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

Options

A. `0.15m//s^(2)`
B. `0.18m//s^(2)`
C. `0.2m//s^(2)`
D. `0.1m//s^(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass from grams to kilograms
Given mass \( m = 10 \, \text{g} \).

To convert grams to kilograms:
\[ m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg} \]

### Step 2: Calculate the distance covered in two revolutions
The radius of the circle is given as \( r = 6.4 \, \text{cm} \).

Convert radius to meters:
\[ r = 6.4 \, \text{cm} = 6.4 \times 10^{-2} \, \text{m} = 0.064 \, \text{m} \]

The distance covered in one revolution is:
\[ \text{Distance for one revolution} = 2\pi r \]
Thus, for two revolutions:
\[ s = 2 \times (2\pi r) = 4\pi r \]
Substituting the value of \( r \):
\[ s = 4\pi(0.064) \approx 4 \times 3.14 \times 0.064 \approx 0.804 \, \text{m} \]

### Step 3: Use the kinetic energy to find the final velocity
The kinetic energy \( KE \) at the end of the second revolution is given as:
\[ KE = 8 \times 10^{-4} \, \text{J} \]

Using the formula for kinetic energy:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ 8 \times 10^{-4} = \frac{1}{2} \times 0.01 \times v^2 \]
Solving for \( v^2 \):
\[ v^2 = \frac{8 \times 10^{-4} \times 2}{0.01} = \frac{16 \times 10^{-4}}{0.01} = 0.016 \]
Thus,
\[ v = \sqrt{0.016} = 0.4 \, \text{m/s} \]

### Step 4: Use the kinematic equation to find tangential acceleration
We know:
\[ v^2 = u^2 + 2as \]
Where \( u = 0 \) (initial velocity), \( v = 0.4 \, \text{m/s} \), and \( s = 0.804 \, \text{m} \).

Substituting the values:
\[ (0.4)^2 = 0 + 2a(0.804) \]
\[ 0.16 = 2a(0.804) \]
Solving for \( a \):
\[ a = \frac{0.16}{2 \times 0.804} = \frac{0.16}{1.608} \approx 0.0994 \, \text{m/s}^2 \]
Rounding off, we get:
\[ a \approx 0.1 \, \text{m/s}^2 \]

### Final Answer
The magnitude of the tangential acceleration is approximately \( 0.1 \, \text{m/s}^2 \).

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