In a conical pendulum arrangement, a string of length 1 m is fixed at one end with a bob of mass 100 g and the string makes `(2)/(pi) mvs^(-1)`around a vertical axis through a fixed point. The angle of inclination of the string with vertical is: (Take `g = 10 ms^(-1)`)

Options

A. (a)`tan^(-1) ((5)/(8))`
B. (b)`tan^(-1) ((3)/(5))`
C. (c)`cos^(-1) ((8)/(5))`
D. (d)`cos^(-1) ((5)/(8))`

Explanation

To solve the problem of finding the angle of inclination of the string in a conical pendulum arrangement, we can follow these steps:

### Step 1: Identify the given parameters
- Length of the string, \( L = 1 \, \text{m} \)
- Mass of the bob, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Linear velocity, \( v = \frac{2}{\pi} \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Relate linear velocity to angular velocity
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
where \( r \) is the horizontal radius of the circular path.

### Step 3: Find the radius of the circular path
In a conical pendulum, the radius \( r \) can be expressed in terms of the angle \( \theta \):
\[ r = L \sin \theta \]
Substituting this into the equation for linear velocity:
\[ v = L \sin \theta \cdot \omega \]

### Step 4: Calculate angular velocity
The angular velocity \( \omega \) can be calculated using:
\[ \omega = \frac{v}{r} = \frac{v}{L \sin \theta} \]
Given \( v = \frac{2}{\pi} \), we can express \( \omega \) as:
\[ \omega = \frac{2/\pi}{L \sin \theta} \]

### Step 5: Write the equations of motion
For the conical pendulum, we have two forces acting on the bob:
1. Tension \( T \) in the string.
2. Weight \( mg \) acting downward.

The vertical component of the tension balances the weight:
\[ T \cos \theta = mg \]
The horizontal component provides the centripetal force:
\[ T \sin \theta = m r \omega^2 \]

### Step 6: Substitute \( r \) and \( \omega \)
Substituting \( r = L \sin \theta \) and \( \omega = \frac{v}{r} \):
\[ T \sin \theta = m (L \sin \theta) \left(\frac{v}{L \sin \theta}\right)^2 \]
This simplifies to:
\[ T \sin \theta = m \frac{v^2}{L} \]

### Step 7: Substitute \( T \) from the vertical balance
From \( T \cos \theta = mg \), we can express \( T \):
\[ T = \frac{mg}{\cos \theta} \]
Substituting this into the horizontal equation:
\[ \frac{mg}{\cos \theta} \sin \theta = m \frac{v^2}{L} \]

### Step 8: Simplify the equation
Dividing both sides by \( m \):
\[ \frac{g \sin \theta}{\cos \theta} = \frac{v^2}{L} \]
This gives:
\[ g \tan \theta = \frac{v^2}{L} \]

### Step 9: Substitute known values
Substituting \( g = 10 \, \text{m/s}^2 \), \( v = \frac{2}{\pi} \, \text{m/s} \), and \( L = 1 \, \text{m} \):
\[ 10 \tan \theta = \frac{\left(\frac{2}{\pi}\right)^2}{1} \]
Calculating \( \left(\frac{2}{\pi}\right)^2 \):
\[ \left(\frac{2}{\pi}\right)^2 = \frac{4}{\pi^2} \]
Thus:
\[ 10 \tan \theta = \frac{4}{\pi^2} \]
\[ \tan \theta = \frac{4}{10 \pi^2} = \frac{2}{5 \pi^2} \]

### Step 10: Find \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{2}{5 \pi^2}\right) \]

### Final Answer
The angle of inclination of the string with the vertical is:
\[ \theta = \tan^{-1}\left(\frac{2}{5 \pi^2}\right) \]