$P_1=$ Preessure of par $M$

$P_2=$ Pressure of part $N$

As temperature is constant

Therefore, we can write $PV =$ constant

For $M : P_0(40 A)=P_1(40-y) A$

$\Rightarrow P_1=\frac{P_0(40)}{(40-y)} \ldots \ldots .(\text { i })$

For $N : P_0(40 A)=P_2(40+y) A$

$\Rightarrow P_2=\frac{P_0 40}{(40+y)}$

Here $A=$ area of cross section of tube

Pressure at lower face $\left(P_1\right)=$ Pressure at upper face $\left(P_2\right)+$ Pressure due to mercury column of height $20 cm$.

$P_1=P_2=20 pg$

Use (i) and (ii) in (iii)

$P_0 \frac{40}{(40-y)}=\frac{P_0 40}{(40+y)}+20 p g$

$\frac{40 P_0}{(40-y)}-\frac{40 P_0}{(40+y)}=20 p g$

$40 \times 76 p g\left[\frac{2 y}{1600-y^2}\right]=20 p g$

$(52)(2 y)=1600-y^2$

$304 y=1600-y^2$

$y^2+304 y-1600=0$

Solving for $y$, we get $y=5.18\; cm$