Question

If the atmospheric pressure is 76 cm of mercury, at what depth of water will the pressure come to 4 atmospheres?
A. $31\,{\text{m}}$
B. $21\,{\text{m}}$
C. $11\,{\text{m}}$
D. $9\,{\text{m}}$

Answer 1

Use the formula for pressure depth relation. This formula gives the relation between the atmospheric pressure, pressure at a depth h, density of liquid and acceleration due to gravity. Convert the given atmospheric pressure in the SI system of units and determine the required depth of the water.

Formula used:
The pressure-depth formula is given by
$P = {P_0} + \rho gh$ …… (1)
Here, ${P_0}$ is the atmospheric pressure, $P$ is the pressure at depth $h$, $\rho$ is density of the liquid and $g$ is acceleration due to gravity.

Complete step by step answer:
We have given that the atmospheric pressure is 76 cm of mercury.
We know that the pressure 76 cm of mercury is equal to one atmospheric pressure.
$1\,{\text{atm}} = {\text{76 cm of mercury}}$
The pressure in one atmosphere is equal to $1.013 \times {10^5}\,{\text{Pa}}$.
$1\,{\text{atm}} = 1.013 \times {10^5}\,{\text{Pa}}$
Hence, the atmospheric pressure $1.013 \times {10^5}\,{\text{Pa}}$.
${P_0} = 1.013 \times {10^5}\,{\text{Pa}}$
Here, we have asked to determine the depth of water at which the pressure of the water becomes 4 atm.
$P= 4atm$
Convert the unit of pressure at depth h in Pascal.
$P = \left( {4\,{\text{atm}}} \right)\left( {\dfrac{{1.013 \times {{10}^5}\,{\text{Pa}}}}{{1\,{\text{atm}}}}} \right)$
$\Rightarrow P = 4 \times 1.013 \times {10^5}\,{\text{Pa}}$
We know that the density of water is $1000\,{\text{kg/}}{{\text{m}}^3}$ and the value of acceleration due to gravity is $9.8\,{\text{m/}}{{\text{s}}^2}$.
$\rho = 1000\,{\text{kg/}}{{\text{m}}^3}$
$g = 9.8\,{\text{m/}}{{\text{s}}^2}$
Let us now determine the depth at which pressure of the water is 4 atm using equation (1).
Substitute $4 \times 1.013 \times {10^5}\,{\text{Pa}}$ for $P$, $1.013 \times {10^5}\,{\text{Pa}}$ for ${P_0}$, $1000\,{\text{kg/}}{{\text{m}}^3}$ for $\rho$ and $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$ in equation (1).
$\left( {4 \times 1.013 \times {{10}^5}\,{\text{Pa}}} \right) = \left( {1.013 \times {{10}^5}\,{\text{Pa}}} \right) + \left( {1000\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)h$
$\Rightarrow 980h = \left( {4 \times 1.013 \times {{10}^5}\,{\text{Pa}}} \right) - \left( {1.013 \times {{10}^5}\,{\text{Pa}}} \right)$
$\Rightarrow h = \dfrac{{\left( {4 - 1} \right) \times 1.013 \times {{10}^5}\,{\text{Pa}}}}{{9800}}$
$\Rightarrow h = \dfrac{{3 \times 1.013 \times {{10}^5}\,{\text{Pa}}}}{{9800}}$
$\Rightarrow h = 31\,{\text{m}}$

Therefore, the depth at which the pressure of water is 4 atm is $31\,{\text{m}}$.

So, the correct answer is “Option A”.

Note:
The students should not forget to convert the unit of the pressure from atmospheric pressure to Pascal.
This is because all the physical quantities used in the formula pressure depth relation are in the SI system of units.
Hence, one should convert the units of both the pressure values 1 atm and 4 atm.