# 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)

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by 2092000

2015-04-08T17:14:31+05:30

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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

what is piston sir?