NEET NEET SOLVED PAPER 2019

  • question_answer
    A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after \[2\pi \] revolutions is- [NEET 5-5-2019]

    A) \[12\times {{10}^{4}}\text{ }Nm\]        

    B) \[2\times {{10}^{6}}\text{ }Nm\]

    C) \[2\times {{10}^{6}}\text{ }Nm\]

    D) \[2\times {{10}^{3}}\text{ }Nm\]

    Correct Answer: C

    Solution :

    \[\omega _{2}^{2}=\omega _{1}^{2}+2\alpha \theta \] \[0={{\left( 2\pi \frac{3}{60} \right)}^{2}}-2\times 2\pi (2\pi )\alpha \] \[\alpha =\frac{9}{60\times 60\times 2}=\frac{1}{800}\] \[\tau =I\alpha =\frac{m{{r}^{2}}}{2}\alpha \] \[\frac{2\times 16\times {{10}^{4}}}{2}\times \frac{1}{800}=2\times {{10}^{-6}}\]            


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