Answer
d) $196000Pa$
Work Step by Step
Knowing that:
$P=\rho gh$
Taking $1$ as the depth of $5m$ and $2$ as the depth of $25m$ and assuming that the density does not change with the depth:
$\Delta P=P_{2}-P_{1}=\rho_{water}gh_{2}-\rho_{water}gh_{1}$
$\Delta P=\rho_{water}g*(h_{2}-h_{1})$
Substituting:
$\Delta P=1000\frac{kg}{m^3}*9.81\frac{m}{s^2}*(25m-5m)$
$\Delta P=196200Pa$
