the pressure at the bottom of a tank is 3P where P is the atmospheric pressure if the water is drawn out till the level of water is lowered by One fifth the pressure at the bottom of the tank will now be
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The hydrostatic pressure due to the water is (4P - P) = 3P
Reduce the depth of water by 3/5 so that it is 2/5 of its original value and the hydrostatic pressure due to the water becomes (2/5) * 3P = 6P/5
But you still have the atmospheric pressure at the surface, so the pressure at the bottom of the tank will be:
(6P/5) + P = (6P + 5P) / 5 = 11P/5
Or in description language
pressure is proportional to density * gravity * height. If the original pressure was 4P than decreasing the water by 3/5ths which would be (1 - 3/5 = 2/5) of the original height than the the new pressure at the bottom of the tank would be 2/5 * 4P = 8P/5. I don't think 11P/5 is the correct answer.
Now if you are getting gauge and absolute suppress mixed up and meant to say the height of the tank was 3/5 of the original height than the answer would be 11P/5
Reduce the depth of water by 3/5 so that it is 2/5 of its original value and the hydrostatic pressure due to the water becomes (2/5) * 3P = 6P/5
But you still have the atmospheric pressure at the surface, so the pressure at the bottom of the tank will be:
(6P/5) + P = (6P + 5P) / 5 = 11P/5
Or in description language
pressure is proportional to density * gravity * height. If the original pressure was 4P than decreasing the water by 3/5ths which would be (1 - 3/5 = 2/5) of the original height than the the new pressure at the bottom of the tank would be 2/5 * 4P = 8P/5. I don't think 11P/5 is the correct answer.
Now if you are getting gauge and absolute suppress mixed up and meant to say the height of the tank was 3/5 of the original height than the answer would be 11P/5
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