2018 WAEC Physics Obj & Theory Questions And Answers – May/June Expo

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2018 WAEC Physics Obj & Theory Questions And Answers – May/June Expo.

 

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VERIFIED WAEC 2018 PHYSICS OBJ AND THEORY ANSWERS FROM EXAMCLASS.NET

 

PHYSICS OBJ:
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1-10 CAABACBADB
11-20 ACCCDABDCC
21-30 DBADCABCAB
31-40 CCBBCCBABD
41-50 BDCDBABBDC
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Physics theory 

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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

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(2)
Material used for making optical fibers
(i) Silicon Dioxide
(ii) Silica Powder
(iii) Germanium Tetrachloride

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3)
-Diamagnetic material
-Paramagnetic material
-Ferromagnetic material

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(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.

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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.

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

(b). A laser is a device that emits a beam of coherent light through an optical amplification process.

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9dii
Q=mlv
Lw=q/m
Q=313885.7
M=1.04kg
Lw=313885.7/1.04
=301013.2j/k

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(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

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(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)
-It is used in production of electricity.
-It is used to study and detect charges in genetic engineering.
-It is used in agriculture.
-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

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