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 :
$(i) \; T$
$(ii) \; T - \frac{m v^{2}}{l}$
$(iii) \; T + \frac{m v^{2}}{l}$
$(iv) \; 0$
$T$ is the tension in the string. [Choose the correct alternative].

(A) The correct answer is $(i) \; T$.
When a particle connected to a string revolves in a circular path on a smooth horizontal table,the centripetal force required for circular motion is provided solely by the tension $T$ in the string.
Since the table is smooth,there is no friction,and the only horizontal force acting on the particle directed towards the centre is the tension $T$.
Therefore,the net force on the particle is $F_{\text{net}} = T = \frac{m v^{2}}{l}$.