WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    If the kinetic energy of a body changes by 20%, then its momentum would change by

    A)  20%                                        

    B)  24%                 

    C)  40%                                      

    D)  44% (e) None of the above

    Correct Answer: D

    Solution :

                    (e) AS   \[KE=\frac{{{p}^{2}}}{2m}\] Percentage change in \[KE=\frac{K{{E}_{f}}-K{{E}_{i}}}{K{{E}_{i}}}\times 100\] \[\frac{\frac{p_{f}^{2}}{2m}-\frac{p_{i}^{2}}{2m}}{\frac{p_{i}^{2}}{2m}}\times 100=20\]                 \[\Rightarrow \]               \[\frac{pf}{{{p}_{i}}}=\sqrt{1.2}=1.095\] \[\Rightarrow \]               \[\frac{{{p}_{f}}-{{p}_{i}}}{{{p}_{i}}}=0.095\] Therefore % increase = 9.5% (no answer correct).


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