the lighter fluid to the right interface (it is $R=0.15$ m lower):

$$p_R=2.453+0.83(9.81)(0.15)=3.674\ \text{kPa}$$

Down 0.20 m of water to the mercury surface:

$$p_{Hg}=3.674+9.81(0.20)=5.636\ \text{kPa}$$

This is balanced by a mercury column $h$ in the open limb:

$$h=\frac{p_{Hg}}{\gamma_{Hg}}=\frac{5.636}{13.6\times9.81}=0.0422\ \text{m}$$

Answer: differential mercury level $h\approx42$ mm (for this reading). The method is the same for other dimensions: add $\gamma h$ going down, subtract going up, and equate to $\gamma_{Hg}h$.