Question

A barrel contains a 0.120 m layer of oil floating on water that is 0.250 m deep. The density of the oil is ${600kg/m}^3$. (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?

Answer 1

Let P be the atmospheric pressure P, be the pressure at oil-water interface and $P_2$ be the pressure at the bottom of the barrel

Let, ${\rho }_o$ be the density of oil and ${\rho }_w$ be the density of water

${\rho }_o=600kg/m^3,{\rho }_w=1000kg/m^3$

$\left(a\right)$ We know that

$P_1=P_0+{\rho }_{o}gh$ - $\left(1\right)$ (h-depth of oil layer)

$\Rightarrow P_1-P_0=600\times 9.8\times 0.12$

$=72\times 9.8Pa$

$=705.6Pa$

$\Rightarrow$ Gauge pressure at the oil-water

interface $=705.6Pa$

($\because$ Gauge pressure at interface $=P_1-P_0$)

$\left(b\right)$ $P_2=P_1{\rho }_{w}g{h}^{\prime }$(h-depth of water layer)

Now, $P_1=P_0+{\rho }_{o}gh$ (from $\left(1\right)$)

$\Rightarrow P_2=P_0+{\rho }_{o}gh+{\rho }_{w}gh$

$\Rightarrow (P_2-P_0)={\rho }_{0}gh+{\rho }_{w}gh$

$=(600\times 9.8\times 0.12)+(1000\times 9.8\times 0.25)$

$=705.6+2450Pa$

$=3155.6Pa$

$\Rightarrow$ Gauge pressure at the bottom of the barrel

$=3155.6pa$

($\because$ Gauge pressure at bottom $=P_2-P_0$)