A small stone of mass `m` is attached to a light string which passes through a hollow tube. The tube is held by one hand and the free end of the string by the other hand. The mass is set into revolution in a horizontal circle of radius `r_(1)` with a speed `v_(1)` . The string is pulled down shortening the radius of the circular path to `r_(2)`. Which of the following is not correct? (`omega_(1)` and `T` have usual meanings)

Options

A. `(omega_(2))/(omega_(1))=(r_(1)^(2))/(r_(2)^(2)`
B. `(E_(k1))/(E_(k2))=(r_(2)^(2))/(r_(1)^(2))`
C. `(E_(k1))/(E_(k2))=(r_(2)^(2))/(r_(1)^(2))`
D. `(T_(2))/(T_(1))=(r_(1)^(3))/(r_(2)^(3))`

Explanation

To solve the problem, we need to analyze the motion of the stone attached to the string as the radius of its circular path changes. We will use the principles of angular momentum and tension in the string.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The stone of mass `m` is revolving in a horizontal circle of radius `r1` with speed `v1`.
- The angular momentum (L) of the stone can be expressed as:
\[ L_1 = m \cdot v_1 \cdot r_1 \]

2. **Changing the Radius**:
- When the string is pulled down, the radius changes from `r1` to `r2`, and the speed changes from `v1` to `v2`.
- Since there is no external torque acting on the system, the angular momentum is conserved:
\[ L_1 = L_2 \implies m \cdot v_1 \cdot r_1 = m \cdot v_2 \cdot r_2 \]
- We can cancel the mass `m` from both sides (assuming it remains constant):
\[ v_1 \cdot r_1 = v_2 \cdot r_2 \]

3. **Relating Speeds and Radii**:
- Rearranging the equation gives us the relationship between the speeds and the radii:
\[ \frac{v_1}{v_2} = \frac{r_2}{r_1} \]
- This implies:
\[ v_2 = v_1 \cdot \frac{r_2}{r_1} \]

4. **Tension in the String**:
- The tension in the string can be expressed in terms of the speed and radius:
\[ T = \frac{m \cdot v^2}{r} \]
- For the two states, we have:
\[ T_1 = \frac{m \cdot v_1^2}{r_1} \quad \text{and} \quad T_2 = \frac{m \cdot v_2^2}{r_2} \]

5. **Finding the Ratio of Tensions**:
- To find the ratio of tensions:
\[ \frac{T_2}{T_1} = \frac{v_2^2 \cdot r_1}{v_1^2 \cdot r_2} \]
- Substituting \( v_2 = v_1 \cdot \frac{r_2}{r_1} \):
\[ \frac{T_2}{T_1} = \frac{\left(v_1 \cdot \frac{r_2}{r_1}\right)^2 \cdot r_1}{v_1^2 \cdot r_2} = \frac{r_2^2 \cdot r_1}{r_1^2 \cdot r_2} = \frac{r_2}{r_1} \]

6. **Angular Velocity Relation**:
- The angular velocity is related to linear velocity and radius:
\[ \omega = \frac{v}{r} \]
- For the two states:
\[ \frac{\omega_2}{\omega_1} = \frac{v_2/r_2}{v_1/r_1} = \frac{v_2 \cdot r_1}{v_1 \cdot r_2} \]
- Substituting \( v_2 = v_1 \cdot \frac{r_2}{r_1} \):
\[ \frac{\omega_2}{\omega_1} = \frac{\left(v_1 \cdot \frac{r_2}{r_1}\right) \cdot r_1}{v_1 \cdot r_2} = \frac{r_2}{r_1} \]

7. **Conclusion**:
- After analyzing the options, we find that the only incorrect statement is option A, which does not conform to the derived relationships.

### Final Answer:
The option that is **not correct** is **A**.