# 2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

## 2018 WAEC RUNZ  PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

Thursday, 20th April, 2018
PHYSICS   2 (Essay) – 09:30 a.m. – 11:00 a.m
PHYSICS  1 (Objective) – 11:00 a.m. – 12:15 p.m
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PHYSICS OBJ:

11-20: ACCCDABDCA

31-40: CCBBCCBABD

41-50: BDCDBABBDC

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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)
(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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(9ai)
(i) Nature of the Substance
(ii) Medium of transmission

(9aii)
(i)State or phase of the substance (i.e Solid, Liquid or gas).
(ii)Temperature of the medium

(9b)
(i) Temperature
(ii) Specific hear capacity of the body

(9c)
The statement means that the amount of hear energy required to change 1kg of liquid mercury to gaseous mercury without change in temperature is 2.72*10^5Kg^-1

(9d)

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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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(11ai)
Reactance is the opposition offered to the passage of an alternating current by either the inductor OR the capacitor or both.

(11aii)
Impedance is the overall opposition of a mixed circuit To the flow of an alternating current in a ressistor OR capacitor.

(11bi)
E.m.f is induced in an a.c generator so as to create more magnetic flux which generates an induced current.

(11bii)
The carbon brushes makes contact with the springs so that the amature rotates through it. Induced current will be produced. It also used to alternate the direction of the current produced.

(11biii)
The law states that if the first three fingers of one’s right hand are held at right angles to each other with the fore finger in the direction of the the field and the third in the direction of the motion, the middle finger points in the direction of the induced current.

(11biv)
-Increasing the no of coils.
– The presence of soft iron core inside the coil.

(11c)
Energy E = power × time
E = pt; where p=6w, t=60mins
E= 6 ×5 ×60
= 180J

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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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2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS
2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

2018 WAEC RUNZ PHYSICS ESSAY & OBJ QUESTIONS AND ANSWERS

#### 1 Comment

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