2. Solved Papers
  3. NEET

NEET NEET SOLVED PAPER 2016 Phase-II

  • question_answer
    A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by\[{{60}^{o}}\]is W. Now the torque required to keep the magnet in this new position is

    A)  \[\frac{W}{\sqrt{3}}\]               

    B)  \[\sqrt{3W}\]

    C)  \[\frac{\sqrt{3}W}{2}\]             

    D)  \[\frac{2W}{\sqrt{3}}\]

    Correct Answer: B

    Solution :

                \[W=PE(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})\]             \[W=PE(\cos 0-\cos {{60}^{{}^\circ }})\]             =\[\frac{PE}{2}\]             \[\Rightarrow PE=2W\]             \[\tau =PE\,\sin \theta =2W\sin {{60}^{{}^\circ }}=\sqrt{3}W\]


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