Answer:
The gauge pressure is [tex]P_g = 2058 \ P_a[/tex]
Explanation:
From the question we are told that
The height of the water contained is [tex]h_w = 30 \ cm = 0.3 \ m[/tex]
The height of liquid in the cylinder is [tex]h_t = 40 \ cm = 0.4 \ m[/tex]
At the bottom of the cylinder the gauge pressure is mathematically represented as
[tex]P_g = P_w + P_o[/tex]
Where [tex]P_w[/tex] is the pressure of water which is mathematically represented as
[tex]P_w = \rho_w * g * h_w[/tex]
Now [tex]\rho_w[/tex] is the density of water with a constant values of [tex]\rho_w = 1000 \ kg /m^3[/tex]
substituting values
[tex]P_w = 1000 * 9.8 * 0.3[/tex]
[tex]P_w = 2940 \ Pa[/tex]
While [tex]P_o[/tex] is the pressure of oil which is mathematically represented as
[tex]P_o = \rho_o * g * (h_t -h_w )[/tex]
Where [tex]\rho _o[/tex] is the density of oil with a constant value
[tex]\rho _o = 900 \ kg / m^3[/tex]
substituting values
[tex]P_o = 900 * 9.8 * (0.4 - 0.3)[/tex]
[tex]P_o = 882 \ Pa[/tex]
Therefore
[tex]P_g = 2940 - 882[/tex]
[tex]P_g = 2058 \ P_a[/tex]
