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

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_WAEC 2018 PHYSICS ANSWERS_

PHYSICS OBJ:

1-10 CAABACBADB

11-20 ACCCDABDCC

21-30 DBADCABCAB

31-40 CCBBCCBABD

41-50 BDCDBABBDC

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 COMPLETE PHYSICS OBJ

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*1a*

Strain is

force (a part of one's body or oneself) to make an unusually great effort.

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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 ) quantity of charge

ii) nature of element

iii ) Time


2bi

The boiling point of a liquid Is the point at which The vapour pressure is equal To atmospheric pressure


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3)

-Diamagnetic material

-Paramagnetic material

-Ferromagnetic material


)4a)

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


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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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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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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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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

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*_Ayostuffs_*

www.ayostuffs.blogspot.com

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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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