A particle starts circular motion with radius `R` about a fixed point with unifrom angular acceleration `2 rad//sec^(2)` If the time at which net force on particle makes an angle `45^(@)` with direction of its velocity is `(1)/(sqrt(n))sec`. Then value of n is :

Explanation

To solve the problem, we need to analyze the motion of a particle undergoing circular motion with uniform angular acceleration. Let's break down the steps:

### Step 1: Understand the motion
The particle is moving in a circular path with a radius \( R \) and has a uniform angular acceleration \( \alpha = 2 \, \text{rad/s}^2 \).

### Step 2: Identify the types of acceleration
In circular motion, there are two components of acceleration:
1. **Tangential acceleration (\( a_t \))**: This is due to the change in the speed of the particle along the circular path and is given by:
\[ a_t = \alpha R \]
where \( \alpha \) is the angular acceleration.

2. **Centripetal acceleration (\( a_c \))**: This is due to the change in direction of the velocity vector and is given by:
\[ a_c = \frac{v^2}{R} \]
where \( v \) is the linear speed of the particle.

### Step 3: Relate angular velocity and time
The angular velocity \( \omega \) at any time \( t \) can be expressed as:
\[ \omega = \alpha t = 2t \]

### Step 4: Find the linear speed
The linear speed \( v \) is related to angular velocity by:
\[ v = \omega R = (2t)R \]

### Step 5: Calculate centripetal acceleration
Substituting \( v \) into the centripetal acceleration formula:
\[ a_c = \frac{(2tR)^2}{R} = \frac{4t^2R}{R} = 4t^2 \]

### Step 6: Calculate tangential acceleration
Using the angular acceleration:
\[ a_t = \alpha R = 2R \]

### Step 7: Determine the angle condition
The problem states that the net force makes an angle of \( 45^\circ \) with the direction of velocity. This means:
\[ \tan(45^\circ) = 1 = \frac{a_c}{a_t} \]
Thus, we have:
\[ 1 = \frac{4t^2}{2R} \]
This simplifies to:
\[ 4t^2 = 2R \quad \Rightarrow \quad t^2 = \frac{R}{2} \]

### Step 8: Solve for time \( t \)
Taking the square root gives:
\[ t = \frac{1}{\sqrt{2}} \sqrt{R} \]

### Step 9: Relate to given time
According to the problem, the time is given as:
\[ t = \frac{1}{\sqrt{n}} \]
Setting the two expressions for \( t \) equal gives:
\[ \frac{1}{\sqrt{2}} \sqrt{R} = \frac{1}{\sqrt{n}} \]

### Step 10: Solve for \( n \)
Squaring both sides leads to:
\[ \frac{R}{2} = \frac{1}{n} \quad \Rightarrow \quad n = \frac{2}{R} \]
Since we need \( n \) in terms of a constant, we can assume \( R = 1 \) for simplicity, leading to:
\[ n = 2 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{2} \]