Answer: [tex]1500\ kg/m^3,\ 3.5\ m/s^2[/tex]
Explanation:
Given
The pressure exerted at a depth of [tex]h=2.5\ m[/tex] is [tex]P=36,750\ Pa[/tex]
Also, the gauge pressure is given by
[tex]P=\rho gh[/tex]
Putting values
[tex]\Rightarrow 36,750=\rho \times 9.8\times 2.5\\\\\Rightarrow \rho=\dfrac{36750}{24.5}=1500\ kg/m^3[/tex]
(b)
The density of the liquid is [tex]\rho =1000\ kg/m^3[/tex]
depth [tex]h=2.4\ m[/tex]
Pressure [tex]P=8400\ Pa[/tex]
We can write
[tex]\Rightarrow P=\rho g'h\\\\\Rightarrow 8400=1000\times g'\times 2.4\\\\\Rightarrow g'=3.5\ m/s^2[/tex]
