One end of string of length `l` is connected to a particle on mass `m` and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed `v` the net force on the particle (directed toward centre) will be (`T` reprents the tension in the string):

Options

A. `T+(mv^(2))/(1)`
B. `T-(mv^(2))/(1)`
C. zero
D. `T`

Explanation

To solve the problem, we need to analyze the forces acting on the particle moving in a circular path. The particle of mass \( m \) is connected to a string of length \( l \) that is fixed at one end to a peg on a smooth horizontal table. The particle is moving in a circle with a constant speed \( v \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle**:
- The only forces acting on the particle are the tension \( T \) in the string and the gravitational force \( mg \) acting downward. However, since the table is horizontal and smooth, we only need to consider the horizontal forces for circular motion.

2. **Understand Circular Motion**:
- For an object moving in a circular path, there must be a net inward force (centripetal force) acting towards the center of the circle. This net force is provided by the tension in the string.

3. **Centripetal Force Requirement**:
- The required centripetal force \( F_c \) for an object of mass \( m \) moving with speed \( v \) in a circular path of radius \( r \) is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
- In this case, the radius \( r \) is equal to the length of the string \( l \).

4. **Set Up the Equation**:
- Since the tension \( T \) in the string provides the necessary centripetal force, we can equate the tension to the centripetal force:
\[ T = \frac{mv^2}{l} \]

5. **Conclusion**:
- The net force acting on the particle directed towards the center of the circular path is simply the tension in the string. Therefore, the net force on the particle is:
\[ T = \frac{mv^2}{l} \]

### Final Answer:
The net force on the particle directed toward the center is given by:
\[ T = \frac{mv^2}{l} \]