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

WAEC/NE2

  Question 7
Explain the rise of water in a narrow tube dipped vertically in a though of water.
Comments
Most candidates were defining capillarity instead of the explanation.  Some could not relate the observation to the concept of capillarity and the forces of adhesion and cohesion.  The question was poorly attempted.
The expected answer is
The rise of water is a result of capillarity.                                                                       
 
The force of attraction/adhesion between the molecules of water and those of the tube is greater than cohesion among water molecules.   So water clings to the wall of the tube causing the rise.
Question 8   List two properties of cathode rays.    State how the intensity and energy of cathode rays may be increased.
_____________________________________________________________________________________________________
Comments
Most respondent candidates were able to state the properties of cathode rays but   could not state how the intensity and energy of cathode rays may be increase.  Performance was fair.
The expected answers are:
(a)     Properties of cathode rays
e.g  Cathode rays            –           are negatively charged
–           travel in straight line (in field free space)
–           are deflected by electric / magnetic field
–           possess (kinetic) energy
–           possess momentum
–           can produce intense heat
–           affect photographic films.  
 (b)(i)   The intensity of cathode rays may be increased by raising the voltage across the 
heater/the temperature of the cathode/increasing the current.
     (ii)     The energy of cathode rays may be increased by increasing the potential 
difference  between the anode and the cathode/the anode potential.
Question 9   Define diffusion.   State two applications of electrical conduction through gases.
_____________________________________________________________________________________________________
Comments
 Most candidates were not able to give an acceptable definition of diffusion though they were able to state the applications of electrical conduction through gases.  Many defined diffusion as movement of particles from region of high concentration to a region of low concentration.  Many candidates still  confuse diffusion with Osmosis. The expected answers are:(a)Diffusion is the process by which substances mix (intimately) with one another                              due to the random motion of their molecules.                                         
OR
[Scattering of a beam of light on reflection from a rough surface / transmission through certain materials]
(b)Any two correct applications  e.g . in
–           advertising industry / Neon signs
–           lighting / fluorescent tubes
–           identification of gases
–           cathode ray oscilloscope / T.V. tube
Question 10       a).State Heisenberg’s Uncertainty Principleb).State one phenomenon that can only be explained in terms of the wave nature of  light.
Comments
 Most candidates were able to state the Heisenberg’s Uncertainty principle though some left out the key word “simultaneously” in their statement.  Majority were also able to state the phenomenon that illustrate wave nature of light.  Generally the question was well attempted.The expected answers are:a)It is impossible to determine with accuracy both the position and momentum of a particle simultaneously                                                                                             
OR
p x  ≥ h/2π
Where all symbols are correctly specified i.e
h = plank’s constant ; p =  uncertainty / error in momentum, x =  uncertainty / error in position.
NOTE:   Accept h, h or
 h in place of  h 
4                 2      
 
(b)       One phenomenon correctly mentioned
e.g.
–           diffraction
–           interference

 

Question 11a.Define:    i.work;ii.energy.b.A pump is used to raise water from a depth of 20 m to fill a reservoir of volume

   1800 m3 in 5 hours. Calculate the power of the pump.
( Density of water = 1000 kg m-3; g = 10 ms-2)

c.(i)What is terminal velocity?

    (ii) Explain the action of a parachute.

d.Why is it important that a defender in football should have a lot of mass?

_____________________________________________________________________________________________________
Comments
1.       Most candidates gave the correct definition of energy but could not define work   correctly.  The phrase “ in the direction of the applied force “ was missing in their definition of work.  Some candidates mistook the definition of energy for work.2.       The calculation of the power of the pump was well tackled by most respondent candidates.  However, very few candidates had problems deriving the mass of water m from volume v and density ρ of water.  Some candidates did not convert the time from hours to seconds before substitution while few other candidates could not relate  Pt  =  mgh.1.       Many candidates were not able to explain the concept of terminal velocity nor the action of a parachute as many of them failed to establish the fact that the amount of air resistance experienced is increased by the umbrella of the parachute. Performance was low. 1.       The reason why a defender in football should have a lot of mass was poorly tackled by many candidates as they could not relate mass to inertia.The expected answers are:
a)(i)     Work is the product of  the force and  displacement/ distance in the
 
direction of the force.                                                                                             
 
 Accept W = Fs  provided  the letters  are correctly defined.     (ii)     Energy is the capacity / ability to do work.                                                          (b)   m  =  D x V                                                                                      
=      1000 x 1800                                                                         
 
=      1,800,000                                                                             
 
Pt            =      mgh                                                                                       
 
P x 5 x 60 x 60   =      1,800,000 x 10 x 20                                                                        
 
P              =      20,000W                                                                               
(c) (i)   The steady/constant speed with which an object/body moves freely 
downward through a fluid when the resultant force on it is zero                                                                                                      
 
OR
The steady/constant speed attained by an object/body falling freely through a fluid when the viscous force and upthrust of the fluid on it are (jointly) equal to the weight of the object/body
(ii)     The umbrella of the parachute increases the amount of air resistance 
experienced  by the sky driver and this in turn slows down the (rate of )fall
 
until terminal velocity is attained.

(d)       An attacker runs towards a defender’s goal with high velocity that 
generates high momentum in order to stop/block the attacker the defender 
 
must have great inertia which is a measure of mass so the defender must
 
have a lot of mass.

ALL QUESTIONS AND COMMENTS PROVIDED BY WAEC

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