A thin tube sealed at both ends is \(100\, cm\) long. It lies horizontally, the middle \(20\, cm\) containing mercury and two equal ends containing air at standard atmospheric pressure . If the tube is now turned to a vertical position, by what amount will the mercury be displaced ? (Given : cross-section of the tube can be assumed to be uniform) ........ \(cm\)

\(P_1=\) 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 \(y=5.18\; cm\)