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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:

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$(\frac{13}{5}) P$

$(\frac{8}{5}) P$

$(\frac{4}{5}) P$

answer is B.

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Detailed Solution

$\begin{array}{l} 2 P = hρg \Rightarrow h = \frac{2 P}{ρg} \\ P^{'} = (\frac{4}{5} h) ρg = (\frac{4}{5} \times \frac{2 P}{ρg}) ρg = \frac{8 P}{5} \end{array}$
Total pressure $= \frac{8 P}{5} + P = \frac{13 P}{5}$

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