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# WAEC 2018 Physics Obj And Essay Answer – May/June Expo

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## WAEC Physics Obj And Essay/Theory Solution Questions and Answer – MAY/JUNE 2018 Expo Runz.

Physics OBJ:
31DCBBCCBABB

10a)
Diffraction is the ability of waves to bend around obstacles in their path

10bi)
Critical angle is the highest angle of incidence in a denser medium when angle of refraction in the less dense mediumis 90 degrees
10bii)
Critical angle=90-44=46degrees
The refractive index of the glass is obtained as follows
The refractive index(n)=sini/sinr=sin46/sin90=0.7193/1
=0.7193
10ci)
fo=200Hz
f1=3v/4l =>closed pipe
f1=v/l=>open pipe
but 3fo=f1=>closed pipe
3*200=f1=>f1=600Hz
Also f1=2fo=>open pipe
2fo=600
fo=600/2=300Hz
10cii)
v=330m/s
L=?
fo=200Hz
fo=v/4l
=>200=330/4l
800l=330
l=330/800
l=0.4124m
10ciii)
fo=v/2l
fo=300Hz
v=330m/s
300=330/2l
l=330/600
l=0.55m

9ai)
i)Nature of surface
ii)medium of transmission
9aii)
i)state or phase of the substance (i:e solid,or gas)
ii)temperature of the meduim

9b)
i)temperature
ii)specific heat capacity of the body
9c)
the statement means that the amount of heat energy required to change 1kg of liquid mercury to gaseous mecury without change in temperature is 2.72*10^5jkg^-1
9di)
Q=MCDSin
V^2/R t=MCDsin
(220)^2*4*60/35=M*4200*(100-28)
331885.7=302400m
m=331885.7/302400 =1.098kg
9dii)
V^2/R t = MLv
(220)^2*5*60/35=0.3*Lv
414857.14=0.3Lv
Lv=414857.14/0.3
Lv=1382857.13JKg^-1

[CLICK HERE And HERE For No.12 Solution]
12a) This is defined as the amount of energy that must be supplied to a nucleus to completely separate it’s nuclear particles (nucleons)

12b)
i) They have short wavelength and high frequency.
ii) They are highly penetrating.
iii) They travel in straight lines.
iv) They don’t require material medium for their propagation.
12c)
i)It is used in production of electricity.
ii)It is used to study and detect charges in genetic engineering.
iii)It is used in agriculture.
iv)It is used in treatment of cancer.
12di)
E = hf-hfo
but f = v/landa
E= v/landa.h – wo
Where wo = hfo = work function
f= frequency
landa = wavelength
Hence
hf = hfo – E
f = hfo – E/h
f = wo – E/h
Recall; that v = f landa
Therefore f = v/landa = 3×10^8/4.5×10-7
=3/4.5 × 10^8+7
=6.6×10^14Hz
f = 6.6×10^14Hz
12dii)
E = hf
=6.6×10^-34 × 6.6×10^14Hz
=43.56×10^-20J
12diii)
Energy of the photoelectron E = hf – vo
=Energy of incident electron – work function
=4.356×10^-19J – 3.0×10^-19J
=1.356×10^-19J

1a)
Strain can be defined as the ratio of extension per unit length
Strain=extension/length

1b)
Strain=extension/length
Let original length=L
Final length=2L
Extension=2L-L=L
Strain=L/L=1

2)
i) Silicon Dioxide
ii) Silica Powder
iii) Germanium Tetrachloride

3)
i) Ferromagnetic material
ii) Diamgentic material
iii) Pamangentic material

4a)
An intrinsic semiconductor is an undoped semiconductor that is a pure semiconductor without any significant dopant species present.

4b)
The P Type semiconductor is a type of semiconductor that carries a positive charge, while the N type semiconductor carries a negative charge .

5)
Range = u²Sin2tita/g
At maximum range
Sin2tita = 1
2tita =sin^-1(1)
2tita = 90dgrees
Tita = 90/2 = 45degree

Maximum height reached = u²sin²tita/2g
=u²(sin45)²/2g
=200²(sin45)²/2(10)
=40000(1/√2)2/20
=40000(1/2)/20
=20000/20
=1000metres.

6)
Given constant = 2.9×10^-3mk
Temperature = 57degreeC = (57+273)k = 330k
Using landamaxT = constant
landamaxT330 = 2.9×10^-3
landamax = 2.9×10^-3/330
landamax = 8.788×10^-6m
The speed of electromagnetic wave, v = 3×10^8m/s
Using V = f landa
f = v/landa
=3×10^8/8.788×10^-6
=3.4×10^13Hz

7a)
LASER stands for Light Amplification by Stimulated Emission of Radiation.

7b)
A laser is a device that emits a beam of coherent light
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