Question

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, what is the pressure at the bottom of the tank now?
A. $2P$
B. $\dfrac{{13}}{5}P$
C. $\dfrac{8}{5}P$
D. $\dfrac{4}{5}P$

Answer 1

Hint
The pressure at the bottom of a tank consists of two components. One is the atmospheric pressure and the other is the pressure exerted by the weight of the liquid.
$P = \rho gh$ , Where, $P$ is the total pressure at the bottom of a column of height $h$ , filled with liquid of density $\rho$ , and $g$ is the gravitational acceleration. The SI unit of pressure is Pascal.

Complete step by step answer
We are provided with the information that the pressure at the bottom of a tank is $3P$ . We also know that the atmospheric pressure is given by $P$ . We are aware that the total pressure at the bottom of a tank can be written as:
${P_t} = {P_{atm}} + {P_{liquid}}$

So, if we try to find the pressure exerted by the weight of water in this case, we get:
${P_w} = {P_t} - {P_{atm}} \Rightarrow {P_w} = 3P - P$
This gives us the pressure due to the weight of water as $2P$ .
We also know that this pressure can be expressed as:
$P = \rho gh$ [Eq. 1]
Now according to the question, we reduce the level of water (i.e. the height of water h) by one fifth. This gives us the new height as:
$h' = h - \dfrac{1}{5}h \Rightarrow h' = \dfrac{{5h - h}}{5}$
Solving for $h'$ , we get the new height = $\dfrac{{4h}}{5}$
Hence, the new pressure will be given as:
${P_{new}} = \rho gh'$
Substituting the value of $h'$ gives us:
${P_{new}} = \rho g \times \dfrac{4}{5}h \Rightarrow {P_{new}} = \dfrac{4}{5}(\rho gh)$
We use Eq. 1 to get the answer in generic terms:
${P_{new}} = \dfrac{4}{5}P$

$\therefore$ The correct answer is option D i.e. $\dfrac{4}{5}P$.

Note
The trick here is to check the point at which we need to measure the pressure. For example, the pressure at the top of a tank would only be equal to the atmospheric pressure, but at the bottom it is different (as we saw in this question). The pressure in between is also calculated by the addition of atmospheric pressure and the pressure of the weight of the liquid above that height.