Answer:
[tex]\rho=2710.957\ kg.m^{-3}[/tex]
Explanation:
Given:
height of the given liquid in the tank, [tex]h=25\ ft=7.62\ m[/tex]
pressure at the surface of the liquid, [tex]P_{surf}=3\ atm[/tex]
pressure at the bottom of the liquid, [tex]P_{botm}=5\ atm[/tex]
So the pressure due to height of the liquid column:
[tex]\Delta P=P_{botm}-P_{surf}[/tex]
[tex]\Delta P=5-3[/tex]
[tex]\Delta P= 2\ atm=202650\ Pa[/tex]
Now as we know that the pressure due to the height of liquid column is given as:
[tex]\Delta P=\rho.g.h[/tex]
where:
[tex]\rho=[/tex] density of the liquid
[tex]g=[/tex] acceleration due to gravity
[tex]202650=\rho\times 9.81\times 7.62[/tex]
[tex]\rho=2710.957\ kg.m^{-3}[/tex]
