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