A ball is suspended by a thread from the ceiling of a tram car. The brakes are applied and the speed of the car changes uniformly from `36 kmh^(-1)` to zero is `5s`. The angle by which the ball deviates from the vertical is `(g =10ms^(-2))` .

Options

A. `tan^(-1)((1)/(3))`
B. `sin^(-1)((1)/(5))`
C. `tan^(-1)((1)/(5))`
D. `cot^(-1)((1)/(3))`

Explanation

To solve the problem of the ball suspended in a tram car that decelerates, we can follow these steps:

### Step 1: Convert the speed of the tram car from km/h to m/s
The initial speed of the tram car is given as \(36 \, \text{km/h}\). To convert this to meters per second (m/s), we use the conversion factor \(1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}\).

\[ V_i = 36 \, \text{km/h} \times \frac{5}{18} = 10 \, \text{m/s} \]

### Step 2: Calculate the acceleration of the tram car
The final speed of the tram car is \(0 \, \text{m/s}\) and the time taken to stop is \(5 \, \text{s}\). We can use the formula for acceleration:

\[ a = \frac{V_f - V_i}{t} \]
Substituting the values:

\[ a = \frac{0 - 10}{5} = -2 \, \text{m/s}^2 \]

### Step 3: Analyze the forces acting on the ball
The ball experiences two forces:
1. The gravitational force acting downward, \(mg\).
2. The tension \(T\) in the thread, which acts at an angle \(\theta\) from the vertical.

In a non-inertial frame (the accelerating tram), we also consider a pseudo force acting on the ball in the opposite direction of the acceleration of the tram car.

### Step 4: Resolve the tension into components
The tension \(T\) can be resolved into two components:
- \(T \cos \theta\) acting vertically upward.
- \(T \sin \theta\) acting horizontally (in the direction opposite to the acceleration).

### Step 5: Set up the equations of motion
For vertical motion (upward balance):
\[ T \cos \theta = mg \]

For horizontal motion (net force equals mass times acceleration):
\[ T \sin \theta = ma \]

### Step 6: Divide the equations to eliminate \(T\)
Dividing the second equation by the first gives:
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{ma}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{a}{g} \]

### Step 7: Substitute the known values
We know \(a = 2 \, \text{m/s}^2\) and \(g = 10 \, \text{m/s}^2\):
\[ \tan \theta = \frac{2}{10} = \frac{1}{5} \]

### Step 8: Calculate the angle \(\theta\)
Now we can find \(\theta\) using the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]

### Final Answer
Thus, the angle by which the ball deviates from the vertical is:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]

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