A stone is tied to one end of a steing and rotated in horizontal circle with a uniform angular velocity. The tension in the string is T, if the lrngth of the string is halved and its angular velocity is doubled , the tension in the string will be

Options

A. 4 T
B. `T//2`
C. `T//4`
D. 2 T

Explanation

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

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- Let the initial length of the string be \( L \).
- Let the initial angular velocity be \( \omega \).
- The tension in the string is given as \( T \).

2. **Identify the Formula for Tension:**
- The tension \( T \) in the string when the stone is rotating in a horizontal circle can be expressed as:
\[ T = M \omega^2 R \]
- Here, \( R \) is the radius of the circular path, which is equal to the length of the string \( L \) in this case.

3. **Substituting Initial Values:**
- Substituting the initial values, we have:
\[ T = M \omega^2 L \]

4. **Change the Conditions:**
- The length of the string is halved, so the new length \( L' = \frac{L}{2} \).
- The angular velocity is doubled, so the new angular velocity \( \omega' = 2\omega \).

5. **Calculate the New Tension:**
- The new tension \( T' \) can be calculated using the new values:
\[ T' = M (\omega')^2 (L') \]
- Substituting the new values:
\[ T' = M (2\omega)^2 \left(\frac{L}{2}\right) \]
- Simplifying this gives:
\[ T' = M (4\omega^2) \left(\frac{L}{2}\right) = 2M\omega^2 L \]

6. **Relate New Tension to Initial Tension:**
- From the initial tension \( T = M \omega^2 L \), we can express \( T' \) in terms of \( T \):
\[ T' = 2(M \omega^2 L) = 2T \]

### Final Answer:
Thus, the new tension in the string when the length is halved and the angular velocity is doubled is:
\[ T' = 2T \]