A particle is kept fixed on a turntable rotating uniformly. As seen from the ground the particle goes in a circle, its speed is `20 cm//s` & acceleration is `20 cm//s^(2)`. The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed & acceleration will be:

Options

A. `10 cm//s, 10 cm//s^(2)`
B. `10 cm//s, 80 cm//s^(2)`
C. `40 cm//s, 10cm//s^(2)`
D. `40 cm//s, 40 cm//s^(2)`

Answer

To solve the problem, we need to analyze the motion of the particle on the turntable in terms of its speed and acceleration as the radius changes.

### Step-by-Step Solution:

1. **Understand the Given Values:**
- Initial speed \( v = 20 \, \text{cm/s} \)
- Initial acceleration \( a = 20 \, \text{cm/s}^2 \)

2. **Relate Speed and Acceleration to Angular Velocity:**
- The speed of the particle in circular motion can be expressed as:
\[ v = \omega r \]
- The centripetal (normal) acceleration can be expressed as:
\[ a = \omega^2 r \]
- Here, \( \omega \) is the angular velocity and \( r \) is the radius of the circular path.

3. **Calculate Angular Velocity:**
- From the speed equation, we can express \( \omega \):
\[ \omega = \frac{v}{r} \]
- From the acceleration equation:
\[ \omega^2 = \frac{a}{r} \]

4. **Using the Given Values:**
- Since both equations involve \( \omega \), we can set them equal:
\[ \frac{v^2}{r^2} = \frac{a}{r} \]
- Plugging in the values:
\[ \frac{(20 \, \text{cm/s})^2}{r^2} = \frac{20 \, \text{cm/s}^2}{r} \]
- This simplifies to:
\[ 400 = 20r \implies r = 20 \, \text{cm} \]

5. **New Radius:**
- The problem states that the radius is now half of the original value:
\[ r' = \frac{r}{2} = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm} \]

6. **Calculate New Speed:**
- The new speed \( v' \) can be calculated using:
\[ v' = \omega r' = \omega \left(\frac{r}{2}\right) = \frac{v}{2} = \frac{20 \, \text{cm/s}}{2} = 10 \, \text{cm/s} \]

7. **Calculate New Acceleration:**
- The new acceleration \( a' \) can be calculated using:
\[ a' = \omega^2 r' = \omega^2 \left(\frac{r}{2}\right) = \frac{a}{2} = \frac{20 \, \text{cm/s}^2}{2} = 10 \, \text{cm/s}^2 \]

### Final Results:
- New speed \( v' = 10 \, \text{cm/s} \)
- New acceleration \( a' = 10 \, \text{cm/s}^2 \)