Question

What will be the pressure due to a water column of height \[10m\]. Take the density of water $ = 1000kg{m^{ - 3}}$, $g = 10m{s^{ - 2}}$. The answer is $p\times{10^5}Pa$ then $p = $

Answer 1

Hint
We can directly substitute the given values in a formula to get what we require
formula used: Hydrostatic pressure formula-
$\Rightarrow P = \rho gh$
Where P is the pressure, $\rho $ is the density, g is the acceleration due to gravity and h is the height of the liquid column.

Complete step by step answer
Hydrostatic pressure is the pressure exerted by a liquid at hydrostatic equilibrium on the contact surface due to gravity. If a liquid is confined in a container, the pressure on the bottom and the walls of the container is due to hydrostatic pressure.
We should be able to find the answer correctly as we already have what is needed to solve this problem.
Therefore, we can substitute the values of density, acceleration due to gravity, and height in the pressure formula, what we’ll get will be as follows.
$\Rightarrow P = \rho gh = 1000kg{m^{ - 3}}\times10m{s^{ - 2}}\times10m$
As we simplify this we’ll get, $p\times{10^5}Pa$
$\Rightarrow P = 1\times{10^5}Pa$.
We have been given the answer as $p\times{10^5}Pa$, if we compare our solution with this, we’ll get the value for p which will be,
$\Rightarrow p\times{10^5} = 1\times{10^5}$
Therefore from above, we can say that $p$ is $1$.

Additional Information
From the formula it is clear that the pressure of a liquid is dependent on height h, which implies that as we go deeper, the pressure increases.
The formula that we have taken is valid only for constant density and gravity. If the density of a liquid at height h keeps varying then the equation fails.

Note
It is to be noted that every value should be substituted in the formula without any change, if we change the unit of one value then we might entirely get the wrong answer.