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Q5.4 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 - \frac{mv^2}{l}

(iii)    T + \frac{mv^2}{l}

(iv)    0

T is the tension in the string. [Choose the correct alternative].

Answers (1)

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When the particle is moving in a circular path, the centripetal force will be :

                                                              F_c\ =\ \frac{mv^2}{r}

This centripetal force will be balanced by the tension in the string.

So,  the net force acting is :   

                                                     F\ =\ T\ =\ \frac{mv^2}{r} 

Posted by

Devendra Khairwa

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