A circular table of radius `0.5 m` has a smooth diametrical groove. A half of mass `90g` is is placed inside the groove along with a spring constant `10^(2)N cm^(-1)` One end of the spring is tied to the table and the other end to the ball The ball is at a distance of `.0.1 m` from the center when the table is at rest On rotating the table with a constant angular frequency of `10^(2)(rad)/s^(-1)` the ball moves away from the center by a distance nearly equal to

Options

A. `10^(-1) m`
B. `10^(-2) m`
C. `10^(-3) m`
D. `2 xx10^(-1) m`

Explanation

To solve the problem step by step, we will analyze the forces acting on the ball when the table rotates and how the spring's compression affects its position.

### Step 1: Understand the Setup
We have a circular table with a radius of 0.5 m and a groove along its diameter. A ball of mass \( m = 90 \, \text{g} = 0.09 \, \text{kg} \) is placed in the groove. The spring constant is given as \( k = 10^2 \, \text{N/cm} = 10000 \, \text{N/m} \). The ball is initially at a distance of \( r_0 = 0.1 \, \text{m} \) from the center.

### Step 2: Identify Forces Acting on the Ball
When the table rotates with an angular frequency \( \omega = 10^2 \, \text{rad/s} = 100 \, \text{rad/s} \), a centrifugal force acts on the ball, which can be expressed as:
\[ F_{\text{centrifugal}} = m \omega^2 r \]
where \( r \) is the distance of the ball from the center.

### Step 3: Set Up the Equilibrium Condition
As the ball moves outward, it compresses the spring. If \( x \) is the distance the spring is compressed, the new position of the ball will be:
\[ r = r_0 + x = 0.1 + x \]
At equilibrium, the force exerted by the spring must balance the centrifugal force:
\[ k x = m \omega^2 r \]
Substituting \( r \) gives:
\[ k x = m \omega^2 (0.1 + x) \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 10000 x = 0.09 \cdot (100^2) \cdot (0.1 + x) \]
Calculating \( 0.09 \cdot 10000 = 900 \):
\[ 10000 x = 900 (0.1 + x) \]
Expanding the right side:
\[ 10000 x = 90 + 900 x \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 10000 x - 900 x = 90 \]
\[ 9100 x = 90 \]
Now, solving for \( x \):
\[ x = \frac{90}{9100} = \frac{9}{910} \approx 0.00989 \, \text{m} \]

### Step 6: Final Result
Thus, the distance the ball moves away from the center is approximately:
\[ x \approx 0.00989 \, \text{m} \approx 10^{-2} \, \text{m} \]

### Conclusion
The final answer is that the ball moves away from the center by a distance nearly equal to \( 10^{-2} \, \text{m} \).

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