What is the mass of the liquid in the vat?

A 0.60-m-diameter vat of liquid is 2.7m deep. The pressure at the bottom of the vat is 1.5atm .

Answer 1

p=p0+ρgh
then
ρ=(p-p0)/(gh)
1 atm=101,325 Pa
ρ=(1.5-1)*101,325/(9.8*2.7)
ρ=1914.7 kg/m^3
m=ρV
V=0.76 m^3
m=1914.7*0.76=1455.2 kg

Answer 2

Get the volume and use the pressure to determine the density.

What is missing is the pressure is absolute or relative. I’ll assume absolute.

V = πr²h = π(0.3)²(2.7) = 0.763 m³

Pfluid = ρgh
Pfluid is pressure in Pa or N/m²
ρ is the density of the fluid in kg/m³
g is the acceleration of gravity 9.8 m/s²
h is the height of the fluid above the object in m
1.5 atm is 1.5•101 kPa

ρ = P/gh = (151.5e3) / (9.8)(2.7) = 5726 kg/m³
(possibly molten tin)

5726 kg/m³ x 0.763 m³ = 4370 kg

Answer 3

the pressure times the area of the bottom of the vat = the total force of gravity on the mass

this is a snap to calculate

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