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 – mv2 / l, (iii) T + mv2 / l, (iv) 0 T is the tension in the string. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-5 Laws Of Motion



Q.4:- 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 – mv2 / l, (iii) T + mv2 / l, (iv) 0
T is the tension in the string.
[Choose the correct alternative].

 

 

Answer:- 



(i) T
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,
F = T = mv2 / l
Where F is the net force acting on the particle.