Question

A cylindrical jar of cross sectional area of 50cm^2 is filled with water to a height of 20cm. it carries a tight-fitting piston of negligible mass. the

pressure at the bottom of the jar when a mass of 1kg is placed on the piston (ignore atmospheric pressure, given:g=9.8)

Answer 1

Area A = 50 cm^2  = 50 /10000 m^2 = 0.005 m^2
height = 20 cm = 0.20 meter

Pressure due to atmospheric air = 1.013 * 10^5 Pa = 101,300 Pa
Pressure on the surface of water = by the mass of 1kg
        =   weight of 1kg/area of piston
       = 1 kg * 9.8 m/sec² / 0.005 m^2 
       = 1,960  Pa

Pressure due to the water column (depth) of water at the bottom of the jar
     = weight of water column / area of cross section
     = ρ g h                (ρ is the density )
     = 1000 kg/m³ * 9.8 m/sec² * 0.20 m
     = 1,960 Pa

If we Ignore atmospheric pressure it is: 3, 920 Pa
 but if we count the atmospheric pressure:  then 105, 220 Pa