A mass is supported on a frictionless horizontal surface. It Is attached to a string and rotates about a fixed center at an angular velocity `omega_(0)`.If the length of the string and angular velocity both are doubled, the tension in the string which was initially `T_(0)` is now

Options

A. `T_(0)`
B. `T_(0)//2`
C. `4T_(0)`
D. `8T_(0)`

Explanation

To solve the problem, we need to analyze the relationship between the tension in the string, the mass of the object, the angular velocity, and the length of the string.

### Step-by-Step Solution:

1. **Understanding the Initial Condition:**
The initial tension in the string is given by the formula:
\[ T_0 = m \omega_0^2 l \]
where:
- \( T_0 \) is the initial tension,
- \( m \) is the mass of the object,
- \( \omega_0 \) is the initial angular velocity,
- \( l \) is the initial length of the string.

2. **Doubling the Length and Angular Velocity:**
According to the problem, both the length of the string and the angular velocity are doubled. Therefore:
- New length \( l' = 2l \)
- New angular velocity \( \omega' = 2\omega_0 \)

3. **Calculating the New Tension:**
We can express the new tension \( T' \) in terms of the new length and angular velocity:
\[ T' = m (\omega')^2 (l') \]
Substituting the new values:
\[ T' = m (2\omega_0)^2 (2l) \]
Simplifying this gives:
\[ T' = m (4\omega_0^2) (2l) = 8m \omega_0^2 l \]

4. **Relating New Tension to Initial Tension:**
Now, we can relate \( T' \) to \( T_0 \):
\[ T' = 8 (m \omega_0^2 l) = 8 T_0 \]

5. **Final Answer:**
Therefore, the new tension in the string when both the length and angular velocity are doubled is:
\[ T' = 8 T_0 \]

### Conclusion:
The tension in the string, after doubling both the length and the angular velocity, is \( 8 T_0 \).

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