A particle of mass 100 g tied to a string is rotated along the circle of radius 0.5 m. The breaking tension of the string is 10 N. The maximum speed with which particle can be rotated without breaking the string is

Options

A. `10 " ms"^(-1)`
B. `9.8 "m s"^(-1)`
C. `7.7 " m s"^(-1)`
D. `7.07 " m s"^(-1)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (conversion from grams to kilograms)
- Radius of the circle, \( r = 0.5 \, \text{m} \)
- Breaking tension of the string, \( T = 10 \, \text{N} \)

### Step 2: Understand the relationship between tension, mass, velocity, and radius
When a particle is rotating in a circle, the tension in the string provides the necessary centripetal force. The formula for centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
Where:
- \( F_c \) is the centripetal force (which is equal to the tension in the string when the particle is at maximum speed),
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( r \) is the radius of the circle.

### Step 3: Set up the equation using the maximum tension
Since the maximum tension \( T \) is equal to the centripetal force at maximum speed, we can write:
\[ T = \frac{mv^2}{r} \]
Rearranging this equation to solve for \( v \):
\[ v^2 = \frac{Tr}{m} \]
\[ v = \sqrt{\frac{Tr}{m}} \]

### Step 4: Substitute the known values into the equation
Now, substitute \( T = 10 \, \text{N} \), \( r = 0.5 \, \text{m} \), and \( m = 0.1 \, \text{kg} \):
\[ v = \sqrt{\frac{10 \times 0.5}{0.1}} \]

### Step 5: Calculate the value
Calculating the expression inside the square root:
\[ v = \sqrt{\frac{5}{0.1}} = \sqrt{50} \]
Now, calculating \( \sqrt{50} \):
\[ v \approx 7.07 \, \text{m/s} \]

### Conclusion
The maximum speed with which the particle can be rotated without breaking the string is approximately \( 7.07 \, \text{m/s} \).