2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    The magnitude of torque on a particle of mass 1 kg is 2.5 N m about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[\frac{\pi }{3}\]                                    

    B) \[\frac{\pi }{8}\]

    C) \[\frac{\pi }{4}\]                        

    D)   \[\frac{\pi }{6}\]

    Correct Answer: D

    Solution :

    As\[\tau =rF\sin \theta \] \[\sin \theta =\frac{\tau }{rF}=\frac{2.5}{5\times 1}=\frac{1}{2},\theta =\frac{\pi }{6}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner