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 π r v s − 1 around a vertical axis through a fixed point. The angle of inclination of the string with vertical is: (Take g = 10 m s − 1 )

Options

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

Explanation

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

### Step 1: Understand the Problem
We have a conical pendulum with:
- Length of the string (L) = 1 m
- Mass of the bob (m) = 100 g = 0.1 kg (conversion to kg)
- Number of revolutions per second (n) = 2π r/s = 2/5 revolutions per second
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Identify Forces Acting on the Bob
The forces acting on the bob are:
1. The gravitational force (mg) acting downwards.
2. The tension (T) in the string acting along the string.

### Step 3: Resolve the Tension into Components
The tension can be resolved into two components:
- Vertical component: \( T \cos \theta \) (balances the weight)
- Horizontal component: \( T \sin \theta \) (provides the centripetal force)

### Step 4: Write the Equations of Motion
1. For vertical motion (balancing forces):
\[ T \cos \theta = mg \]
2. For horizontal motion (centripetal force):
\[ T \sin \theta = m \frac{v^2}{r} \]
where \( v \) is the linear velocity and \( r \) is the radius of the circular path.

### Step 5: Express Linear Velocity in Terms of Angular Velocity
The linear velocity \( v \) can be expressed in terms of angular velocity \( \omega \):
\[ v = r \omega \]
Given \( n = \frac{2}{5} \) revolutions per second, we can find \( \omega \):
\[ \omega = 2 \pi n = 2 \pi \left(\frac{2}{5}\right) = \frac{4\pi}{5} \text{ rad/s} \]

### Step 6: Substitute for \( v \) and \( r \)
The radius \( r \) can be expressed in terms of \( L \) and \( \theta \):
\[ r = L \sin \theta = 1 \cdot \sin \theta = \sin \theta \]
Thus, the centripetal force equation becomes:
\[ T \sin \theta = m \frac{(r \omega)^2}{r} = m r \omega^2 \]
Substituting \( r = \sin \theta \) gives:
\[ T \sin \theta = m \sin \theta \left(\frac{4\pi}{5}\right)^2 \]

### Step 7: Simplify the Equations
From the vertical force equation:
\[ T = \frac{mg}{\cos \theta} \]
Substituting this into the horizontal force equation:
\[ \frac{mg}{\cos \theta} \sin \theta = m \sin \theta \left(\frac{4\pi}{5}\right)^2 \]
Cancelling \( m \sin \theta \) (assuming \( \sin \theta \neq 0 \)):
\[ \frac{g}{\cos \theta} = \left(\frac{4\pi}{5}\right)^2 \]

### Step 8: Solve for \( \cos \theta \)
Rearranging gives:
\[ \cos \theta = \frac{g}{\left(\frac{4\pi}{5}\right)^2} \]
Substituting \( g = 10 \):
\[ \cos \theta = \frac{10}{\left(\frac{16\pi^2}{25}\right)} = \frac{250}{16\pi^2} \]

### Step 9: Calculate \( \theta \)
Now, we can find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{250}{16\pi^2}\right) \]

### Step 10: Final Calculation
Using \( \pi \approx 3.14 \):
\[ \theta \approx \cos^{-1}\left(\frac{250}{16 \cdot (3.14)^2}\right) \]
Calculating this gives the angle of inclination.