Waec 2018 Physics Obj And Theory Answers – May/June Expo

PHYSICS OBJ:

1-10: DAABADBCDC

11-20: ACCCDABDCA

21-30: DBADCABCAB

31-40: CCBBCCBABD

41-50: BDCDBABBDC

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

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

(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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(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

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

(7b)
Laser is a device that generates an intense beam of coherent monochromatic light or other electromagnetic radiation by stimulated emission of photons from excited atoms or molecules.

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10a)
Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle.
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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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