A particle of mass 10 g moves along a circle of radius `6.4` cm with a constant tangential acceleration. What is the magnitude of this acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to `8xx10^(-4)`J by the end of the second revolution after the beginning of the motion?

Options

A. `0.15ms^(-2)`
B. `0.18ms^(-2)`
C. `0.2ms^(-2)`
D. `0.1ms^(-2)`

Explanation

To solve the problem, we need to find the magnitude of the tangential acceleration of a particle moving in a circular path, given its mass, radius, and kinetic energy after two revolutions.

### Step-by-Step Solution:

1. **Convert Units**:
- The mass of the particle is given as 10 g. Convert this to kilograms:
\[ m = 10 \, \text{g} = 0.01 \, \text{kg} \]
- The radius is given as 6.4 cm. Convert this to meters:
\[ r = 6.4 \, \text{cm} = 0.064 \, \text{m} \]

2. **Calculate the Distance for Two Revolutions**:
- The distance covered in one revolution is given by the circumference of the circle:
\[ \text{Distance for one revolution} = 2\pi r \]
- Therefore, the distance for two revolutions is:
\[ s = 2 \times (2\pi r) = 4\pi r \]
- Substituting the value of \( r \):
\[ s = 4 \pi (0.064) \approx 0.804 \, \text{m} \]

3. **Use Kinetic Energy to Find Final Velocity**:
- The kinetic energy (KE) at the end of the second revolution is given as \( 8 \times 10^{-4} \, \text{J} \). The formula for kinetic energy is:
\[ KE = \frac{1}{2} mv^2 \]
- Rearranging this to find \( v^2 \):
\[ v^2 = \frac{2 \times KE}{m} = \frac{2 \times (8 \times 10^{-4})}{0.01} \]
- Calculating \( v^2 \):
\[ v^2 = \frac{16 \times 10^{-4}}{0.01} = 0.016 \, \text{m}^2/\text{s}^2 \]

4. **Calculate Tangential Acceleration**:
- We can use the equation of motion for tangential acceleration:
\[ v^2 = u^2 + 2as \]
- Since the particle starts from rest, \( u = 0 \):
\[ v^2 = 2as \]
- Rearranging to find \( a \):
\[ a = \frac{v^2}{2s} \]
- Substituting the values of \( v^2 \) and \( s \):
\[ a = \frac{0.016}{2 \times 0.804} \approx \frac{0.016}{1.608} \approx 0.00994 \, \text{m/s}^2 \]

5. **Final Result**:
- The magnitude of the tangential acceleration is approximately:
\[ a \approx 0.01 \, \text{m/s}^2 \]