91Ó°ÊÓ

The thickness of the oil layer is $h_1=0.120\mathrm{m}$.

The thickness of water layer is $h_2=0.250\mathrm{m}$.

The density of oil is $\mathrm{ÒÏ}=600\mathrm{kg}/{\mathrm{m}}^3$.

In order to evaluate the gauge pressure at the bottom of the barrel, it is required to add the gauge pressure at the bottom of oil layer and the gauge pressure at the bottom of the water layer.

The relation of gauge pressure can be written as:

$P_g=\mathrm{ÒÏ}g{h}_1$

Here, gis the gravitational acceleration.

Substitute 0.120 m for$h_1$, $600\mathrm{kg}/{\mathrm{m}}^3$for $\mathrm{ÒÏ}$, and $9.80\mathrm{m}/{\mathrm{s}}^2$for $g$in the above relation.

$$\begin{gathered} P_g=\left(600\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(0.120\mathrm{m}\right) \\ {P}_g=705.6\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the required gauge pressure is $705.6\mathrm{N}/{\mathrm{m}}^2$.

The relation of gauge pressure can be written as:

$P_g\text{'}=P_g+{\mathrm{ÒÏ}}_{w}g{h}_2$

Here,${\mathrm{ÒÏ}}_w$ is the density of water.

Substitute 0.250 m for $h_2$, $1000\mathrm{kg}/{\mathrm{m}}^3$for $\mathrm{ÒÏ}$, $705.6\mathrm{N}/{\mathrm{m}}^2$for $P_g$and $9.80\mathrm{m}/{\mathrm{s}}^2$for $g$in the above relation.

$$\begin{gathered} P_g\text{'}=705.6\mathrm{N}/{\mathrm{m}}^2+\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(0.250\mathrm{m}\right) \\ {P}_g\text{'}=3155.6\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the required gauge pressure is $3155.6\mathrm{N}/{\mathrm{m}}^2$.