WAEC/NECO PHYSICS PAPER 2 GCE PAST QUESTIONS AND ANSWERS AS REQUESTED 1

WAEC/NECO PHYSICS PAPER 2 GCE PAST QUESTIONS AND ANSWERS AS REQUESTED 1

Question 1          A ball is projected horizontally from the top of a tall building.  State the factors on           which the distance at which it strikes the ground from the foot of the building depends.
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Comments
 This was a very popular question among the candidates however, most respondent candidates were only able to state only one of the factors (speed of projection) while majority left out the second factor (time of flight).  Some clever candidates list more than two factors expecting the examiner to pick the correct two for them.  Among the factors listed are:  speed/velocity, angle of projection, acceleration due to gravity “g” height of the building, time of flight.  Candidates just need recall the formula for horizontal range  R  =  Uxt.
The expected answer  is

  •    Its initial speed
  •    Time (of flight)
 Question 2
The upward acceleration of a lift of total mass 2500 kg is 0.5 ms-2.  The lift is
 
supported by a steel cable which has a maximum safe working stress of
1.     x 107 Nm-2.  Determine the cross sectional area of the cable.( g = 10ms-2 )
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Comment
            This question was not popular among the candidates and it was poorly attempted  
Most candidates did not realize that for a body inside a upward moving lift it
experiences a weight increase hence they were unable to determine the tension
 
in the cable.  Some candidates stated that stress =   instead of stress =       
          Some candidates used  ‘g’ instead of (a+g).
The expected answer is
Tension in the cable = (mass x acceleration) + (weight of the mass)                         
 
Stress, S  =            Tension
Cross-sectional area                                                               
 
OR
Cross-sectional
 
area                  =          mass (acceleration + acceleration of free fall)
stress
OR
A                     =          m (a+g)
A
=          2500 (0.5 + 10)
1.0 x 107                     
                        =          26.25 cm2        OR   2.625 x 10-3 m2    
Question 3      Differentiate between plane polarization and interference as applied to waveso      List two uses of polaroids.
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Comments
 This was a popular question. Candidates were able to us the definitions of the two concepts to distinguish between them.  They were also able to state correctly two uses of polaroids.  Performance was commendable.
The expected answers are:
Plane polarization                                                      Interference    
 
is the confinement of vibration           is the effect produced when two waves
(of waves) to one plane                       of the same frequency / wave length and
 
amplitude travelling in the same direction in a medium are super imposed
 
is an attribute of transverse                 is an attribute of both transverse and
 
waves only                                          longitudinal waves.
  (b)       Two uses of Polaroid’s Production/determination of
–          3-dimensional films
–           concentration of sugar solution/saccherometry
–           polaroid cameras
–           sun glasses

 

Question 4
A wire of length 55 m and force constant 700 Nm-1 extends to 55.06 m when loaded.  Calculate the work done by the load.
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Comments
 This question was well understood and also well attempted by most candidates as they were able to use the relation w = ½ ke2 correctly.  However, very few use wrong formula.  That is w = ½ ke and arithmetic blunder thus  = 55.06 – 55 = 0.6.Instead of 0.06           
The expected solution is
Work done      =          ½ Ke2                                                                                     
 
=          ½ x 700 x (55.06 – 55)2                                                         
 
=          1.26 J      
  Question 5
A needle floating on the surface of water sinks when a drop of detergent is added to the water.  Explain this observation.      
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Comments
   Most candidates could not relate the observation ( i.e sinking of the needle) to the concept of surface tension.  They were using “upthrust” to explain the observation which was wrong.  Another group of candidates lost marks for not mentioning net/resultant downward force (since there are more than one force acting) which was the cause of the needle sinking.
The expected answer is
The needle floats on the surface of water due to surface tension . The addition of detergents reduces / weakens surface tension of the water; hence there is a resultant downward force causing the needle to sink.
  Question 6
When a steady current of 1.57A is passed through a copper voltameter for 20.0 minutes, a mass of 0.625 x 10-3 kg of copper is deposited on the cathode.  Calculate the relative atomic mass of copper.
(1 F  96500 C)
Comments
    This question was poorly attempted as most respondent candidates were not able to relate the mass of copper deposited to the quantity of charge passed through the voltameter and then the amount of charge required to deposited a mole of copper.  ml   =  mr =
Q l     Q r
[Q1  =  lt]
Q r =  2  x  96500 C
mr =  0.625  x  10-3  x  2  x  96500
1.57  x  20  x  60
 
=      0.064 kg
 The expected answer is
Q         =          It                                                                                                        
 
=          1.57 x 20 x 60
=          1884C                                                                                                
 
OR
1884 C            0.625 x 10-3
12 x 96500 C             
 2 x 96500 x 0.625 x 10-3                                                           
1884
                                    0.064 kg                                                                                 

TO BE CONTINUED

ALL QUESTIONS AND COMMENTS PROVIDED BY WAEC

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