Question

The pressure on a swimmer 10 m below the surface lake is: (Atmospheric pressure = $1.01\times {10}^5\text{Pa}$, Density of water = $1000kg/m^3$).
(A) 10 atm
(B) 5 atm
(C) 15 atm
(D) 2 atm

Answer 1

Hint: Here we need to find the pressure on the swimmer at a depth $h$below the surface by using the formula $P=P_0+\rho gh$. We can substitute all the values that are given in the question to get the final pressure.
$P=P_0+\rho gh$
Where $P$ is the pressure at depth below the surface,
$P_0$is the pressure at the surface,
$\rho$is the density of the liquid,
$g$ is the gravitational constant and
$h$ is the depth below the surface at which the pressure is being measured.

Complete step by step answer:
The formula used to calculate the absolute pressure, the pressure that varies with the depth of the fluid is given as follows:
$P=P_0+\rho gh$
We need to substitute the given parameters in the above equation as,
$P_S=P_A+\rho gh$
In the question we are provided that the pressure exerted by the air is equal to $1.01\times {10}^5\text{Pa}$, the depth at which the swimmer is swimming is 10 m and the density of the water is $1000kg/m^3$.
Upon substituting the above values in the above equation, we get,
$P_S=1.01\times {10}^5+1000\times 10\times 10$
On calculating these values we get,
$P_S=2\times {10}^{5}Pa$
Now we need to perform the unit conversion, that is, the value of the pressure we have obtained is in terms of the Pascal, but in the options, the unit used is atm, so we will convert the unit to the same.
The conversion factor from Pascal to atmosphere is , $1\text{Pa}={10}^{-5}\text{atm}$
Using this relation, we will convert the unit of the pressure value obtained.
So, we get,
$P_S=(2\times {10}^5)\times {10}^{-5}$
So on doing the calculation we have,
$P_S=2\text{atm}$
Hence the pressure on the swimmer is 2 atm.
$\therefore$ The pressure on a swimmer 10 m below the surface lake is 2 atm, thus, option (D) is correct.

Note:
We can see from the answer that the pressure at a depth of 10m is much more than the pressure at the surface of the lake. So as the depth under the water surface increases the pressure also increases. This is because, as the depth increases, the amount of water over the swimmer increases. So the pressure also increases.