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
Diffcult
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If we ignore the atmospheric pressure, the pressure at the bottom is $2 P$
We know that the pressure by a liquid column is given by $h \rho g$
$\therefore h \rho g=2 P$
After the height getting lowered by one fifth, the height becomes four fifth.
$\therefore \frac{4}{5} h \rho g=\frac{2 \times 4}{5} P=\frac{8}{5} P$
Now including the atmospheric pressure it becomes
$\left(\frac{8}{5}+1\right) P=\frac{13}{5} P$
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