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)

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-(mv2/1) (C) T+(mv2/1) (D) 0

Tardigrade Admin · Accepted Answer

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. (T=( mv 2) / I )

Q. 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)

T

$T - \frac{m v ^{2}}{1}$

$T + \frac{m v ^{2}}{1}$

0

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 $T$ . $(T = (m v^{2}) / I)$

Q. 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)

T

T−1mv2​

T+1mv2​

0

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. (T=(mv2)/I)

Q. 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)

$T - \frac{m v ^{2}}{1}$

$T + \frac{m v ^{2}}{1}$

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 $T$ . $(T = (m v^{2}) / I)$