Find the pressure on the container having height 4cm density of water is 1000kg\m³
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Given:
Atmospheric pressure, pa=1.0×105 N/m2Density of water, ρW=103 kg/m3Acceleration due to gravity, g=10 m/s2Volume of water, V=500 mL≈500 g≈0.5 kg
Area of the top of the glass, A = 20 m2
Height of the glass, h = 20 cm
(a) Force exerted on the bottom of the glass = Atmospheric force + Force due to cylindrical water column or glass
=pa×A+A×h×ρw×g=Aρa+hρwg=20×10-4105+20×10-2×103×10=204 N
(b) Let Fs be the force exerted by the sides of the glass. Now, from the free body diagram of water inside the glass, we can find out the resultant force exerted by the sides of the glass.
Thus, we have:
pa×A+mg=A×h×ρw×g+Fs+pa×A⇒mg=A×h×ρw×g+Fs⇒0.5×1=20×10-4×20×10-2×10-3×10+Fs⇒Fs=5-4=1 N (upward)
Atmospheric pressure, pa=1.0×105 N/m2Density of water, ρW=103 kg/m3Acceleration due to gravity, g=10 m/s2Volume of water, V=500 mL≈500 g≈0.5 kg
Area of the top of the glass, A = 20 m2
Height of the glass, h = 20 cm
(a) Force exerted on the bottom of the glass = Atmospheric force + Force due to cylindrical water column or glass
=pa×A+A×h×ρw×g=Aρa+hρwg=20×10-4105+20×10-2×103×10=204 N
(b) Let Fs be the force exerted by the sides of the glass. Now, from the free body diagram of water inside the glass, we can find out the resultant force exerted by the sides of the glass.
Thus, we have:
pa×A+mg=A×h×ρw×g+Fs+pa×A⇒mg=A×h×ρw×g+Fs⇒0.5×1=20×10-4×20×10-2×10-3×10+Fs⇒Fs=5-4=1 N (upward)
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