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

Options

A. 2T
B. 4T
C. T
D. 8T

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial conditions
Let’s denote:
- The mass of the stone as \( m \).
- The initial length of the string as \( r \).
- The initial angular velocity as \( \omega \).
- The initial tension in the string as \( T \).

In circular motion, the tension in the string provides the necessary centripetal force to keep the stone moving in a circle. The centripetal force is given by the formula:
\[ F_c = m \omega^2 r \]
Since the tension \( T \) provides this centripetal force, we can write:
\[ T = m \omega^2 r \]

### Step 2: Analyze the new conditions
Now, according to the problem:
- The length of the string is halved, so the new length \( r' = \frac{r}{2} \).
- The angular velocity is doubled, so the new angular velocity \( \omega' = 2\omega \).

### Step 3: Calculate the new tension
Using the same formula for centripetal force, the new tension \( T' \) can be expressed as:
\[ T' = m (\omega')^2 r' \]
Substituting the new values:
\[ T' = m (2\omega)^2 \left(\frac{r}{2}\right) \]
Calculating this gives:
\[ T' = m \cdot 4\omega^2 \cdot \frac{r}{2} \]
\[ T' = 2m \omega^2 r \]

### Step 4: Relate the new tension to the initial tension
From our initial condition, we know that:
\[ T = m \omega^2 r \]
Now we can relate \( T' \) to \( T \):
\[ T' = 2(m \omega^2 r) = 2T \]

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

### Final Answer
The tension in the string will be \( 2T \).

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