Question 12(a)   State the principle of conservation of energy.
(b)   State
 three effects of heat on a substance.
(c)    Explain why when the bulb of thermometer is dipped into ice-cold water the
mercury level first rises before falling.
(d)    Water of volume 200 cm3 in an aluminum container at 20o C is cooled to – 10oC
when put in a freezer.
1.     Calculate the quantity of heat energy extracted from the water.2.     If heat is extracted at a rate of 120 Js-1 by the freezer, calculate the time taken to produce this ice.3.     State a reason why the calculated time will be less than the actual time.(Density of water = 1000 kg m-3; specific heat capacity of water = 4200 J kg-1; specific latent heat of fusion of ice = 336000\j kg-1; specific heat capacity of ice = 2100J kg-1 K-1)
Commentsa)     This was well answered by most candidates however, few candidates omitted the  phrase “ in a closed system or “isolated system” in their statement.(b)   The three effects of heat on a substance fell an easy prey to most candidates.Performance was satisfactory.

  • Explanation of how mercury level changes when dipped in ice-cold water proved         difficult to most respondent candidates as they could not provide satisfactory explanation.
  •   The numerical section was poorly tackled by most respondent candidates.  The 

   formula m = v eluded most of the candidates.

   Also applying the formula Q = mc  appropriately was another problem due to their poor  understanding of the concepts of specific heat capacity and latent heat.

The expected answers are:

(a) In an isolated / closed system, the total amount of energy is (always) constant  (although energy may be transformed from one  form to another)
(b)       The effect of heat on a substance
e.g      Change
–           in temperature
–           in size (expansion/contraction)
–           of  phase/state
–           in resistance
–           in colour
(c) The bulb comes in contact with the cold water before the mercury, hence the glass cools first, contracts and squeezes the mercury up the stem until (thermal) equilibrium is reached.  The mercury being a better conductor of heat cools faster than the glass; hence the fall in mercury level

(d)(i)    m         =           x V                                                                                       
=          1000 x       200
=          0.2 kg                                                                                    
Q         =          MwCw
Өw + Mwlf + Mi Ci Өi                                     
=          0.2 x 4200 x (20-0) + 0.2 x 3.36 x 105 + 0.2 x 2100 x [0-(-10)
=          16800 + 67200 + 4200
=          88200J                                                                                              

 (ii)     Q      = Pt    or P = Q

t =

                            =    88200
=  735s  or 12.25 minutes

 (iii)      Why the calculated time is less than the actual time

e.g (No consideration is given to)

  • Heat absorbed by the aluminum container
  • Heat lost / absorbed from the surroundings
  • Effect of impurities on freezing point of liquid.
 Question 13
(a)   Define:
 principal focus for a diverging lens;
 principal axis.
(b)    State a situation under which the speed of light can change.
(c)      The separation between the lenses of an astronomical telescope in normal
  adjustment is 35.0 cm and the angular magnification is 8 Calculate the focal lengths of the lenses.The lenses separation are adjusted to produce the final image 30 cm from the eye piece lens.  Calculate the new separation of the lensesSketch a ray diagram to illustrate the image formed in 13(c)(ii)above.
(a)    Most respondent candidates were not able to define correctly principal focus for a 
diverging lens and principal axis.
(b)    Relatively few candidates succeeded in stating the required situation under which 
the speed of light can change.  Many of them failed to realize that the speed
changes as light travels from one medium to another of different optical densities.
(c)    Most candidates messed up with their responses on this numerical on astronomical telescope in normal adjustment.  This state of affairs could readily be ascribed to the effect of rote-learning as candidates reactions showed lack of proper understanding of the concept. The sketching of the ray diagram was a mirage.The expected answers are:(a)(i)    This is a point on the principal axis to which all rays parallel and close to the
principal axis appear to diverge after refraction / passing through the lens.
    (ii)    This is a line joining the centre of curvature (principal focus) of the lens to
the optical centre.   
(b)        When the light travels from one medium to another of different refractive
index/optical density.
(c)       Angular magnification=     
fo + fe              =          35
8fe + fe                            =          35
9fe                    =          35
fe                     =          3.89 cm                                                         
fo                     =          35 – 3.89
=          31.11 cm                                                       
 1                      =          1       +    1                                                      
fe                                 u               v
 1                      =          1       +        1                                                  
3.89                              u                -30
                                    1                      =          1       +   1       =     30 + 3.89                        
u                                3.89         30               116.7

u                      =          116.7                                      

                                                            =          3.44cm                                                                      

New separation of lenses = 3.44 + 31.11 = 34.55cm                             

Parallel rays incident on objective lens
Correct position of real image formed by objective lens
Virtual image formed close to the objective lens   

Question 14
(a)   Define magnetic flux.(b)    Sketch a diagram to illustrate magnetic lines of force for a bar magnet placed in a uniform earth’s magnetic field  with its north pole pointing north.Indicate the neutral points.(c)    State two ways in which the earth’s magnetic field is important to life on the earth surface.(d)    List two applications of electromagnets(e)    In a certain house, three ceiling fans each of 80 W, an air-conditioner rated 1500 W are operated for 6 hours eachday.  Seven lamps each rated 40 W are switched on for 10 hours each day.  The home theatre 100 W and television set 80 W are also operated each day for 5 hours after which they are kept in the standby mode.  In the standby mode, the power consumption is 5% of the power rating.  Calculate the:

1.     Total energy consumed in the house for 30 days in kWh;

2.     Cost of operating all the appliances for 30 days at N10.00 per kWh.


(a) The description of magnetic flux posed some problem to candidates who failed to see it as basically a collection of (imaginary) lines of force in a magnetic field.(b)This part called for a sketch of a given magnetic field.  It was poorly reacted to  by candidates.  Some sketches were rough while some did not indicate the correct direction of the arrows on the magnetic lines.(c)Most candidates were able to state ways in which magnetic field is important to life on earth.  Performance was fairly alright.(d)Listing of applications of electromagnets was well answered by virtually all the respondent candidates.  Performance was good.(e)The numerical part was poorly tacked most respondent candidates mishandled the standby mode part of the question.  Performance was poor.

 The expected responses are:
(a)       Magnetic flux consists of lines which represent the direction and strength of a magnetic field at any point in the field.

(b)correct shape of lines of force correct arrows / directions two  neutral points P correctly indicated             

 (c)Importance of earth’s magnetic field to life on earth  e.g
–           Navigation by sailors and pilots
–           Plotting of contour lines by surveyors
–           shielding of dangerous radiations / cosmic rays from the sun

(d) Application of electromagnetic e.g
–           loud speakers
–           door bells
–           automatic switches / relays
–           separation of magnetic materials from non magnetic materials
–           lifting of heavy magnetic materials

Appliance Quantity Power rating / W Standby power / W Total Watt KW KWh
Ceiling fan      3     80         – 240 0.24 1.44
Air conditional      1   1500         – 1500 1.50 1.50
Lamps      7     40         – 280 0.28 2.80
Home theatre      1    100         5 100  5 0.100     0.005 0.5   0.095
Television set      1    80         4 80        4 0.080 0.004 0.4   0.076

NOTE: Standby mode was for 19 hours

1.      Energy consumed in a day               =  14.311 kWh

Energy consumed in 30 days           =   14.311 x 30
=   429.33 kWh

                 (ii)   Cost of operating for 30 days
1 KWh                         N10.00
429.33KWh               429.33 x 10

                                                =          N 4293.30                         

 Question 151.       Explain radioactivity under the following headings:2.      meaning;3.      types.1.      State two:

2.      advantages of nuclear fusion over nuclear fission;

3.     methods of detection of radioactivity.

1.       A nuclear fusion equation for generation of power is shown below

    +                              +   
[  H = 1.00728u;  H = 2.00141u He = 3.00160u He = 4.00150u;
c  = 3.0 x 108 ms-1; Iu = 1.7 x 10-27 kg]               

1.       An electron of mass 9.1 x 10-31 kg moves with a speed of 4.8 x 108 ms-1 in a   vacuum.  Calculate the wavelength of the associated wave.

[h = 6.6 x 10-34 Js]

1.       Many candidates gave a fair attempt at this question as they found it very tractable.  However, few candidates were still using the word “spontaneous” in defining radioactivity.

1.     Advantages of nuclear fusion over nuclear fission was poorly attempted.  However, they performed well in their attempt to state two methods of detecting radioactivity.

.In the calculation of the energy released in the reaction, some candidates made themselves creditably equal to the required task.  However, few candidates found the conversion of the unified atomic mass to kilogram difficult..Very few candidates did the correct thing while most of other respondent candidates derailed by starting their responses on a wrong note; i.e with wrong formula.  Generally, performance was fair.The expected responses are:(a)(i)  Radioactivity is the decay/disintegration of the nucleus of an unstable 
nucleus/atom (of an element) with the release of any one, two or all of
 alpha, beta and gamma radiations.       (ii)    Types of radioactivity

  • Natural and
  • Artificial

(b)(i)   Advantages of nuclear fusion over nuclear fission 

–           greater energy is released
–           less dangerous radiations are released
–           raw materials required are cheaper
–           raw materials are readily available.

    (ii)    Methods of detecting radioactivity
Any correct 2                                                            
–           photographic films
–           Geiger muller (G-M) tube
–           (Electron) – e loud chamber
–           semiconductor detectors

(c)        m =   mass defect   =          (mass of products ) – (mass of reactants)
=          4.00150                                                          2.00141
 1.00728                                  –                   +  3.00160
 5.00878                                                          5.00301                          

            Mass defect   =          0.00577u
=          5.77 x 10-3u
=          5.77 x 10-3 x 1.7 x 10-27 kg                                                  
=          9.81 x 10-30 kg                                              
E         =          mc2                                                                                                  
=          9.81 x 10-3 x ( 3 x 108 )2
=          8.83 x 10-13 J

(d)       λ          =          h
mV                                                                                                                             =                     6.6 x 10-34
9.1 x 10-31 x 4.8 x 108                                                                           

                        =          1.51 x 10-12 m                                                                                          



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