The variation of absolute pressure P with depth h below the free...

The absolute pressure $P$ at a depth $h$ is given by the equation:
$P = P_{atm} + \rho g h$

This represents a linear equation of the form $y = c + mx$, where the slope of the $P$ versus $h$ graph is $m = \rho g$.

From the given data, the slope is:
$\dfrac{\Delta P}{\Delta h} = \dfrac{2.2 \times 10^5 - 1 \times 10^5}{10 - 0}$
$\dfrac{\Delta P}{\Delta h} = \dfrac{1.2 \times 10^5}{10} = 1.2 \times 10^4$ Pa/m

Equating this slope to $\rho g$:
$\rho g = 1.2 \times 10^4$
$\rho \times 10 = 12000$
$\rho = 1200$ kg/m$^3$

Using $P/h$ at $h=10$ without subtracting the intercept leads to the incorrect $2200$ kg/m$^3$.

Answer: $1200$ kg/m$^3$