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 of lower the level of water by one fifth then, the pressure at the bottom of the tank will be
A. $2P$
B. $\dfrac{{13}}{5}P$
C. $\dfrac{8}{5}P$
D. $\dfrac{4}{5}P$

Answer 1

Hint:In this question, we are required to find the pressure at the bottom for that we need to know about the Bernoulli theorem for the initial as well as the final situations of the liquid.Bernoulli theorem states that increase in speed of a fluid is proportional to the decrease in pressure.

Formula used:
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = constant$

Complete Step by Step Answer:
Firstly, we are assuming $h$be the initial height.Now ${h_1}$be the water level after decreasing by height by $\dfrac{1}{5}h$.
Then, ${h_1} = h - \dfrac{1}{5}h = \dfrac{4}{5}h$
We are given that initially pressure is $3P$.If we ignore the atmospheric pressure $\left( P \right)$, the pressure at the bottom becomes $3P - P$$ = $$2P$
We know that pressure on the liquid is given by $h\rho g$.
Using Bernoulli’s Theorem,
$h\rho g = 2P$ $ \ldots \ldots \left( 1 \right)$
After the lowering of liquid, we got ${h_1}$so here we are replacing the height $h$ by $\dfrac{4}{5}h$
$
\Rightarrow\dfrac{4}{5}h\rho g
$
Using equation $\left( 1 \right)$ putting the value of $h\rho g$
$\dfrac{4}{5}(2P)$
$\Rightarrow\dfrac{8}{5}P$
Now, for calculating the pressure at the bottom we need to add the atmospheric pressure too i.e.
$\dfrac{8}{5}P + P\\
\Rightarrow (\dfrac{8}{5} + 1)P$
$\therefore\dfrac{{13}}{5}P$
This is our required solution.

Hence the correct option is B.

Note:Here in our particular question, we had ignored the atmospheric pressure which is somewhat the same as the acceleration due to gravity while atmospheric pressure doesn’t even change the outcomes of most problems. So, it’s better to ignore the atmospheric pressure.Pressure on the top of the object is mostly greater than the below of the objects.