#### Answer

The correct answer is:
C. $P_a = \frac{5}{4}~P_i$

#### Work Step by Step

We can find the gauge pressure at a depth of 4 cm below the surface of the water before the liquid is added:
$P_g = \rho_w~g~h_w$
$P_g = (1000~kg/m^3)(10~m/s^2)(0.040~m)$
$P_g = 400~Pa$
The density of the liquid is $500~kg/m^3$. We can find the gauge pressure at a depth of 2 cm below the surface of the liquid:
$P_g = \rho_l~g~h_l$
$P_g = (500~kg/m^3)(10~m/s^2)(0.020~m)$
$P_g = 100~Pa$
After the liquid is added, the new gauge pressure at the base of the column is $400~Pa+100~Pa$ which is $500~Pa$.
The correct answer is:
C. $P_a = \frac{5}{4}~P_i$