91Ó°ÊÓ

There are four different situations for liquids in the U-tube.

Using the concept of absolute pressure, we can get the equilibrium state. Liquids in the U- tube will be in static equilibrium when absolute pressure on both sides will be the same.

Formula:

The absolute pressure is equal to the sum of the atmospheric pressure and gauge pressure, $p_{abs}=p_0+\mathrm{ÒÏ}gh$ (i)

a)

Pressures on both sides of the U-tube must be the same for the liquids to be in equilibrium. For situation (2), the grey liquid is at the same level, implying that the pressure on the grey level is the same. But this cannot be true.

Because, on the left hand side, there is additional weight of red liquid; this would cause some extra pressure along with the atmospheric pressure. Again on the right hand side, there is only the pressure becauseof the atmosphere. Hence for equilibrium,the levelon the right hand side should move up to have the same pressure on both the sides. Therefore, situation (2) is not in static equilibrium.

There is more pressure on the left side, so the liquids cannot be in static equilibrium.

b)

Consider the lower line as the reference line. For equilibrium, the pressure on both thesidesshouldbe the same. Thus, it can be given using equation (i) as follows:

${\mathrm{ÒÏ}}_{red<}{\mathrm{ÒÏ}}_{grey}$

$$\begin{gathered} {\mathrm{ÒÏ}}_{0+}{\mathrm{ÒÏ}}_{red}g{h}_{red}={\mathrm{ÒÏ}}_0+{\mathrm{ÒÏ}}_{grey}g{h}_{grey} \\ {\mathrm{ÒÏ}}_{red=}{\mathrm{ÒÏ}}_{grey}\left(\frac{h_{grey}}{h_{red}}\right) \end{gathered}$$

For situation (1),$h_{grey<}{h}_{red}$the density relation is given by,

${\mathrm{ÒÏ}}_{red<}{\mathrm{ÒÏ}}_{grey}$

For situation (3),$h_{grey<}{h}_{red}$the density relation is given by,

${\mathrm{ÒÏ}}_{red<}{\mathrm{ÒÏ}}_{grey}$

For situation (4),$h_{grey<}{h}_{red}$the density relation is given by,

${\mathrm{ÒÏ}}_{red<}{\mathrm{ÒÏ}}_{grey}$

Hence, the density relations can be given in the other three situations.