A particle is moving in a circle of radius `R` in such a way that at any instant the normal and tangential component of the acceleration are equal. If its speed at `t=0` is `u_(0)` the time taken to complete the first revolution is

Options

A. `(R)/(v_(0))`
B. `(R)/(v_(0))e^(-2pi)`
C. `(R)/(v_(0))(1-e^(-2pi))`
D. `(R)/(v_(0))(1+e^(-2pi))`

Explanation

To solve the problem, we need to find the time taken for a particle moving in a circular path of radius \( R \) to complete one full revolution, given that the normal and tangential components of acceleration are equal.

### Step-by-Step Solution:

1. **Understanding the Components of Acceleration**:
- The tangential component of acceleration \( a_t \) is given by \( \frac{dv}{dt} \).
- The normal (centripetal) component of acceleration \( a_n \) is given by \( \frac{v^2}{R} \).
- According to the problem, these two components are equal:
\[ a_t = a_n \Rightarrow \frac{dv}{dt} = \frac{v^2}{R} \]

2. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ \frac{dv}{v^2} = \frac{dt}{R} \]

3. **Integrating the Equation**:
- Integrate both sides. The left side integrates from the initial velocity \( u_0 \) to \( v \), and the right side integrates from \( 0 \) to \( t \):
\[ \int_{u_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R} \]
- This results in:
\[ -\frac{1}{v} + \frac{1}{u_0} = \frac{t}{R} \]

4. **Solving for Velocity \( v \)**:
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{u_0} - \frac{t}{R} \]
- Taking the reciprocal leads to:
\[ v = \frac{R u_0}{R - u_0 t} \]

5. **Finding the Distance for One Revolution**:
- The distance covered in one complete revolution is \( 2\pi R \).
- The relationship between distance, velocity, and time is given by:
\[ dx = v dt \]
- Substituting for \( v \):
\[ dx = \frac{R u_0}{R - u_0 t} dt \]

6. **Integrating to Find Time \( t \)**:
- Integrate from \( 0 \) to \( 2\pi R \) for \( x \) and from \( 0 \) to \( t \) for \( t \):
\[ \int_{0}^{2\pi R} dx = R u_0 \int_{0}^{t} \frac{dt}{R - u_0 t} \]
- The integral on the right side can be solved using the natural logarithm:
\[ 2\pi R = R u_0 \left[-\frac{1}{u_0} \ln(R - u_0 t)\right]_{0}^{t} \]

7. **Evaluating the Integral**:
- Evaluating the limits gives:
\[ 2\pi R = -R u_0 \left( -\frac{1}{u_0} \ln(R - u_0 t) + \frac{1}{u_0} \ln(R) \right) \]
- Simplifying leads to:
\[ 2\pi = \ln\left(\frac{R}{R - u_0 t}\right) \]

8. **Exponentiating to Solve for \( t \)**:
- Exponentiating both sides:
\[ \frac{R}{R - u_0 t} = e^{2\pi} \]
- Rearranging gives:
\[ R - u_0 t = \frac{R}{e^{2\pi}} \]
- Therefore:
\[ u_0 t = R \left(1 - \frac{1}{e^{2\pi}}\right) \]
- Finally, solving for \( t \):
\[ t = \frac{R}{u_0} \left(1 - e^{-2\pi}\right) \]

### Final Answer:
The time taken to complete the first revolution is:
\[ t = \frac{R}{u_0} \left(1 - e^{-2\pi}\right) \]