A beaker filled with water, is accelerated at a rate a `m/s^(2)` in forward direction. The surface of water shall make an angle

Options

A. a. `tan^(-1)((a)/(g))` backward
B. b. `tan^(-1)((a)/(g))` forward
C. c. `cot^(-1)((g)/(a))` backward
D. d. `cot^(-1)((g)/(a))` forward

Answer

To solve the problem of determining the angle θ that the surface of the water makes when the beaker is accelerated forward at a rate of a m/s², we can follow these steps:

### Step 1: Understand the Forces Acting on the Water
When the beaker is accelerated forward, the water inside experiences a pseudo force acting in the opposite direction (backward). This is due to the acceleration of the beaker itself. The gravitational force acts downward on the water.

### Step 2: Set Up the Problem
We can visualize the situation as follows:
- The beaker is accelerating to the right with acceleration \( a \).
- The gravitational force \( g \) acts downward.
- The pseudo force acts to the left with the same magnitude as the acceleration \( a \).

### Step 3: Analyze the Resultant Forces
The surface of the water will tilt due to the balance of these forces. The angle θ that the surface makes with the horizontal can be determined by considering the forces acting on the water:
- The horizontal component of the pseudo force is \( a \).
- The vertical component due to gravity is \( g \).

### Step 4: Use Trigonometric Relationships
From the geometry of the situation, we can relate the angle θ to the forces:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{a}{g} \]

### Step 5: Solve for θ
To find the angle θ, we take the arctangent of both sides:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]

### Final Answer
Thus, the angle θ that the surface of the water makes with the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{a}{g}\right) \]