The bob of a 0.2 m pendulum describs an arc of circle in a vertical plane. If the tension in the cord is `sqrt(3)` times the weight of the bob when the cord makes an angle `30^(@)` with the vertical, the acceleration of the bob in that position is `g//n`. Find value of `n`.

Options

A. 1
B. 2
C. 3
D. 5

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Identify the forces acting on the bob
The forces acting on the bob of the pendulum are:
1. The weight of the bob, \( W = mg \) (acting downwards).
2. The tension in the cord, \( T \) (acting along the cord).

### Step 2: Resolve the weight into components
When the pendulum makes an angle of \( 30^\circ \) with the vertical, we can resolve the weight \( mg \) into two components:
- The component along the direction of the tension (vertical component):
\[ W_{\text{vertical}} = mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2} \]
- The component acting tangentially (horizontal component):
\[ W_{\text{tangential}} = mg \sin(30^\circ) = mg \cdot \frac{1}{2} \]

### Step 3: Set up the equation for tension
According to the problem, the tension \( T \) in the cord is given as:
\[ T = \sqrt{3} \cdot mg \]

### Step 4: Apply Newton's second law in the radial direction
In the radial direction, the net force is given by:
\[ T - W_{\text{vertical}} = m \cdot a_c \]
Substituting the values we have:
\[ \sqrt{3} \cdot mg - mg \cdot \frac{\sqrt{3}}{2} = m \cdot a_c \]
Factoring out \( mg \):
\[ mg \left( \sqrt{3} - \frac{\sqrt{3}}{2} \right) = m \cdot a_c \]
This simplifies to:
\[ mg \cdot \left( \frac{2\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \right) = m \cdot a_c \]
\[ mg \cdot \frac{\sqrt{3}}{2} = m \cdot a_c \]
Cancelling \( m \) from both sides gives:
\[ g \cdot \frac{\sqrt{3}}{2} = a_c \]

### Step 5: Calculate the tangential acceleration
Using Newton's second law in the tangential direction:
\[ mg \sin(30^\circ) = m \cdot a_t \]
Substituting the values:
\[ mg \cdot \frac{1}{2} = m \cdot a_t \]
Cancelling \( m \):
\[ g \cdot \frac{1}{2} = a_t \]

### Step 6: Find the net acceleration
The net acceleration \( a \) is the vector sum of the centripetal and tangential accelerations:
\[ a = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{\left( g \cdot \frac{\sqrt{3}}{2} \right)^2 + \left( g \cdot \frac{1}{2} \right)^2} \]
Calculating each term:
\[ a = \sqrt{g^2 \cdot \frac{3}{4} + g^2 \cdot \frac{1}{4}} = \sqrt{g^2 \cdot 1} = g \]

### Step 7: Relate the net acceleration to \( g/n \)
Given that the acceleration is expressed as \( \frac{g}{n} \):
\[ g = \frac{g}{n} \]
This implies:
\[ n = 1 \]

### Final Answer
The value of \( n \) is \( 1 \).