Answer:
The answer is "[tex]1155\ \frac{kg}{m^3}[/tex]"
Explanation:
Please find the complete question in the attached file.
[tex]p = p_0 + ?gh[/tex]
pi = pressure only at two liquids' devices
PA = pressure atmosphere.
1 = oil density
2 = uncertain fluid density
[tex]h_1 = 11 \ cm\\\\h_2= 3 \ cm[/tex]
The pressures would be proportional to the quantity [tex]11-3 = 8[/tex] cm from below the surface at the interface between both the oil and the liquid.
[tex]\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\[/tex]
[tex]= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}[/tex]