2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    A thin metal disc of radius of \[0.25\text{ }m\]and mass \[2\text{ }kg\]starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is \[4\text{ }J\]at the foot of inclined plane, then the linear velocity at the same point, is in m/s

    A)  \[2\]            

    B)  \[2\sqrt{2}\]

    C)  \[2\sqrt{3}\]          

    D)  \[3\sqrt{2}\]

    Correct Answer: B

    Solution :

    Rotational kinetic energy \[=\frac{1}{2}I\,\,{{\omega }^{2}}\] \[\therefore \]   Rotational \[KE=\frac{1}{2}\left[ \frac{1}{2}m{{r}^{2}} \right]\frac{{{v}^{2}}}{{{r}^{2}}}\] \[\left( where\,\,\,\,I=\frac{1}{2}\,m{{r}^{2}} \right)\] \[4=\frac{1}{2}\left[ \frac{1}{2}(2){{r}^{2}} \right]\frac{{{v}^{2}}}{{{r}^{2}}}\] \[\Rightarrow \] \[{{v}^{2}}=8\] \[v=2\sqrt{2}m/s\]


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