2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q is shot with speed 2v, the closest distance of approach would be

    A)  \[\frac{r}{4}\]                  

    B)         \[\frac{r}{2}\]                  

    C)  \[2r\]                   

    D)         \[r\]

    E)  \[\frac{3}{2}r\]

    Correct Answer: A

    Solution :

    \[\frac{1}{2}m{{v}^{2}}=\frac{kQq}{{{r}_{1}}}\]                         ...(i) \[\frac{1}{2}m.4{{v}^{2}}=\frac{kQq}{{{r}_{2}}}\]                                               ?. (ii) From Eqs. (i) and (ii), we can say                 \[\frac{{{r}_{1}}}{{{r}_{2}}}=4\]                 \[{{r}_{2}}=\frac{r}{4}\]


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