Given:
Length of the wagon, L = 5 m
Acceleration, a = 1.96 m/s2

To find:
The level below the open top upto which water may be filled without spilling.

Concepts used:
The force acting on the water in the wagon is due to the acceleration of the wagon. This force is given by the product of the mass of water and the acceleration. If this force exceeds the weight of the water, it will spill out of the wagon.

Solution:
Let us assume that the level upto which water can be filled without spilling is h meters from the bottom of the wagon.

The force acting on the water due to the acceleration of the wagon is given by:
F = ma
where
m = mass of the water in the wagon
a = acceleration of the wagon

The weight of the water is given by:
W = mg
where
g = acceleration due to gravity (= 9.81 m/s2)

For the water to not spill out of the wagon, the force F must be less than or equal to the weight W. Thus, we have:
ma ≤ mg
or
a ≤ g

Substituting the given values, we get:
1.96 ≤ 9.81
This condition is satisfied.

Now, let us find the maximum height upto which water can be filled without spilling.

Consider a small element of water of height dh at a depth h from the top of the wagon. The force acting on this element due to the acceleration of the wagon is:
dF = ρgdh
where
ρ = density of water (= 1000 kg/m3)
g = acceleration due to gravity (= 9.81 m/s2)

The weight of this element is:
dW = ρgdh

For the element to not spill out of the wagon, the force dF must be less than or equal to the weight dW. Thus, we have:
dF ≤ dW
or
ρgdh ≤ ρgdh0
where
h0 = the maximum height upto which water can be filled without spilling.

Simplifying, we get:
h ≤ h0

Integrating both sides, we get:
∫0h d(h) ≤ ∫0h0 d(h)
or
h2/2 ≤ h0

Substituting the given values, we get:
h2/2 ≤ h0
or
h ≤ √(2h0)

Substituting h0 = 0.8 m, we get:
h ≤ √1.6 ≈ 1.265 m

Thus, the maximum height upto which water can be filled without spilling is 1.265 m.

Answer:
The level below the open top upto which water may be filled without spilling is less than or equal to 1.265 m. Therefore, option (d) 0.8 m is correct.

Tan @=a/g =H/(B/2)
1.96/9.81=H/2.5
H=.5m