Answer
b) $19.6kPa$
Work Step by Step
Knowing that:
$P=\rho gh$
Taking $1$ as the top of the pool and $2$ as the bottom of the pool and assuming that the density does not change with the depth:
$\Delta P=P_{2}-P_{1}=\rho_{water}gh_{2}-\rho_{water}gh_{1}$
$\Delta P=\rho_{water}g*(h_{2}-h_{1})$
Assuming $h_{1}=0m$ (Reference):
$\Delta P=1000\frac{kg}{m^3}*9.81\frac{m}{s^2}*(2m-0m)$
$\Delta P=19.62kPa$
