The pressure at the bottom of a tank of water | vedclass

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(B) The total pressure at the bottom of the tank is given by $P_{total} = P_{atm} + P_{gauge} = P + h\rho g = 3P$.Therefore,the gauge pressure due to

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$(13/5)P$

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(B) The total pressure at the bottom of the tank is given by $P_{total} = P_{atm} + P_{gauge} = P + hho g = 3P$.Therefore,the gauge pressure due to the water column is $hho g = 3P - P = 2P$.When the water level is lowered by one-fifth,the new height of the water column becomes $h' = h - (1/5)h = (4/5)h$.The new gauge pressure is $P'_{gauge} = h'ho g = (4/5)hho g$.Substituting the value of $hho g$,we get $P'_{gauge} = (4/5) 	imes 2P = (8/5)P$.The new total pressure at the bottom is $P'_{total} = P_{atm} + P'_{gauge} = P + (8/5)P = (13/5)P$.

The pressure at the bottom of a tank of water is $3P$,where $P$ is the atmospheric pressure. If the water is drawn out until the level of water is lowered by one-fifth,the pressure at the bottom of the tank will now be:

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