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The pressure at the bottom of a tank of water 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

  • a)
    2 P

  • b)
    (13/5) P

  • c)
    (8/5) P

  • d)
    (4/5) P

Correct answer is option 'B'. Can you explain this answer?
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Solution:

Given:
Pressure at the bottom of the tank = 3P (where P is the atmospheric pressure)

To find:
The pressure at the bottom of the tank when the water level is lowered by one fifth.

Assumption:
We assume that the density of water remains constant throughout the process.

Explanation:
When a liquid is in a container, the pressure at any point at the same horizontal level is the same. This is known as Pascal's law. In this case, the pressure at the bottom of the tank is the same as the pressure at the water level.

Let's consider a small column of water in the tank. The pressure at the bottom of this column is given by the formula:

Pressure = Density × Acceleration due to gravity × Height

Since the density of water remains constant, we can say that the pressure at the bottom of the tank is directly proportional to the height of the water column.

When the water level is lowered by one fifth, the new height of the water column is 4/5 of the original height.

To find the new pressure at the bottom of the tank, we can use the formula:

New Pressure = Old Pressure × New Height / Old Height

Substituting the given values:
New Pressure = 3P × (4/5) / 1

Simplifying the expression:
New Pressure = (12/5)P / 1
New Pressure = (12/5)P

Therefore, the pressure at the bottom of the tank when the water level is lowered by one fifth is (12/5)P or (13/5)P.

Hence, the correct answer is option B, (13/5)P.
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