An oil tanker is partially filled with oil and moves forward on a level road with uniform acceleration. The free surface of oil then

In a uniformly accelerating reference frame the fluid experiences an effective gravity equal to the vector sum of real gravity (downwards) and the inertial pseudo-acceleration (opposite the tank’s acceleration); the free surface orients perpendicular to this resultant [1]. The surface therefore tilts; its slope obeys tan(θ) = a/g, where a is the forward acceleration and g is gravity. For a tank accelerating forward the pseudo-force is directed backward, so the resultant leans backward and the liquid level rises at the rear (larger depth at the rear end) and falls at the front, as shown in standard analyses and examples. Hence option 3 is correct.

Sources

1. [1] https://eng.libretexts.org/Bookshelves/Civil_Engineering/Fluid_Mechanics_(Bar-Meir)/04%3A_Fluids_Statics/4.4%3A_Fluid_in_an_Accelerated_System/4.4.1%3A_Fluid_in_a_Linearly_Accelerated_System