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)
Answers
Answered by
39
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
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
rajusetu:
what is piston sir?
Similar questions