Fig. shows a U-tube of uniform cross- sectional area A accelerated with acceleration a as shown. If d is the separation between the limbs, then the difference in the levels of the liquid in the U-tube is

mass of the liquid in the tube  Ad$\rho$

pseudo force acting on it Ad$\rho$a     

Where A is the area of cross section of the tube and $\rho$is the density of the liquid

Force due to pressure difference= $\left(h_1-h_2\right)\rho gA$

equating the forces 

$\left(h_1-h_2\right)\rho gA$=$Ad\rho a$

$h=\frac{ad}{g}$

Hence the correct answer is $\frac{ad}{g}.$