waecE-LEARNING
Physics Paper 2, MAy/June. 2013  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main


General Comments
Weakness/Remedies
Strength

























Question 8

  1. The accelerating potential in a cathode ray oscilloscope is 2.5kV.  Calculate the maximum speed of the accelerated electrons.
    [e=1.6 x 10-19C; me = 9.1 x 1031kg]


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Observation

Few candidates attempted this question.  Majority of those who attempted it could not apply the correct formula.  Some even mixed up or could not distinguish correctly some of the terms involved e.g. velocity (v) and potential difference (V).  Those that got the formula correct failed due to poor arithmetric.
The expected answer is
 ½ mev2  =   eV                                                                             
v2    =   2eV
              me
=   2x1.6x10-19x 2500
           9.1 x 10-31       
=  8.79 X 1014                                                                                                        
         v =  2.965 x 107 ms-1  

 

 

 

 

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