CORRECT: 2016/2017 WAEC GCE PHYSICS OBJ Essay Theory Questions And Answer Expo/ Runs

PHY.

1.ABDCCCDBCA

11.DBCAABBCAD

21.BBDCDBABDC

31.ABABDDDCBD

41.AACDABDAAD

(2)
(i) The position of the ion of the atom of the element in the electrochemical(activity)series.
(ii) the nature of the electrode.
(iii) the concentration of the electrolyte

3)
l=6.5m
d=3.0mm=0.003m
e=0.20mm=0.0002m
F=60N
stress=F/A
A=pie*D^2/4
=3.142*0.003^2/4
=3.142*9*10^-6/4
=7.065*10^-6/4m^2
stress=60/7.065*10^-6
=8.49*N/m^2

4a)
i)They are used in the fields of communications
4b)
ii)warfare
ii)in exploring space
+++++++++++++++++++++++++++++++++++++++++++++++
(5) Applications of permanent magnet:
(i) in telephone earpiece
(ii) in electrical bell
(iii) in generators
(iv) in electrical motors
(v) in induction coil systems
++++++++++++++++++++++++++++++++++++++++++++++++
(7)
Doping of semiconductor increases the
number of holes and electrons in the semiconductor,thereby making it impure
whch makes it to be useful or active.

SECTION B:
(Answer three)
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(10a)
-Sound wave is mechanical wave while light wave is electromagnetic
-Sound wave is longitudinal wave while light wave is transverse wave
-Sound wave cannot be plane polarized while light wave can be plane polarized
(10bi)
DRAW THE DIAGRAM
(10bii)
Nature=>It is a real image and inverted
Position=>It is at 2f on the other side
Size=>It has the same size of the object
(10c)
Because the light ray travel on a straight line ie rectilinear propagation of light
rays
(10d)
u=36m=3600cm
hi=2.5cm
v=0.30cm
(i)m=v/u
m=0.3/3600
m=0.000083
(ii)m=hi/ho
ho=hi/m
ho=2.5/0.000083
ho=30120cm
ho=301.2m
(iii)
1/f=1/u+1/v
1/f=1/3600+1/0.3
1/f=0.00028+3.33333
1/f=3.33361
f=0.3cm
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(11a)
-Number of turns in the coil
-Area of coil
(11b)
Farad is a capacitor of a unit charge when a unit of voltage drop is applied
(11ci)
The brightness of the bulb remains the same since the same current will flow
through them
(11cii)
(a)The resistor should be connected in parallel with the galvanometer
(b)VR=Vg
I*R=0.05I*Rg
IR=0.05I*100
IR=5I
R=5I/I
R=5ohm
(11di)
I=40sin5w(pie)t
but I=Iosin2(pie)ft
therefore 2(pie)ft=500(pie)t
2f=500
f=500/2
f=250Hz
(11dii)
Io=40amp
11diii)
Ir.m.s=Io/sqroot2
=40/sqroot2
Ir.m.s=28.3amp
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(12ai) Decay constant is defined as the instantaneous rate of decay per unit
atom of a substance.
(12aii)
Halflife ,T=0.693/decay contant. Hence, decay constant,llamda =0.693/T ;
=0.693/3days; =0.182day-1 i.e per day.
(12bi) the moderator of a nuclear fission reactor is made up of graphite. Its
function is that slow down the movement of a neutron,so that the chain
reaction will not dry out.
(12bii) the control rods is made of Boron steel.its function is to ensure
movement of the floor,so that it can shut off the reactor automatically in case
of electrical failure