Question

A liquid of density $12{\text{ kg}}{{\text{m}}^{ - 3}}$ exerts a pressure of 600 Pa a point inside a liquid. What is the height of the liquid column above that point? $\left( {g = 10{\text{ m}}{{\text{s}}^{ - 2}}} \right)$
A. 4 m
B. 5 m
C. 6 m
D. 7 m

Answer 1

In the question, we need to determine the height of the liquid column where the pressure is 600 Pascal inside the liquid column having the liquid density of $12{\text{ kg}}{{\text{m}}^{ - 3}}$. For this, we will use the relation between the pressure, the density of the liquid, and the height of the column, which is given as $P = \rho gh$.

Complete step by step answer:

The product of the density of the liquid, acceleration due to gravity, and the height of the liquid column gives the pressure of the liquid at that point. Mathematically, $P = \rho gh$ where $\rho$ is the density of the liquid in $kg{m^{ - 3}}$, ‘g’ is the acceleration due to gravity in $m{s^{ - 2}}$ and ‘h’ is the height of the liquid column where pressure is to be determined in Pascal.

So, substitute $P = 600{\text{ Pa}}$, $\rho = 12{\text{ kg}}{{\text{m}}^{ - 3}}$ and $g = 10{\text{ m}}{{\text{s}}^{ - 2}}$ in the formula $P = \rho gh$ to determine the height of the liquid column above the point where the pressure is given.

$P = \rho gh \\\ \Rightarrow 600 = 12 \times 10 \times h \\\ \Rightarrow 600 = 120h - - - - (i) \\\$

Divide the term with the unknown quantity ‘h’ to both sides of the equation (i) as:

$\dfrac{{600}}{{120}} = \dfrac{{120h}}{{120}} \\\ \Rightarrow h = 5{\text{ m}} \\\$

Hence, the height of the liquid column above the point where the pressure is 600 Pascal inside the liquid column having the liquid density of $12{\text{ kg}}{{\text{m}}^{ - 3}}$ is 5 meters.

Option B is correct.

Note: Students should be aware while using the value of the acceleration due to gravity. Here in the question, it is already mentioned to use the value as $\left( {g = 10{\text{ m}}{{\text{s}}^{ - 2}}} \right)$.