Question

One end of a string of length ‘ l ’ is connected to a particle of mass ‘m’ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed ‘v’, the net force on the particle (directed towards the centre) is : (T is the tension in the string)

• A. T
• B. \(T-\frac{mv^2}{l}\)
• C. \(T+\frac{mv^2}{l}\)
• D. 0

Answer 1

The Correct Option is A

For a particle moving in a circle, the net force acting on the particle is the centripetal force, which is given by: \[ F_{\text{centripetal}} = \frac{mv^2}{l} \] Here, \(m\) is the mass of the particle, \(v\) is its speed, and \(l\) is the radius of the circular path.
In this case, the net force on the particle is provided by the tension \(T\) in the string.
The tension in the string provides the centripetal force needed for circular motion.
Therefore, the net force directed towards the center of the circle is equal to the tension in the string.

The correct answer is (A) : T.

Answer 2

The particle is moving in a circle, so the force directed towards the center of the circle is the centripetal force. For an object of mass \( m \) moving with speed \( v \) in a circle of radius \( l \), the centripetal force \( F_c \) is given by: \[ F_c = \frac{mv^2}{l} \] This force is provided by the tension \( T \) in the string. Therefore, the net force on the particle (directed towards the center) is the tension in the string, \( T \), since it exactly provides the centripetal force. Thus, the correct answer is \({A} \).