Question

The pressure of water on the ground floor in a water pipe is $150000Pa$, whereas pressure on the fourth floor is $30000Pa$. The height of fourth floor is (take $g = 10m{s^{ - 2}}$)
(A) $10m$
(B) $9m$
(C) $12m$
(D) $15m$

Answer 1

If you consider the water pipe containing water in it, it will be a vertical pipe. When you deal with pressure, it varies according to height, as in this case. You can see that the pressure on the fourth floor is lesser than the pressure at the ground floor. This increase in pressure as you go down the pipe is due to the pressure exerted by the weight of the water. Derive pressure as a function of height and then you will be able to find the height of the fourth floor.

Complete step by step answer:
Let us take a point in the pipe and denote the pressure at that height level as $P$. If you go down an infinitesimal distance equal to $dh$ below that point, the pressure will increase, the pressure at this point will be $P + dP$. Consider the cylinder in which the water is trapped. Let the upper and lower surface area of the cylinder be $A$.

If you imagine the free body diagram of this trapped water, you can observe three forces on it: (1) ${F_1}$ - vertically downward force due to water above the imaginary cylinder (2) $W$ - weight of the trapped water itself and (3) ${F_2}$ - vertically upward force due to water below the imaginary cylinder.

Pressure is defined as force per unit area, therefore, we have ${F_1} = PA$ and ${F_2} = \left( {P + dP} \right)A$. The weight of water in the cylinder will be $W = mg = \rho Vg$ where $\rho$ is the density of water and $V$ is the volume of the cylinder.
As $V = Adh$, $W = \rho gAdh$.
The water is supposed to be in equilibrium and therefore the net force on the water trapped in the cylinder of height $dh$ should be zero.
$- {F_1} + {F_2} - W = 0$
Let us substitute the values of ${F_1},{F_2}\& W$ that we have obtained.
\- \left( {PA} \right) + \left( {P + dP} \right)\left( A \right) - \rho gAdh = 0 \\\ \- PA + PA + AdP - \rho gAdh = 0 \\\ dP = \rho gdh \\\ dP = - \rho gdh \\\
The negative sign is due to decrease in height.
If we integrate the equation, we get
\int\limits_{{P_1}}^{{P_2}} {dP} = - \int\limits_{{h_1}}^{{h_2}} {\rho gdh} \\\ {P_2} - {P_1} = \rho g\left( {{h_1} - {h_2}} \right) \\\
Now, comparing it with our case, ${P_2}$ is the pressure at the ground floor, ${P_1}$ is the pressure at the fourth floor, ${h_2}$ is the height of ground floor and ${h_1}$ is the height of fourth floor. Now, let us substitute the values in the above obtained equation.
$150000 - 30000 = \left( {1000} \right)\left( {10} \right)\left( {{h_1} - 0} \right) \\\ \therefore {h_1} = 12m$
Therefore, the height of the fourth floor is $12m$.

Hence, option C is correct.

Note: You have to consider the negative sign, because as you go down, the pressure increases but the height is going to decrease. Keep in mind that the excess pressure we obtain is due to the weight of water. Remember the method we used to find the weight of the water. While solving for questions, always check the unit of mass, volume and density given to you. Remember the value of density of water.