A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal. The angle  of thread deflection in the extreme position will be - 2

Options

A. `tan^(-1)(2)`
B. `tan^(-1)(sqrt(2))`
C. `tan^(-1)(1/2)`
D. `2tan ^(-1)(1/2)`

Answer

To solve the problem, we need to analyze the forces acting on the ball in both the extreme position and the lowest position of the swing.

### Step 1: Analyze the forces in the lowest position
When the ball is at the lowest position, the forces acting on it are:
- The tension \( T \) in the thread acting upwards.
- The weight \( mg \) acting downwards.

At this position, the ball is in circular motion, so the net centripetal force is provided by the difference between the tension and the weight:
\[ T - mg = \frac{mv^2}{L} \]
where \( L \) is the length of the thread and \( v \) is the speed of the ball at the lowest position.

### Step 2: Analyze the forces in the extreme position
In the extreme position (where the thread makes an angle \( \theta \) with the vertical), the forces acting on the ball are:
- The tension \( T \) in the thread acting along the thread.
- The weight \( mg \) acting vertically downwards.

The components of the forces can be resolved:
- The vertical component of tension must balance the weight:
\[ T \cos(\theta) = mg \]
- The horizontal component provides the centripetal force:
\[ T \sin(\theta) = \frac{mv^2}{L} \]

### Step 3: Set the accelerations equal
According to the problem, the acceleration in the extreme position and the lowest position are equal. The acceleration at the lowest position can be expressed as:
\[ a_{lowest} = \frac{v^2}{L} \]
At the extreme position, the radial acceleration can be expressed as:
\[ a_{extreme} = g \sin(\theta) \]
Setting these equal gives us:
\[ \frac{v^2}{L} = g \sin(\theta) \]

### Step 4: Substitute for \( T \)
From the equation \( T \cos(\theta) = mg \), we can express \( T \) as:
\[ T = \frac{mg}{\cos(\theta)} \]
Substituting \( T \) into the horizontal force equation:
\[ \frac{mg}{\cos(\theta)} \sin(\theta) = \frac{mv^2}{L} \]
This simplifies to:
\[ mg \tan(\theta) = \frac{mv^2}{L} \]

### Step 5: Combine the equations
Now we have two equations:
1. \( \frac{v^2}{L} = g \sin(\theta) \)
2. \( g \tan(\theta) = \frac{v^2}{L} \)

Setting these equal gives:
\[ g \sin(\theta) = g \tan(\theta) \]
This implies:
\[ \sin(\theta) = \tan(\theta) \cos(\theta) \]
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can simplify to:
\[ \sin(\theta) = \sin(\theta) \]
This is always true, but we need to find the angle \( \theta \).

### Step 6: Solve for \( \theta \)
From the relationship \( \tan(\theta) = 1 \), we find:
\[ \theta = 45^\circ \]

### Final Answer
The angle \( \theta \) of thread deflection in the extreme position will be \( 45^\circ \).