A stone of mass 1 kg is tied to end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is:

Options

A. `40ms^(-1)`
B. `20ms^(-1)`
C. `400ms^(-1)`
D. `10ms^(-1)`

Answer

To solve the problem, we need to determine the maximum linear velocity of a stone tied to a string, given the mass of the stone, the length of the string, and the breaking tension of the string.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the stone, \( m = 1 \, \text{kg} \)
- Length of the string (radius of the circular path), \( r = 1 \, \text{m} \)
- Breaking tension of the string, \( T = 400 \, \text{N} \)

2. **Understand the Forces Acting on the Stone:**
- When the stone is rotating in a horizontal circle, the tension in the string provides the necessary centripetal force to keep the stone moving in a circular path.
- The centripetal force \( F_c \) required for circular motion is given by the formula:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the linear velocity.

3. **Set Up the Equation:**
- For the stone to rotate without breaking the string, the tension in the string must equal the centripetal force:
\[ T = \frac{m v^2}{r} \]

4. **Substitute the Known Values:**
- Substitute \( T = 400 \, \text{N} \), \( m = 1 \, \text{kg} \), and \( r = 1 \, \text{m} \) into the equation:
\[ 400 = \frac{1 \cdot v^2}{1} \]

5. **Solve for \( v^2 \):**
- Rearranging the equation gives:
\[ v^2 = 400 \]

6. **Calculate \( v \):**
- Taking the square root of both sides:
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer:
The maximum linear velocity the stone can have without breaking the string is \( 20 \, \text{m/s} \).

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