TOTAL SOLUTIONS IN

PHY113
CALCULATIONS AND THEORY
QUESTIONS AND ANSWERS INCLUSIVE
(INCLUDING SOLVINGS OF ALL EXERCISES IN YOUR TEXTBOOK)

BY KAYMATH
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Calculations and theoretical aspects inclusive (strictly exam focus)

All texts and calculations typed for easy assimilation
Over 100 Theoretical questions and answers.
By Kaymath (call 08068552755)
TABLE OF CONTENT
WHAT YOU NEED TO KNOW ABOUT APPROXIMATIONS—page 1

VIBRATIONS—page 2

DAMPED OSCILLATIONS—page 3

FORCED DAMPED OSCILLATIONS—page 5

WAVES—page 6

INTERFERENCE OF WAVES—page 8

NORMAL MODES—page 10

OPTICS—page 13

REFLECTION AT PLANE AND CURVED SURFACES—page 14

REFRACTION THROUGH PLANE SURFACES—page 16

REFRACTION THROUGH CURVED SURFACES (LENSES)—page 17

OPTICAL INSTRUEMENTS—page 18

DISPERSION AND ABBERATION—page 18

THEORITICAL ASPECT OF PHY113 (INCLUDES—page 19

QUESTION, ANSWERS AND EXPLANATION.
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By kaymath (call 08068552755) 1

WHAT YOU NEED TO KNOW ABOUT APPROXIMATIONS

Approximations are very significant in uniben physics and chemistry exams
because if you approximate wrongly you will get a wrong answer but in
this material I will Teach you how to approximately correctly. Note that
you approximate A number only when the decimal number is in the range
of 0-2 and when the decimal number is in the range of 7-9, you don’t
approximate when the decimal number is in the range of 3-6 or else u get
a wrong number which will be either bigger or lesser than answers in the
options instead write your figures and solve like that. I will Explain what I
mean in details .for instance you are solving and you Get a number like
2.1112 If you must approximate you get 2.1 or 2.11 because decimal range
is 0-2,if you get a number 3.7 or 3.79 or 3.668 you can approximate to
4,3.8 and 3.67 respectively because the decimal numbers you
approximated ranges from 7-9 .but note when u have a number
2.555,2.554,2.556 or 2.443 or 2.553 you don’t approximate instead solve
with the figures the Way you see it in the calculator because the decimal
number ranges from 3-6.This aspect is very important when solving
questions else you get a number slightly different from the ones in the
options please take note. More examples, if you get a number
2.2275,7.1284,6.394,2.3844,3.293, you get 2.23,7.13,6.4,2.4,3.3,
respectively .it doesn’t matter where the decimal number is as far as it
follows the rules. also if u get 2.335,2.445,2.333333,2.66666,2.43434, do
not approximate solve with it the way you see it. TAKE NOTE!!!!!!!!!!!
( let’s begin see page 19 for the theory questions and answers but let’s
deal with the calculation aspect first)
 By kaymath (call 08068552755) 2

Amplitude of oscillation Velocity of the

VIBRATIONS (CHAPTER 1) particle as it passes the equilibrium position?
Period of oscillation is given by T= SOLUTION
Where r=radius, v=speed of oscillation Period =3cm =4cm, A=?, from A=√
is also given by T= ,w=angular Frequency, A=√ ,A=5cm. The velocity will be given
by the Pythagoras theorem of the two
NOTE period can also be given by T= ,where
velocities i.e V=√ V=5cm/s
t=time, n=no of oscillations. Hooke’s law is
1.2 A punch bag of mass 0.6kg is struck so
given by F=KX k=force constant,
that it oscillates with SHM. The oscillation has
x=displacement equation for simple harmonic
a frequency of 2.6Hz and an amplitude of
motion is given by + =0 angular 0.45m what is (a) the maximum velocity of
the bag? (b) The maximum kinetic energy of
velocity In S.H.M is given by w=√ m=mass the bag? (c) what happens to the energy as
Amplitude is given by A=√ and the oscillation dies away
are equation constants other equations for SOLUTION
S.H.M are m=o.6kg, f=2.6Hz, a=0.45m, from w=2πf,
w=2x3.142x2.6=16.3384rad/s. in uniben
( )
And ( ) ( ) a=amplitude exams you don’t approximate anyhow that’s
why “w” was not approximated. See first page
t=time, phase angle difference, which is
on hints on approximations. (a) from
Given by Ф= ,maximum velocity is =WA =16.3384x0.45=7.35m/s
Given by = , maximum acceleration is (b) The maximum kinetic energy will be given
given by = ,velocity at any point of By KE= = x0.6x =16.2J. (c) The
oscilliation is given by = √ where punch bag will transfer heat to the
x=displacement, a=amplitude, w=angular surrounding as it stops gradually
velocity, period of oscillation in 1.3 A light spiral spring is loaded with a mass
S.H.M is given by = √ l=length, g=gravity of 50g and extends by 10cm. What is the
period of small vertical oscillations if the
Note also this important formular = & acceleration due to gravity is 10m/s
√ √
SOLUTION
√ = √ ,energy in S.H.M given by
L=0.1m (converted to m), g=10m/s, from
E= ,energy per mass is given by
= √ =2X3.142√ = 0.63s.
= , period is also given by T=
SOLUTION TO EXERCISE 1 1.4 How much would the time keeping of a
1.1 A particle moving with simple harmonic pendulum clock be affected by taking it to the
motion has velocities 4cm/s and 3cm/s at moon? Gravity on the moon is 1.6m/ ,
distances of 3cm and 4cm respectively from Compared to with 10m/ on the earth.
the equilibrim position. What is the SOLUTION
We are asked to find the ratio of the two
By kaymath (call 08068552755) 3

Period from √ = √ , =1.6m/ Force constant,k=force constant, m=mass. For

(gravity of moon), =10m/ (gravity of critically damped motion b= √ . equation
For damped oscillation is given by
earth), √ = √ , =√ =2.5s, means
X= =exp ( ) ( ) please this
period is 2.5 times slower in the moon formula is important don’t ever forget it
1.5 A simple pendulum has a period of 7s. X=damping amplitude, =initial amplitude,
When the length was shortened by 1m, the t=time. Note that equation can be broken in
period is 6s. find the original length of the
two when solving --> X= exp ( ) This
pendulum
SOLUTION one is mainly used to solve questions but know
=6s, =7s, it was stated that the length Both of them. Note that when
was shortened by 1m meaning the original exp(exponenetial) crosses the ’=’ sign it
becomes ’ln’. don’t worry you will understand
length-1,form = , = , =
√ √ √ √ Better when we start solving exercises. is
49L-49=36L, 13L=49, L=3.77m
1.6 A clown is on a rocking chair in the dark Given by = and =√ , relaxation time
His glowing red nose moves back and forth is given by = , damping time is given by
with a distance of 0.42m exactly 30 times a
minute in a simple harmonic motion (a) = , damped angular frequency is also
what is the amplitude of the motion? (b) ( )
given by =√ , nature of
what is the period of this motion? (c) what is
the frequency of this motion? ( )
oscillation is positive if > , quality
SOLUTION
(a) If the total distance back and forth is half factor is given by Q= , Energy in damped
the distance back and forth is 0.42m then oscillation is given by E= , period of
amplitude is half the distance a= =0.21m damped oscillation is given by T=
(b) t=1x60=60secs,n=30,from T= = =2sec SOLUTION TO EXERCISE 2
(c) from T= ,F= = =0.5Hz. 2.1 A simple pendulum of length 22m is set
Into oscillation with amplitude 0.05m,after
DAMPED OSCILLATION(CHAPTER 2)
5min it has fallen to 0.025m.calculate the
The identity equation for a damped
relaxation time
oscillation is given by ̈ + ̇ + =0 m=mass
SOLUTION
B0damping force constant, k=force constant Pendulum was set into oscillation that means
The three angular frequencies in damped The amplitude is initial =0.05m, t=5x60
oscillation are; t=300secs,pendulum fell to 0.025m,x=0.025m
= natural angular frequency
= damped angular frequency from = , we find to get relaxation time.
or( )=damping force angular frequency X= =exp ( ), 0.025= exp )
is given by =√ ,b=damping = exp- , 0.5=exp 300,
Exp will cross the ‘=’ sign and change to ‘ln’
By kaymath (call 08068552755) 4

‘ln’ is in your calculator, ln0.5= 150 question. It is same equation. Hence we find
0.693= 150, =4.62x . from equation =0.25, =0.5
= =432.9secs. ( )
= = =1.571rad/s.from =√
2.2 The equation given ̈ + + =0
represents (a) critically damped (b) lightly ( )
1.571=√ , 2.468= ,
damped (c) simple harmonic motion (d)
forcedly damped (e) none of the above =2.5305, =1.59rad/s. from T=
SOLUTION
T=3.95s. (b) from E= , =
The equation given represents an equation
of damped oscillation.answer is E.note that =5m, = =31.6J/kg.(c) from =
equation for simple motion is = =2secs.(d) nature of oscillation rules are;
2.3 A 0.04kg mass is moving on the end of
>0 or positive+ ( it is lightly damped)
a spring with force constant k=300N/m and
=0 (it is critically damped)
is acted upon by a damping force F= bv (a)
<0 or negative- (it is heavily damped)
if b=9.00kg/s. What is the angular
thus from equation =1.571rad/s is greater
frequency of the mass? (b) for what value
of b will the than 0 so it is lightly damped (e) from Q=
motion be critically damped. Q= =3.142.
SOLUTION 2.5 The equation for damped oscillation is given
We are asked o find the damped angular by X=3exp( ) . Calculate the particle
Frequency ( ) because it was acted upon velocity (a) 2.08m/s (b) 8.43m/s (c) 1.25m/s (d)
by a force m=0.4kg, k=300N/m, b=9kg/s none of the above
(a) From =√ , =√ SOLUTION
when given an equation and asked to find the
=√ =25rad/s. particle velocity( ) use this shortcut;
(b) for critically damped motion b= √ for equation of motion ending with sin
b= √ =21.9kg/s. = . For equation of motion ending
2.4 The equation for motion of an with cos = take note hence
oscillation is given by X=5exp ( ) =πx3=3.142x3=9.43m/s. answer is E
(a) calculate the natural angular frequency 2.6 The equation of a damped harmonic
of the oscillation and it’s period. (b) what is oscillation is given by X=3exp( ) ,
The initial energy per unit mass of the What is the nature of oscillation?
damped oscillator? (c) what is the damped SOLUTION
time? (d) what is the nature of oscillation In equation =π=3.142 is greater than 0 it is
(e) what is the quality factor? lightly damped
SOLUTION
2.7 =5 and =0.25 we use our shortcut
The equation in the question conforms to
here. We have cos hence from
X= =exp ( ) . I.e ( φ)
= =5x0.25=1.2m/s.
was broken . Note also cos is sin in this
By kaymath (call 08068552755) 5

FORCED DAMPED (d) 14.32rad/s.

SOLUTION
OSCILLATION (CHAPTER 3) from ̈+ ̇+ =f ( ) m=2, k=7 ,b=4
Formulas for this topic in your textbook are
very complicated but with this material you from =√ , = =3.5 . from =
will be able to tackle any question on this ( )
topic with easy formulas. Let’s begin. = =2 .from =√ ,
The three frequencies for of forced damped ( )
oscillation are given by =√ =1.22rad/s. A is the answer
= natural angular frequency 3.2 The equation of motion of a particle is
=damping force angular frequency given as ̈ + ̇ + =6 ( ) . The
=resonance angular frequency maximum displacement of the periodic
The general equation for forced damped is motion is obtained when the parameter r
given ̈ + ̇ + =f ( ) ,m=mass, take the value?
b=damping force constant, k=force constant SOLUTION
f=driving force frequency, p=resonance we are asked to find ‘r’ from equation p=πr.
frequency (related to driving force Note at maximum displacement system is at
frequency) . Resonance angular frequency is resonance that means =p, b=4, m=2, k=7
( ) from = = =3.5 , from =
given by =√ ,at steady state
( )
period T= ,p=resonance frequency . Note = =2 =√ =1.225rad/s
that at steady state or at resonance. but p= , hence p=1.225rad/s from equation
Resonance angular frequency ( ) becomes p=πr, 1.225=3.142r, r= =0.3898Hz.
the driving force frequency.Hence at 3.3 The equation of motion of a point of mass
resonance =p , we used ‘p’ as driving is ̈ + ̇ + =5 ( ) . find the
force frequency here because p is related to maximum value of r at maximum displace-
‘f’ at resonance. Mechanical impedance in ment of the particle.
forced damped oscillation is given by SOLUTION
=√ ( ) , =mechanical same as question 3.2. b=5, m=3, k=12 , p= .
= = =4 , = = =1.667
resistance(same as damping force constant
‘b’) , m=mass, p=resonance frequency ( )
=√ =1.61555rad/s. p= ,
=√ , = . 1.61555=3.142r , r= =0.514Hz.
SOLUTION TO EXERCISE 3 3.4 The equation of motion of a point is given
3.1The equation of motion of a particle of by ̈ + ̇ + =20 ( ) . Find the
mass 2kg is given by ̈ + ̇ + =6 ( ) resonance frequency.
Determine the resonance frequency (a) SOLUTION
1.22rad/s (b) 4.22rad/s (b) 8.25rad/s m=3, b=7, k=11. We are asked to find
By kaymath (call 08068552755) 6

= == =3.66667 , = . from when solving questions using the 2nd equation

always use =A ( ) .Note the –‘’
( )
= = =2.33333 , =√ sign .phase velocity is given by V= . speed of
( ) a transverse wave (wave on a string) is given
=√ =0.971rad/s.
by V=√ , T=tension given by T=mg μ=mass
be careful about approximations.
3.5 The equation of motion of a particle is per unit length given by μ= l=length, m=mass
given by ̈ + ̇ + =5 ( ). Deter-
mine the type of motion and determine the speed of longitudinal wave is given by V=√ B=
frequency at steady state.
bulk modulus given by B= k=compressibility ,
SOLUTION
At steady state system is at resonance that p=density. Speed of longitudinal wave in a
is =p. In this question ‘’p’’ is ‘’w’’ so that solid is given by V=√ =young modulus, p=
p=2 , f= , hence f= . Note that another
density. The equation of wave can also be
letter can be used to represents p e.g. a,z,c
so it depends on the question. given by y=A ( ) λ=wavelength,
3.6 Referring to question (3.5) above, find t=time, T=period. x=displacement
the angular frequency of the motion when SOLUTION TO EXERCISE 4
the amplitude is maximum 4.1 The equation of a certain transverse wave
SOLUTION is y=4cm ( ). Determine the
Note at maximum amplitude system is wave’s (a) amplitude (b) wavelength (c) frequ-
at resonance, so we asked to find the reson- ency (d) speed of propagation
ance frequency b=6, k=27, m=1. = = , SOLUTION
=27rad/s. = = =6rad/s. from (a) A=4cm. (b) in equation is , = ,λ=50cm
( ) ( )
to m λ=0.05m. (c) according to the equation
=√ =√ =3rad/s. T=0.03 but from T= . f= = =33.3Hz. (d)
WAVES (CHAPTER 4) from V=fλ=33.3x0.5=16.65m/s.
angular frequency is given by W=2 . 4.2 A wave moving along the x-axis is defined
period is given by T= , f=frequency . speed by =5exp ( ) where x is in metres
and t is in seconds . Determine (a) the
of propagation is given by V=f direction of the wave motion (b) The speed of
λ=wavelength. Equations of wave are given the wave
by =A & =A ( ) SOLUTION
t=time , a=amplitude k=wave number which from =A ( ) k=1. A=5, W=5
is given by k= for the 2nd equation of wave (a) This is where we use our rule because
(+) represents a wave travelling in the has a positive sign .It progresses to the
negative x-axis and (-) represents wave negative direction of the x-axis .so answer is
travelling in the positive x-axis -x-y6x+ left (b) from V= = =5m/s
this is simply graph knowledge.
By kaymath (call 08068552755) 7

4.3 Transverse waves on a string have wave v=fλ=1.591x1.571=2.5m/s.

speed of 12m/s, amplitude of 0.05m and A=1m.V=WA=10X1=10m/s. ratio= =0.25.
wavelength 0.4m. The wave travels in the 4.6 A string has a total length of 5m and a total
positive x-direction and at t=0, the x =0 end mass 0.01kg.If the string has a tension of 10N
of the string has zero displacement and is applied to it. What is the speed of a transverse
moving upward (a) find the frequency, wave on the spring.
period and wave number of these waves (b) SOLUTION
write a wave function describing the wave
(c) find the transverse displacement of a l=5m, m=0.01kg, T=10N from V=√ ,μ=
wave at x=0.25 at a time t=0.15sec
SOLUTION μ= =0.002kg/m. V=√ =70.7m/s.
V=12m/s, A=0.05m, λ=0.4m (a) from v=fλ, 4.7 A string of length 10m and total mass
f= = =30Hz. & period T= = =3.33x s. 0.001kg is connected to a mass ‘m’ suppose
& wave number k= = =15.7rad/s. that the string has a very high elastic limit
(meaning that it takes lots of pressure on the
(b) From equation y=A ( ) note string before it will stretch). How much mass
that we substituted ‘’f’’ for . Equation is must you place on the string in order to
produce a wave speed of 200m/s.
y=0.055m ( )
SOLUTION
4.4 Two wave sources separated by 2.0m
apart vibrate in phase with frequency l=10m, m=0.001kg, v=200m/s from V=√ ,μ=
200Hz and velocity 800m/s . Calculate the μ= = =0.0001kg/m.The mass given is to
phase difference at a point midway
find ‘μ’ the mass we are looking for is from
between them. (a) 0 (b) 2π (c) π (d) none of
the above T=mg hence V=√ , 200=√ , squaring
SOLUTION
note that the phase difference at any point both sides = ,m= =0.408kg.
midway between two wave is zero. 4.8 In a liquid with density 900kg/
Answer is ‘’0’’. longitudinal waves with frequency 250Hz are
4.5 The equation of a transverse wave found to have wavelength 8.00m. calculate the
travelling along a stretched string is given as bulk modulus of the liquid.
= ( ), if the displacement at SOLUTION
appoint is zero. What is the ratio of the
phase velocity of the wave to the particle From V=√ , B=?, B= . f=250Hz, λ=8m.from
velocity at the same point? v=fλ=250x8=2000m/s. p=900kg/
SOLUTION B= X900=3.6X pa.
Here we are simply to find the ratio of v=fλ 4.9 The linear mass density of a string is
& v=WA from =A ( ) w= 1.6x kg/m. A transverse wave is propagat-
10rad/s, from w=2πf, f= = =1.591Hz ed on the string and is described by the
k=4. from k= ,λ= = =1.571m following equation =0.0 ( )
By Kaymath (call 08068552755) 8

(a) What is the wave number? (b) What is is 2.7x Pa. What is the maximum speed at
the wave speed? (c) what is the tension on which transverse wave pulses can propagate
the string? along this wire before this stress is exceeded?
SOLUTION (The density of steel is 7.86x kg/ ).
(a) from equation =A ( ) SOLUTION
the wave number k=2rad/m. (b) from v= P=7.86x kg/ , young modulus from
w=30, v= =15m/s.(c)μ=1.6X kg/m from question is γ=2.7x Pa from V=√ =√
V=√ , T= xμ ,T= X1.6X =0.036N V=586m/s.
4.10 A stretched string has a mass per unit INTERFERENCE OF WAVES
length of 5g/cm and a tension of 10N. A (CHAPTER 5)
sinusoidal wave on this string has a an Equation for standing wave is given by
amplitude of 0.12mm and frequency of y=2A , x=displacement, t=time
100Hz and is travelling towards the w=angular frequency, a=amplitude. Node of a
decreasing .write an equation for this wave.
standing wave is given by n= , λ=wavelength.
SOLUTION
A=0.012cm(converted to cm) μ=0.005kg/cm The nodes of a standing wave is a positive
(converted to kg). y=A ( ) , =F , number e.g. (1,2,3,4,5,6 etc.). n= , n= , n=
expanding we have y=A etc. The antinode of a standing wave is given
by = . The antinodes of a standing wave is
,T=10N, f=100Hz. we find λ, V=√ =√
an odd number e.g. (1,3,5,7,9 ) = , =
v=44.72m/s. from v=fλ λ= = =0.4472m. = etc. path length difference of a wave is
y=0.012 ( ) given by ∆L= = distance from 1st
y=0.012 ( ). We used (+) object, = distance from 2nd object. Path
sign because it was stated that it was difference of constructive interference is zero
travelling in decreasing direction of X and integral multiple of the wavelength e.g.
4.11 what is the difference between speed ∆L=0,1,2,3,4 etc. and ∆L=0,λ,2λ,3λ,4λ etc. but
of longitudinal wave in air at C and their λ= , hence ∆L can also be ∆L=0, ,
speed at C? etc. The path difference of a destructive
SOLUTION interference is an odd number of half the
From = , =17 273=290K, =57 273, wavelength given by ∆L=1,3,5,7 etc. but λ= ,
=330K. =340m/s (340 Is a constant hence ∆L can also be ∆L= , etc. Note
value for speed of wave in air). =?
√ that the path length difference can also be
= , = =362.69m/s. they gotten using Pythagoras theorem Formula for
√ √ √
asked for difference. =362.69 340 beats frequency is given by = , where
difference =22.7m/s. =frequency of one (e.g. speaker),
4.12 The elastic limit of a piece of steel wire =frequency of two (e.g. speaker).
By Kaymath (call 08068552755) 9

beat period is given by = , = beat function of position=Find the speed of

propagation of a transverse wave in the string
frequency . Doppler effect formulas are
SOLUTION
given by case(i) when the source is moving
from y=2Asinkxcoswt, adjacent antinodes are
towards a stationery observer, frequency
12cm apart that means the node is between
heard by observer is given by =( )
the antinode An n An n=12cm, from n= =,
=frequency of source , =speed of source
λ=12x2=24cm. from k= = = . A=2.5cm,
V=speed of sound. Case (ii) when the source
is moving away from the observer T=0.5sec , F= = =2Hz (a) y=5sin
frequency heard by observer is given by Same as y=5 . (b) from v=fλ ,
=( ) Case (iii) When the observer is λ=0.24m (converted to m) ,V=2X0.24=0.48m/s.
moving towards a stationery source 5.3 A person stands between two
frequency heard by observer is given by loudspeakers driven by an identical source .
=( ) where =speed of observer. Each speaker produces a tone with a
frequency of 155Hz on a day when the speed
Case (iv) when observer is moving away of sound is 341m/s. The person is 1.65m from
from a stationery source frequency heard one speaker and 4.95m from the other
by observer is given by =( ) speaker. What is the path length difference
SOLUTION TO EXERCISE 5 produced
5.1 Standing waves on a wire of length 4m SOLUTION
described by y=( )coswt with F=155Hz ,V=341m/s, λ= = =2.2m. from
=3cm, w=628rad/s. k=1.25πrad/m and
∆L= , =1.65m , =4.95m,
with the left end of the wire at x=0. At what
∆L= =3.3m. using and testing
distance from the left end are (a) The nodes
of the standing wave? (b) The antinodes of formulas from ∆L= = =3.3m. it gave us the
the standing wave. ∆L as 3.3m. It is a destructive interference
( )
SOLUTION answer is ∆L= = .
K=1.25πrad/m, from k= ,1.25π= ,λ= 5.4 Two point loudspeaker are a certain
λ=1.6m, n= = = 0.8m. (n=0,1,2,3,4 etc. we distance apart and a person stand 12m in front
of one of them on a line perpendicular to the
were simply asking to state the integers of
baseline of the speakers. If the speakers emit
nodes here). (b) = = =0.4m (n= identical 1000Hz tones, what is their minimum
1,3,5,7,9) non-zero separation so the observer hears
5.2 Adjacent antinodes of a standing wave little or no sound? (take speed of sound as
on a string are 12cm apart. A particle at an exactly 340m/s.
antinodes oscillates in simple harmonic SOLUTION
motion with amplitude 2.5cm and period ( ) (B) , = speaker one ,
0.5s. The string lies along the πx axis and is (A) =speaker two . f=1000Hz. They
fixed at X=0 (a) find the equation giving the are simply asking for the path
displacement of a point on the string as a 12cm (C) difference given by pythagoras
By Kaymath (call 08068552755) 10

theorem . we find λ, from λ= , v=340m/s from =( ) =500Hz, =30m/s,

λ= = =0.34m. B=( )=12.17m. v=340m/s. =( ) =544Hz.
∆L=√ √ 2.03m. 5.9 Two loudspeakers are placed side by side
5.5 What are the possible frequencies of a and driven by the same frequency of 500Hz. If
player’s note if a first note is produced the distance from a person to one of the
simultaneously by a first player is exactly speakers is 5m and the person detects little or
440Hz and 2.6 beats per second are heard. no sound, what is the possible distance from
SOLUTION the person to the other speakers? (the speed
=2.5Hz from = , we assume of sound is 340m/s)
=440Hz, we solve separately SOLUTION
2.6= , =437.4Hz and assuming possible distances are 5+ , 5+ , 5+ , 5+ .
again =440Hz, 2.6= , =442.6Hz.
but 5+ gives us best result v=340m/s,
possible frequencies will be 437.4Hz and
442.6Hz. f=500Hz from v=fλ, λ= =0.68m,
5.6 Two tones have frequencies of 300Hz 5+ = 8.06m.
and 298Hz. What is the beat period?.
SOLUTION 5.10 Two waves =A ( ),
=300Hz, =298Hz from = =A ( ). Are travelling in
= =2Hz. From = =0.5secs. opposite directions. Find the amplitude of the
5.7 The frequency of a train horn is 500Hz resulting stationary wave.
assume the speed in air in 340m/s. What is SOLUTION
the frequency heard by the observer if (a) In the equation amplitude can be given by
The observer is moving away from the A for one of them . For left and right
stationary train with a speed of 30m/s? we have two amplitude so we now
(b) The train is approaching the stationary
observer with a speed of 30m/s? have Y=2A
SOLUTION NORMAL MODES
(a) from =( ) =500Hz, v=340m/s, (CHAPTER 6)
=30m/s, =?, =( ) =456Hz. Fundamental frequency for a string and for a
(b) from =( ) , =30m/s, v=340m/s pipe open at both ends is given by =
can also be called zero overtone and
=( ) =548Hz.
their length is given by L= , λ=wavelength,
5.8 A sound of source has a frequency of
l=length, v=speed. First harmonics(zero
500Hz if a listener moves at the speed of
overtone) is same as the formula for the
30m/s toward the source what is the
fundamental frequency .2nd harmonics or first
frequency heard by the listener (the speed
of sound is 340m/s). overtone is=2xfundamental frequency (f= )
SOLUTION 3rd harmonics or 2nd overtone is f=
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i.e. (1st,2nd,3rd,4th,5th ) and so on SOLUTION

shortcut; when asked to calculate frequency m=0.008kg, L=1m, m of load=1.92kg g=10m/
of any harmonics . simply find the from T=mg=1.92x10=19.2N. from μ= =
fundamental frequency and multiply it by
the number of harmonics you are asked to μ=0.008kg It was a string so from = √
find e.g. 2x , 3x , 4x , 5x . Same
shortcut goes for length , you find ‘L’ from = √ =24.5Hz.
L= , and multiply by the number of 6.2 Calculate the frequency of the first
harmonics you are to find e.g. 2XL, 3XL, 4XL overtone of a stretched string of length 60cm ,
etc. you will understand better when you go if the velocity of sound produced of 330m/s
through the exercises. SOLUTION
For a closed pipe at one end the Frequency of first overtone is same as 2nd
fundamental frequency and the length is harmonics. Using the shortcut ,we find the
given by = and L= , It’s harmonics is an fundamental frequency and multiply by it by 2
odd number i.e. 1,3,5,7,9 etc. Note that the v=330m/s, l=0.6m from = = =275Hz.
first harmonics or zero overtone here is hence F=2X =2X275=550Hz.
same as the formula for the fundamental 6.3 A violin string with length of 5m between
frequency same shortcut goes for this one fixed points has a linear mass density of
too . find = and multiply by the 40g/m and a fundamental frequency of 20Hz
(a) calculate the tension in the string (b)
harmonics you are asked to find e.g. 1x ,
calculate the frequency and wavelength of the
3x , 5x , 7x etc. Note that here
second harmonic (c) calculate the frequency
1=first harmonics(zero overtone)
and wavelength of the second overtone .
3=2nd harmonics (first overtone)
SOLUTION
5= 3rd harmonics (2nd overtone) and so on.
F=20Hz , L=5m, μ=0.04kg/m(from g to kg)
End correction for an open pipe is given by
= , =fundamental frequency (a) we find T from = √ , 20= √
( )
l= length of pipe , c=distance from open end. = , T=400X100X0.04=1600N. (b)
End corrections for a closed pipe is given by nd
frequency of 2 harmonics is F=2X20=40Hz.
= . Natural frequency of a taut string
( ) from L= , we find λ .we multiply equation by 2
nd
is given by = √ ,μ=mass per unit length then solve for λ because it in 2 harmonics
5= , λ=5m. (c) 2nd overtone is same as 3rd
μ= , T=tension, L=length .
SOLUTON TO EXERCISE 6 harmonics , F=3X =3X20=60Hz. , from L= we
6.1 A string of mass 8g and length 1m is multiply 3 and find λ because it’s a 3rd
fixed at both ends. If the string is stretched harmonics 5= , λ= = 3.3m.
by a load of 1.92kg and then released, find 6.4 Middle C string on a guitar has a
the fundamental frequency of the stationary fundamental frequency of 200Hz, and the first
wave produced (g=10m/ ). A string above the middle C string has a
By Kaymath (call 08068552755) 12

frequency of 350Hz .if the strings linear

= √ = √ =10Hz. We multiply
mass densities are equal, but the length of
the A string is only 64 percent of the length answer by 1,2,3,4 ,
of the C string , what is the ratio of their 2nd harmonics F=2X10=20Hz, 3rd harmonics
tensions? In other words find : . Where F=3x10=30Hz, 4th harmonics F=4X10=40Hz.
and string is the tension in string C 6.7 The longest pipe found in most medium –
SOLUTION sized pipes is 16ft long. What is the frequency
we are to find the ratio of from the of the note corresponding to the fundamental
node if the pipe is open at both ends? (take
formula = √ , =350Hz , =200Hz , L speed of sound in air= 340m/s)
SOLUTION
of C=L , length of A is 64% of C , μ is constant
1m―>3.28feet 16=3.28X
for A and C hence length of A= XL=0.64L. X ―>16feet X=4.878m
= √ , 350= √ , v=345m/s, =? = = =35.36Hz.
6.8 An organ pipe has a length of 0.75m.what
=122500X4X0.4096 μ , = √
would be the length of a pipe closed at one
end whose third harmonic is the same as the
200= √ , =40000X4X Xμ
fundamental frequency of the open pipe ?
= =1.2544. SOLUTION
its an open pipe so from = , L=0.75m ‘’ ’’
6.5 will a standing wave be formed in a 4m
length stretched string that transmits waves is same for both and 2nd pipe is closed at one
at a speed of 12m/s if it is driven at a end in the 3rd harmonics hence from =
frequency (a) 15Hz or (b) 20Hz? Give = , = , L= =1.1m
reasons for your answers.
6.10 A pipe of length 57cm has a fundamental
SOLUTION
frequency of 224Hz when open at both ends .
We find ‘’ ’’ to know the harmonics
If the displacement antinodes occur at a
v=12m/s , L=4m, from f= = =1.5Hz. but distance of 10cm from the open ends .
F=10X =10X1.5=15Hz. It’s a harmonics 10th calculate the velocity of sound.
harmonics (b) no matter how we multiply SOLUTION
1,2,3,4,5,6 etc. by .5Hz it can’t give 20Hz . it L=0.57m(converted to m), F=224Hz , C=0.1m
is not an harmonics it was an open pipe. From = ,
6.6 find the first four harmonics for a string ( )

that is 2m long, and has a linear mass 224= , V=224x2x0.77=345m/s.

( )
density of 2.5x kg/m, and is under 6.11 An open pipe 30cm long a closed pipe
tension of 40N. 23cm long, both of the same diameters, have
SOLUTION the same frequencies when each of other of
T=40N ,μ=2.5x kg/m, L=2m using the them is sounding it’s first overtone. What is
shortcut we find of the string using the end-correction of these pipes?
SOLUTION
By Kaymath (call 08068552755) 13

We look for the End correction of the closed is critical. Any time you think critical angle
and open pipes as frequency is the same think about total internal reflection. Total
,note that both are in first overtone internal reflection is given by n= where
open pipes ―> 2nd harmonics ―2
=critical angle .
closed pipes―>2nd harmonics ―3 . first
EXAMPLE 7.2: a beam of light is incidental on a
overtone is same as 2nd harmonics .we have
plane mirror at an angle of relative to the
= , = , normal. What is the angle between the
( ) ( ) ( ) ( )
4(23+C)=3(30+2C), 92+4C=90+6C, C=1m. reflected rays and the surface
OPTICS (CHAPTER 7) SOLUTION
since = then = but
I=incident ray ( ) , R=reflected ray ( )
Ө this is what we find. and Ө are (it’s a
formula for refraction from Snell’s law is
right angle . 32+Ө= ,Ө= .
given by = =refractive
EXAMPLE 7.3: Two plane upright mirrors touch
index of ‘’a’’ medium , =angle of ‘’a’’
along one edge where their planes make an
=refractive index of ‘’b’’ =angle of ‘’b’’
angle of . If a beam of light is directed onto
note this formula is very b important in
one of the mirrors at an angle of incidence of
refraction . let’s do a small review on
and is reflected onto the other mirror,
refraction
what will be the angle of reflection of the
normal when ray of light strikes glass
beam from the second mirror.
block at the ( ) it is
SOLUTION
medium(a) reflected through ( ) the
if b is the reflected ray angle of a
medium (b) refracted ray transmitted ray from
c d b= , but to find c, b+c=
medium (a) changes direction as it enters
a b , c= . but we are to find the
another medium (i.e.) medium ‘b’ This is
called refraction . note that angle of (reflected ray) . + +d=
incidence equals angle of refraction . d= = , but d+ = ,
+ = , = .
Refracted index is also given by n= ,
SOLUTION TO EXERCISE 7
c=speed of light in that material (it’s a 7.1 An optical fiber is made of clear plastic
constant given by 3x m/s. note also this with index of refraction n=1.50. What is the
formula = =speed of medium 1 minimum angle of incidence so that the total
=speed of medium 2 , =wavelength of internal reflection can occur?
1 , =wavelength of 2, refractive index is SOLUTION
n=1.5 , =? , from n= , = ,
also given by = here =refracted
=
angle . At critical angle refracted angle is
7.2 Two plane mirror, and are placed
notice that the
critical angle  together with edges touching each other at
refracted angle ( ) incident angle is angle ⍺. If the light ray is incidental on at
the critical angle we are talking about. When angle what is the angle of reflection from
it is critical the refracted angle becomes the second mirror (take ⍺= ).
Total internal reflection occurs when angle SOLUTION
By Kaymath (call 08068552755) 14

from since = ,b= and = , = ,

b+c= . +c= , c= . = .
+ +d= , d= 7.7 The speed of light in water is 75% of the
a d but d+ = . + = speed of light in a vacuum . What is the value
b = of it’s refractive index?
7.4 What is the critical angle of light passing SOLUTION
from a material of index of refraction 1.54
from n= , = = . n= =1.33
to a material of index of refraction n=1.33
SOLUTION 7.8 What is the refractive index of a material if
=1.54 , =1.33 , =?, from n= , we the critical angle of light passing from the
material to air is .
a given two refraction hence we use SOLUTION
= , = , = = . = , from n= = =2.42.
7.5 A glass plate of thickness 0.6cm has a 7.10 At what angle to the surface must a diver
refractive index of 1.55, calculate the time submerged in a lake look towards the surface
taken for the ray of light to pass through it. to see the setting sun?
SOLUTION SOLUTION
n=1.55, c=3x m/s (constant) from n= for the diver to see the setting sun it has to be
at critical angle . note refractive index of water
= = m/s. from v=
is 1.33(constant) n= , = = .
distance=0.006m. time=
time= . REFLECTION AT PLANE AND
7.6 A layer of oil (n=1.45) float on water CURVED SURFACES (CHAPTER 8)
(n=1.33). A ray of light shines onto the oil CHARACTERISTICS OF MIRROR
with an incident angle of .Find the angle AND LENSES( THE SHORTCUT)
the ray makes in the water. i. focal length (F) and radius of curvature (R) is
SOLUTION positive for a concave mirror (+F and +R) while
we have three medium here, air, oil and it is negative for a convex mirror (-F and –R)
water air we find the refracted ii. image distance (V) is positive (+V) when
oil
angle of oil, then image is formed in front of mirror and ‘’+V’’
water
refracted angle of water from means REAL IMAGE while V is negative (-V)
= =1 (note that it is when image is formed at the back of the
constant for air) , =1.45, = , =? mirror (-V means a VIRTUAL IMAGE )
iii. when magnification (M) is positive (+M)
= , = ,
image is UPRIGHT OR ERECT but when it is
= .( this is the refractive angle of oil) negative (-M) IMAGE IS INVERTED
we now look for the refractive angle of iv. when M<1, IMAGE IS DIMINISHED OR
water taking = as incident angle of REDUCED but when M>1 IMAGE IS ENLARGED
oil. =1.45, =1.33, = , =? OR MAGNIFIED and when M=1 IS A REAL
IMAGE
By Kaymath (call 08068552755) 15

MIRROR FORMULAS ARE GIVEN BELOW F= = = 22.5cm , U=15cm, from V=

image distance is given by V= where ( )
V= = 9cm. M= = =0.6
( )
f=focal length , u=object distance , v=image
8.4 What is the focal length of a convex
distance . object distance is given U= , spherical mirror which produces an image
focal length is given by F= ,magnification one sixth the six of an object located 12cm
is given by M= , focal length is also given from the mirrors
SOLUTION
by F= where R=radius of curvature. M= , U=12cm , we are to find the focal length
length of image is given ∆V=
from M= , = , V= cm , from F=
SOLUTION TO EXERCISE 8
8.1 Describe the image of a candle flame F= = 2.4cm.
( )
located 40cm from a concave spherical 8.5 A mirror forms an erect image 30cm from
mirror of radius 64cm. the object and twice it’s height (a) where must
SOLUTION the mirror be situated (b) what is it’s radius of
R=64cm , by F= = =32cm. U=40cm ,we find curvature.
V and M to know the description of the SOLUTION
image from V= = =160cm. V is (+) (a) M=2 (because twice was stated) ,let U=x ,
hence IMAGE IS REAL, we find M from let V=X 30 , from M= ,2= , 2X= X+30,
M= = =-4 . m is negative hence IMAGE X=10cm. (b) V will be V=10 30= 20cm. we
IS INVERTED ANSWER IS REAL AND find focal length before getting ‘R’ from F=
INVERTED F= =20cm, from F= , R=2X20=40cm.
( )
8.2 A rod 10m long is placed along the 8.6 A convex mirror has a radius of curvature
principal axis of a convex mirror of focal of 0.55m. calculate the position of the image
length 4cm, if the side nearer the mirror is of a man 10cm from the mirror.
6cm from it, find the length of the image. SOLUTION
SOLUTION we are to find V, R= m(convex mirror)
It was a convex mirror hence F(-) ,F=-4cm
from F= = = 0.275m.U=10cm from V=
, =6cm, =10cm, V=? ,we find the length
from ∆V= but = = V= = 0.267m.
( ) ( )
2.4cm. we add and and find f 8.7 An object is placed 2.0cm in front of a
, U=10+6=16cm. = = 3.2cm. concave mirror whose radius of curvature is
( ) 8.0cm ,find the position of the image, size and
∆V = ( )=0.8m. it’s orientation.
8.3 An object 7cm high is placed 15cm from SOLUTION
a convex spherical mirror of radius 45cm . To find the position ,size and orientation we
Describe it’s image and give the value for find V and M .U=2cm , R=8cm from
the image distance V and magnification M.
F= = =4cm. (F IS (+) IT’S A CONCAVE MIRROR )
SOLUTION
R= 45cm (convex mirror) from
By Kaymath (call 08068552755) 16

from V= = = 4cm (V IS (-) IMAGE IS depth of object in three media is given by

VIRTUAL ,ERECT AND FORMED BEHIND(OR = + + where , , are real depth in
( ) each medium where =total displacement
BACK) OF MIRROR), from M= = =2cm. of (1) e.g. oil, (2) e.g. water,(3) e.g. glass.
(M>1 ,IMAGE IS ENLARGED (OR MAGNIFIED)
, , are given by = ( ),
so image is 4CM BEHIND THE MIRROR,
VIRTUAL, ERECT & ENLARGED(MAGNIFIED) = ( ) and = ( ) .Refraction
8.8 A child looked at a reflecting Christmas ( )
tree that has a diameter of 9.0cm and sees through prism is given by n= ,
an image of her face that is half the real size.
=minimum deviation , A=refractive index
How far is the child’s face from the ball.
angle or apex angle. Deviation of blue and red
SOLUTION
light of small angular prism are given by
diameter= 9cm, radius in math is given by
=( 1)A , =( 1)A . angular
R= = =4.5cm. from F= = =2.23cm. deviation is given by ∆d=( )A
8.9 In a particular store truck mirror , there PAST QUESTION 2014/2015
is a warning object in the mirror appear to Find the displacement of the bottom of a
be more closer than they appear ,what kind swimming pool if the apparent depth is 1.2m
of mirror must that be and why? ( =1.33).
SOLUTION SOLUTION
IT’S A CONVEX MIRROR. from A=T D and from n= , n=1.33 , A=1.2 ,
REFRACTION THROUGH T=nxA=1.33X1.2=1.596. ,Displacement=?
PLANE SURFACES (CHAPTER 9) D=T A=1.596 1.2=0.396m.
Make sure you memorize characteristics of PAST QUESTION 2014/2015
mirror and lenses in page 14 (very A certain glass prism has a refractive index of
important for your exam) . Refractive index 1.61 for red light and 1.66 for violet light if
‘’n’’ for object in a liquid (e.g. in water) is both colours pass through symmetrically and if
( ) the apex angle is .Find the difference
given by n= = ,Apparent
( ) between the angles of minimum deviation of
depth of an object is given by A=T D , the two colours.
where D=displacement of an object in water SOLUTION
,displacement of object given by We are asked to find the change in minimum
D= ( ) where T=Real depth deviation from ∆ =
A= , =1.66 , =1.61 ,from
, =refractive index of object or substance
.Determination of displacement of an object n= ( )
, for red light 1.61=
( )

in three media e.g. in oil ( ), water ( ),

and glass ( ) is given by = + + 1.61x0.5= = ,
where , , are displacement in each ( )
medium where =total displacement of of red= . for violet 1.66= ,
(1) e.g. oil, (2) e.g. water,(3) e.g. glass. Real
1.66x0.5= =
By Kaymath (call 08068552755) 17

of violet= ∆ = , vertically above is .

∆ = = . SOLUTION
SOLUTION TO EXERCISE 9 We find the refractive index of both first,
If the thickness and refractive index of oil T=20m , A=14cm, from = = =1.42857, we
,water and glass together in a set up are
respectively 4cm,6cm,5cm and 1.26,1.33, . we find the incident ray( ) from n= , ,
Find the value of if the apparent position = , 1.42857= , =0.588x1.42857
of an object at the bottom is 12cm. = = .
SOLUTION Note that in this material is strictly exam
=4cm , =6cm, =5cm, =1.26, =1.33 focus
( )=? .we calculate the value of
REFRACTION THROUGH
before finding from = ( )
CURVED SURFACES (LENSES)
= ( )=0.8254, , = ( )
(CHAPTER 10)
= ( )=1.4887, from A=T D, we Power of lens is given by P= (in metres ‘’m’’)
find D , 12=(4+6+5) D,D=3. From also given by P= (in ‘’cm’’ or dioptre ‘’D’’)
= + + ,3=0.8254+1.4857+ ,
=3 0.8254 1.4857=0.6859. from where F=focal length . Lens maker equation is
= ( ), 0.6859= ( ) given by =(n 1) ( ), =Radius of

0.6859 =5 5 , =1.16. curvature of 1st lens , =Radius of curvature

9.3 The difference between the refractive of 2nd lens ,n=refractive index . For concave
indices of carbon bisulfide for blue and red mirror Note that is negative for ( ) and
light is 0.48 while the critical angle for red is positive ( ) but for convex mirror
light at carbon bisulfide air interface is . is positive for ( ) and is negative
Calculate the critical angle range for carbon ( ) please take note . Focal length of two
sulfide for blue light. lenses in contact is given by = ,
SOLUTION =Focal length of 1st lens , =Focal length of
Difference of two refractive index that of 2nd lens
blue and red will be =0.48 and the PAST QUESTIONS 2012/2013(Q 26)
critical angle( ) of = ,from = The near point of a certain hyperopic eye is
,
100cm in front of the eye .What would be the
= =1.3250 , =0.48 power of the lens that would permit the
, =1.805 . Angle which blue will be critical wearer to see clearly an object that is 25cm in
will be given by = ,1.805 = , front of an eye
= . SOLUTION
9.4 An object at the bottom of a pool 20m For a hyperopic eye image is formed behind
deep was observed to be at a 14cm position. the retina , hyperopic eye means
Find the angles of incidence of the object ,if longsightness problem , for an image behind
the angle of refraction of an observer ‘’V’’ will be negative . check lens laws in page
By Kaymath (call 08068552755) 18

14. V= 100cm, U=25cm, we find focal iv. The difference between the microscope
length first from F= = =33.33. P and telescope is that telescope is used to view
( )
objects of large distance while microscope is
in diopters from P= = = +3D. used to view small objects at close distance.
PAST QUESTION 2012/2013(Q47) Length of a terrestrial telescope is given by
Suppose the absolute values of the radii of L= Where =focal length of
curvature of a double convex lens are both objective lens, =focal length of middle
equal to 10cm and the refractive index of lens, =focal length of eye piece
the glass is 1.52 , what is the focal length of PAST QUESTION 2014/2015(Q 12)
the lens What is the length of a terrestrial telescope
SOLUTION made of and objective lens of focal length
From =(n 1) ( ) ,it was a convex 0.5m , a middle lens of focal length 4cm and
eye piece of focal length of 6cm?
lens hence = 10cm & = 10cm ,
SOLUTION
n=1.52 =(1.52 1) ( ) , =0.104 =50cm(converted to cm) =4cm, =6cm
( )
F= = 9.6cm. From L= =50 ( ) =72cm.
PAST QUESTION 2016/2017 (Q 34) DISPERSION AND ABBERATION
Two thin lenses of focal lengths 9cm and (CHAPTER 12)
6cm are placed in contact . Calculate the
i. The separation of white light into it’s
focal length of the combination?
component colours is called DISPERSION
SOLUTION
ii. The spectrum of visible light is called
, =9cm , = 6cm, From = ‘ROYGBIV’ contains and stands for the
= , = = , F= = 18cm. red,orange,yellow,green,blue,indigo and violet
iii. ‘ROYGBIV’ is in the order of decreasing
EXERCISE 10.5
wavelength, decreasing speed and increasing
Two thin lenses of focal length of focal
angle of deviation (REFRACTION) .
length 12cm and 30cm are in contact
PAST QUESTION 2012/2013(Q 22)
compute the focal length and power of the
A Ray of white light incident upon a glass prism
combination
is dispersed into it’s various colour
= 30cm , = 12cm, From = components, which one of the following
= , = = , F= =20cm colour experiences the least refraction (a)
violet (b) yellow (c) blue (d) green (e) none of
OPTICAL INSTRUMENTS the above
(CHAPTER 11) SOLUTION
i. The least distance of the eye is 25cm . This is a dispersion question. From ‘ROYGBIV’
ii. A simple microscope makes use of a deviation or refraction increases from red to
converging lens to further increase the violet but in question red wasn’t given but
apparent angular size of an object. yellow which is next was given. So yellow has
iii. A compound microscope consists of two the least refraction here.
converging lens. ANSWER IS B  YELLOW
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By Kaymath (call 08068552755) 19

THEORITICAL ASPECT OF point of suspension because the ___

ANSWER- ACCELERATION IS DIRECTLY
PHY113 (INCLUDES PROPORTIONAL TO THE DISPLACEMENT. Note
QUESTION, ANSWERS AND that the acceleration and restoring force are
directly proportional to the displacement.
EXPLANATIONS) 7. Which of the following is NOT a mechanical
Some students think that the theory aspect wave? (a) Waves propagated in stretched
of physics in exams are not too important string (b) waves in closed pipes (c) radio waves
but the truth is that in uniben physics exam (d) water waves (e) sound waves
you will be given at least 10 theory question ANSWER IS C (RADIO WAVES)
out of 50 question , the remaining 40 will be note that all waves in the electromagnetic
calculations. If you are able to get up to 10 spectrum are not mechanical and radio wave
question correctly in theory and able to get is part of the spectrum (RIVUXG Radio
25 question out of 40 in calculations , that is waves, infrared ray, visible light, ultraviolet
an ‘A’ already because you need to get at ray, x-ray and gamma ray.)
least 35 correctly to get an ‘A’ , just make 8. Which of the following is NOT an example of
sure you take the theory aspect serious as longitudinal wave? (a) electromagnetic waves
well as the calculations if you want an easy (b) sound waves
‘A’.In this material I was able to put together 9. A transverse wave and a longitudinal wave
104 theory questions, Let’s begin. travelling in the same direction in a medium
1. In simple harmonic motion period differ essentially in their?
depends on? ANSWER- DIRECTION OF VIBRATION OF THE
ANSWER-MASS OF OBJECT PARTICLES OF THE MEDIUM.
2. The product of the period and frequency 10. The fundamental property of a
of a harmonic oscillator is always equal to? propagating wave which depends only on the
ANSWER- 1 source and not the medium of propagation is
3. The amplitude of a simple harmonic the ?
oscillator may be defined as? ANSWER- FREQUENCY. Two properties are
ANSWER-THE MAXIMUM DISPLACEMENT affected by the medium of propagation of a
4. The period of oscillation of a simple wave: VELOCITY AND THE WAVELENGTH. The
pendulum is independent of ? frequency of the wave depends solely on the
ANSWER-MASS OF THE BOB. The mass of source that generates the wave.
the pendulum doesn’t affect it’s period. 11. Which of the following statements about
5. Forced oscillation is when an external wave is/are correct ?
force maintains vibrating system. At I. A wavefront is a line which contains all
resonance ,the amplitude of the vibrating particles whose vibrations are in phase
body II. The direction of propagation of a wave is
ANSWER-MAXIMIZES. Note that amplitude the line drawn parallel to the wavefront
is maximum at resonance. III. A wavefront is a circle which is common to
6. If freely suspended object is pulled to one all particles that are to be in the same state of
side and released , it oscillates about the disturbances
By Kaymath (call 08068552755) 20

(a) I only (b) II only (c) III only (d) I and II only
progressive mechanical wave is correct?
(e) I and III only. (a) it can be plane polarized (b) it’s energy is
ANSWER- ANSWER IS E (I AND III) localized at specific points of it’s profile
A wavefront is a line or surface that joins (c) it does not require a material medium for
all particles in phase or in same state of it’s propagation (d) it’s frequency is constant
disturbance in a travelling wave .NOTE as it travels between different media.
THAT WAVEFRONT DESCRIBES THE STATE ANSWER- D. Mechanical waves cannot be
OF VIBRATING PARTICLES IN PHASE NOT polarized .It is the electromagnetic waves
THEB WAVE ITSELF that can be plane polarized . Travelling waves
12. A A A transfers energy. It’s energy is therefore, not
N N N N localized to a point. Mechanical waves
require a material medium for propagation.
from the diagram above , the type of wave Wave frequency depends only on source of
obtained is? . ANSWER- STANDING WAVE. wave thus, it’s remains constant through
This diagram shows an incident and different media. However, it’s wave and
reflected wave that superimposed to form velocity changes with medium.
a standing wave . Even though the waves 16. Which of the following characteristics of a
are transverse, The diagram illustrates a wave is used in measuring the depth of sea?
standing wave. (a) diffusion (b) interference (c) refraction (d)
13. If sound wave goes from a cold air reflection
region to a hot air region , it’s wavelength ANSWER- D . Echoes, which can be used in
will?. determining the depth of sea , occur due to
ANSWER- DECREASE. Speed of sound in air reflection of sound.
increases with increase in temperature . 17. Which of the following properties is/are
Thus , when sound travels from cold air to common to all waves? I. diffraction II.
hot air medium (temperature of air refraction III. Reflection
increased ), it’s speed increases accordingly ANSWER- I,II,III (THE THREE ARE COMMON
while it’s wavelength decreases. TO ALL WAVES).
14. The property that is propagated in a 18. A wave that travels through stretched
travelling wave is? strings is known as ?
ANSWER- ENERGY. Travelling waves ANSWER- MECHANICAL WAVE. It requires a
transfers energy. material medium.
15. Which of the following about a 19. Ripples on water and light waves are
progressive mechanical wave is correct? similar because both ?
(a) it can be plane polarized (b) it’s energy is ANSWER- CAN BE REFRACTED AND
localized at specific points of it’s profile DIFFRACTED.
(c) it does not require a material medium for 20. Which of the following is characteristic of
it’s propagation stationary wave? (a) the antinode is a point of
ANSWER- ENERGY. Travelling waves minimum displacement (b) the distance
transfers energy. between two successive nodes is one
15. Which of the following about a wavelength (c) they are formed by two
By Kaymath (call 08068552755) 21

identical waves travelling in opposite and III. ANSWER-A.

direction(d) they can be transverse or 24. Two boys communicating with each other
longitudinal . ANSWER- C. by stretching a string passing through a hole
21. A wave disturbance travelling in air punched at the bottom of each of two
enters a medium in which it’s velocity is less tin cans. The physical principle employed is
than that in air . Which of the following is that sound travels?
true about a wave in the medium? (a) both ANSWER- WITH GREATER EASE THROUGH A
the frequency and wavelength of the wave STRING THAN IN AIR. Speed of sound
are decreased (b) the frequency of the increases with density , thus sound travels
wave is unaltered while the wavelength is faster in string than in air.
increased (c) the frequency of the wave is 25.The difference between sound waves and
unaltered while the wavelength is decreased light waves is that sound waves_____
(d) the frequency of the wave is decreased ANSWER- REQUIRE A MATERIAL MEDIUM TO
while the wavelength is unaltered TRAVEL WHILE LIGHT WAVES DO NOT. Sound
ANSWER- B . This wave was refracted. A waves are mechanical while light waves are
decrease in velocity is accompanied by an electromagnetic.
increase in wavelength and unaltered 26.During a thunderstorm , the sound is heard
frequency. over a long time. This phenomenon is referred
22. From the statements below ,the to as____
conditions for two waves to interfere are ANSWER- REVERBERATION. Reverberation is
I. They should be identical II. They should a multiple reflection of sound waves causing
originate from the same source III. They it’s persistence over a long time.
should be coherent IV. They should be 27. The physical properties of sound waves can
monochromatic . (a) I,II and IV (b) I,II, and III be described by ____
only (c) I,III and IV only (d) II,III and IV only ANSWER- REFLECTION AND DIFFRACTION.
ANSWER- C . For two waves to interfere Sound waves are longitudinal ,thus they cant
constructively , they should be identical be polarized . polarization is exclusive to
and have same frequency , However they transverse waves.
may also have the same wavelength or be 28. Metal cables are used as telephones wires
different by an integral number of because ____
wavelengths. Coherent and ANSWER- THE SPEED OF SOUND IN THEM IS
monochromatic describes electromagnetic VERY HIGH. Speed of sound is highest in
waves that have the same wavelength and solids.
fixed phase relationship. Monochromatic 29. Which of the following affect the velocity
describe lights of same colour and of sound? (a) an increase in the pitch of sound
wavelength. (b) an increase in the loudness of sound (c)
23. Which of the conditions below are wind travelling in the same direction of sound
necessary to produce interference fringes ? (d) a change in the atmospheric pressure at
I. coherence II. Same frequency III. Same constant temperature.
wavelength IV. same intensity ANSWER- C .
(a) I,II and III (b) I and II (c) I,II and IV (d) II 30. Which of the following factor affects speed
By Kaymath (call 08068552755) 22

of sound in air? inversely proportional to length . Thus to

I. temperature II. pressure III. Frequency restore the original frequency, we decrease
(a) I only (b)II only (c)I and II only (d) I and the length of wire or it’s unit mass per unit
III only . ANSWER- A. length
31. The speed of sound travelling in the 35. When the bottom of a turning fork is held
media in contact with a wooden box , a louder sound
listed below increases in the order. is heard , this is phenomenon is known as ?
(a) air, iron bar, water (b) air, water, iron ANSWER- RESONANCE.
bar (c) iron bar, water , air (d) water, air, 36. A note is called an octave of another if___
iron bar (e) water, iron bar, air. ANSWER- IT’S FREQUENCY IS TWICE THAT OF
ANSWER- B. THE FIRST.
32. If a source of sound is moving, a 37. The characteristics of vibration that
stationary listener will hear a sound of determines it’s intensity is (a) an increase in
different frequency , this is called? the pitch of the sound (b) an increase in the
ANSWER- DOPPLER EFFECT. Doppler effect loudness of the sound (c) wind travelling in
is an alteration in the frequency of sound the same direction of the sound (d) a change
heard by the listener when there is a in the atmospheric pressure at constant
relative motion between listener and temperature
source of sound. ANSWER- B . Amplitude of the vibrating
33. If a sound wave goes from a cold-air medium would have been more ideal but in
region to a hot air region , it’s wavelength the absence , loudness is the nearest.
will ____ ANSWER- DECREASE .Going from 38. The characteristics of vibration that
cold-air (denser) to hot-air (less-dense) , determines it’s intensity is____
the speed of a wave increases accompanied ANSWER- AMPLITUDE. Intensity depends on
by a decrease in wavelength also cold-air to amplitude of sound source.
hot-air means an increase in temperature, 39. Musical instruments playing in the same
which increases the speed of sound. note can be distinguished from one another
34. If the load at the end of a sonometer is owing to the difference in their____
immersed in a bucket of water, the original ANSWER- QUALITY . quality depends on
fundamental frequency of the wire could be harmonics and overtone.
restored by ____ 40. The characteristic which differentiate a
ANSWER- DECREASING THE LENGTH OF THE high note from a low note is_____
WIRE. The fundamental frequency is given ANSWER- PITCH
41. The pitch of an acoustic device can be
by f= √ because sonometer is like a
increased by _____
string , immersing a load in water reduces ANSWER- INCREASING TH FREQUENCY
it’s weight and hence reduces the tension REFLECTION AT PLANE AND CURVED
in string . Being that frequency is directly SURFACES, REFRACTION THROUGH PLANE
proportional to tension, this reduces the SURFACES, REFRACTION THROUGH CURVED
frequency also. However , frequency is SURFACES (LENSES), OPTICALINSTRUEMENTS,
DISPERSION AND ABBERATION
By Kaymath (call 08068552755) 23

42. Light is considered as a transverse wave Diffraction is due to spreading out of light
because it travels_____ rays as they pass through an aperture ,such
ANSWER- IN A DIRECTION PERPENDICULAR as the pin-hole camera.
TO THE PLANE CONTAINING THE ELECTRIC 48. Shadows and eclipses result from the
AND MAGNETIC FIELD. Light is a transverse ANSWER- RECTILINEAR PROPAGATION OF
wave because it moves a direction LIGHT. They result from rectilinear
perpendicular to the direction of the propagation of light.
electric and magnetic field. It’s doesn’t 49. The eclipse of the sun occurs when the
have a material medium for propagation ___
hence it is an electromagnetic wave. ANSWER- MOON IS BETWEEN THE SUN AND
43. Which of the following characteristic of THE EARTH. Eclipse literally means
light wave determines it’s colour? obstruction . Eclipse of the sun means the sun
(a) velocity (b) wavelength (c) amplitude (d) is not seen from the earth , if so, then the
intensity. ANSWER- B . Wavelength moon lies between sun and earth blocking
determines colour of light. Each colour of light from the sun from getting to the earth.
spectrum of white light (ROYGBIV) 50. Which of this below is a phenomenon of
represents one wavelength. total solar eclipse? I. total internal reflection
44. non-luminous object can be seen of light II. conservative of light energy III.
because they ___ Relative motion of the earth, sun and moon
ANSWER- EMIT LIGHT. They reflect light IV. rectilinear propagation of light. (a) I and II
into the eye and thus they are seen. only (b) II and IV only (c) I and III only (d) III
45. Which of the following is a non- and IV only . ANSWER- D. The two
luminous body ? (a) candle flame (b) lit bulb phenomena explain eclipse.
(c) moon (d) star. 51. light travelling through a small pinhole
ANSWER- C. The moon has no light on it’s usually does not make a shadow with a
own (it’s non-luminous ) . it reflects the distinct sharp edge because of_____
sun. ANSWER- DIFFRACTION. Diffraction is the
46. The sharpness of the boundary of property of light, like all waves to bend round
shadow is determined by the ____ obstacles or spread out through gaps. It is the
ANSWER- RAYS OF LIGHT PASSING reason behind blurred images seen in pin
THROUGH THE OBJECT. The sharpness of holes that are not small enough.
the boundary is due to rectilinear 52. In daytime , it is possible to see under
propagation light i.e. because rays travel in shady areas such as under a tree because
straight lines . It has nothing to do with the light has undergone _____
intensity of light or nature of the object. It ANSWER- DIFFRACTION.
is all because the light rays move in straight 53. A man standing between two parallel
lines. mirrors in a barber’s shop will see the
47. Which of the following phenomenon is following number of his own image ____
NOT a direct consequence of rectilinear ANSWER- INFINITE. Two parallel mirrors
propagation of light? produce infinite number of images.
ANSWER- DIFFRACTION OF LIGHT. 54. The instrument used by designers to
By Kaymath (call 08068552755) 24

obtain different colour patterns is called principal focus to produce parallel beam of
ANSWER- KALEIDOSCOPE. light . It’s distance is focal or half the radius of
55. Which of the following optical curvature.
instruments does not depend on the use of 64. An object O lies at a distance m in front of
plane mirrors? a concave mirror of focal length f. if m ▪ f, then
ANSWER- SIMPLE MICROSCOPE. Simple the final image obtained will be ______
microscope make use of a single convex ANSWER- MAGNIFIED AND ERECT.
lens. 65. Convex mirrors are used as driving mirrors
56. The plane mirror in a kaleidoscope are because images formed are______
usually placed ANSWER- ERECT, VIRTUAL AND DIMINISHED
ANSWER- AT AN ANGLE OF . 66. In the microscope , the eyepiece lens
57. Which of the following statements merely acts as _____
is/are correct about the image formed by a ANSWER- A MAGNIFIER. While the object lens
palne mirror?. I. The magnification produces a magnified , the eye lens multiplies
produced is 1 II. The image distance is the this magnification producing a highly
same as object distance III. The image is magnified image. It’s acts as a magnifier.
real IV. the image is laterally inverted. (a) I 67. The following optical instruments make
only (b) II only (c) III only (d) I and III only use of lenses in their modes of operation
ANSWER- C. Plane mirror images are virtual EXCEPT a (a) camera (b) microscope (c)
images. periscope (d) projector (e) telescope
60. Which of the following is true for the ANSWER- C. Periscope make use of plane
image formed by a convex mirror?. I. The mirrors in simple periscopes and right-angled
image is always virtual II. the image is triangular prisms in prismatic periscopes .
always erect III. The image lies between They do not uses lenses.
focus and pole IV. the focal length is 68. The terrestrial telescope has one extra
negative. ANSWER- THEY ARE ALL lens more than astronomical telescope. The
PROPERTIES OF CONVEX MIRROR IMAGES. extra lens is for
61. Which of the following statements is ANSWER- ERECTION OF THE IMAGE.
FALSE about parabolic mirrors?. (a) They 69. In a compound microscope , the objective
are preferred in car headlamps (b) They and the eye piece focal length are
exhibit spherical aberration (c) They focus ANSWER- SHORT. In a compound microscope
both paraxial and marginal rays on the , objective and eye piece lenses are convex
principal focus (d) They are improved form lenses of short focal lengths. They are not the
of concave mirrors. same ;the objective’s focal length is shorter
ANSWER- B. Parabolic mirrors prevent than eye len’s focal length
spherical aberration. 70. An astronomical telescope is said to be in
63. In order to produce a parallel beam of normal adjustment when the_____
light , a lamp is placed at a distance from ANSWER- FINAL IMAGE IS AT INFINITY. Note
the concave mirror equal to. that at normal adjustment
ANSWER- HALF THE RADIUS OF (i) the erect image coincides with the
CURVATURE . A lamp is placed at the principal focus of the eye lens which produces
By Kaymath (call 08068552755) 25

a final ERECT, MAGNIFIED AND VIRTUAL 76. A projection lantern of focal length of ,
IMAGE AT INFINITY. (ii) the distance the object distance u, Is such that ____
between the object and eye lenses and ANSWER- . In a projection lantern
where is the focal length the object is placed between the center of
extra lens. curvature and the focus of the lens.
71. A telescope is said to be in normal 77. The eye controls the amount of light
adjustment when the _____ reaching the retina by adjusting the____
ANSWER- OBJECTIVE FOCAL POINT ANSWER- PUPIL. The pupil is a space by
COINCIDES WITH THAT OF THE EYEPIECE . within the iris through which light rays reach
In normal adjustment the objective focal the retina . The iris controls the amount of
length COINCIDES with the eyepiece but light reaching the retina by adjusting the
both focal lengths are NOT EQUAL. diameter of the pupil.
72. What optical instrument can best be 78. The still pictures in films appear to have a
constructed with converging lenses of focal continuity owing to ______
lengths 50cm and 5cm? ANSWER-PERSISTENCE OF VISION
ANSWER- ASTRONOMICAL TELESCOPE. 79. In comparing the camera to the human eye
Terrestrial telescope use 3 converging the film of the camera functions as the ___
lenses . Galileo’s telescope use one ANSWER- RETINA.
converging lens and one diverging lens. 80. Which of the following defeats of vision is
ASTRONIMICAL TELESCOPE use 2 as a result of the eyeball being too long___
converging lenses, one with a long focal ANSWER- SHORT SIGHT. While a short
length (e.g. 50cm) and the other with short eyeball causes long-sight.
focal length (e.g. 5cm). COMPOUND 81. For a short sighted person , light rays from
MICROSCOPE use 2 converging lenses both a point on a very distant object is focused __
of short focal lengths. ANSWER- IN FRONT OF THE RETINA.
73. In a compound microscope____ 82. For correction of the myopic defects in the
ANSWER- THE OBJECTIVE LENS HAS A human eye we require______
SHORTER FOCAL LENGTH THAN THE ANSWER- A CONCAVE LENS.
EYEPIECE WHICH IS CONVEX. In a 83. Presbyopia is a defect of the eye resulting
compound microscope the objective lens from ___. ANSWER- LOSS OF SPHERICITY OF
and eyepiece lenses are convex and the THE LENS. Presbyopia is a loss of lens
objective’s focal length is shorter than the elasticity and increased sphericity due to
eyepiece’s focal length. hardening seen in old age. Presbyopia literally
74. The Galilean telescope has an advantage means ageing.
over all other telescope in that it_____ 84. ASTIGMATISM IS USED IN CORRECTING
ANSWER- IS SHORTEST IN LENGTH. EYE DEFECTS USING CYLINDRICAL LENS
75. A convex lens of long focal length and 85. The angular dispersion of a prism depends
another convex lens of short focal length on__. ANSWER- THE INDEX OF REFRACTION
was used to make an optical instrument, ONLY. Angular dispersion in a prism depends
The distance between the lenses is_____ only on the refractive index or index of
ANSWER- . refraction of the prism.
By Kaymath (call 08068552755) 26

86. A narrow beam of white light can be Thus violet has the least wavelength.
split up into different colours by a glass 93. A spectrum of light is said to be impure
prism . The correct explanation is that___ when__ANSWER- THE DIFFERENT COLOURS
ANSWER- DIFFERENT COLOURS OF WHITE IN IT OVERLAP.
LIGHT TRAVELS WITH DIFFERENT SPEED IN 94. Production of pure spectrum could easily
GLASS. Dispersion results from the be achieved using a _____
different speeds of the components of ANSWER- TRIANGULAR PRISM WITH TWO
white light due to the different CONVEX LENSES
wavelengths. Keep in mind that one 95. In sunlight a blue flower looks blue
wavelength represents one colour of light. because we see the flower by the light it ___
87. The angle of deviation of light of various ANSWER- REFLECTS. Non-luminous objects
colours passing through a triangular prism are seen by the light they reflect.
increases in the order____ (a) green 96. What is the apparent colour of a RED SHIRT
―>violet―>blue (b) blue ―>red―>green when viewed in pure green light?
(c) blue―>green―>red (d) red ANSWER- BLACK. The red colour absorbs the
―>green―>blue green colour with nothing to reflect , the shirt
ANSWER- RED ―>GREEN―>BLUE. The appears black. Red + green―>yellow ,
angle of deviation increases in the order of red + blue―> magenta(purple) , blue + green
the spectrum from ROYGBIV. Take note. ―>cyan(bluish green), red + green + green
88. The spectrum of white light consists of ―>white.
colouring lights arranged in the following 97. When light is incident on an object which is
order; . ANSWER- RED, ORANGE, GREEN, magenta in colour ,which of the following
BLUE, INDIGO, VIOLET. From ROYGBIV colours would be absorbed?
89. Dispersion of white light is the ability of ANSWER- GREEN ONLY
white to___ 98. The electromagnetic waves that are
ANSWER- SEPARATE INTO IT’S sensitive to temperature changes are____
COMPONENT. Dispersion is the separation ANSWER- INFRARED RAYS
of the components of white light after 99. INFRARED RAYS ARE THE LEAST
refraction through a prism. ENERGETIC RAYS
90. The colours seen in soap bubbles are 100. Of the following, which is different from
due to______ the others? (a) X-rays (b) gamma rays (c)
ANSWER- DISPERSION cathode rays (d) ultraviolet rays (e) infrared
91. Which of the following pairs of light rays rays. ANSWER- C. Cathode rays are electrons .
shows the widest separation in the They are not part of the E-M spectrum.
spectrum of white light? 101. Which of the following are true about
ANSWER- BLUE AND RED. infrared radiation? I. It is invisible II. It is called
92. When white light is dispersed by a heat ray III. It’s frequency is higher than that of
spectrometer the components having the blue light IV. It travels as a transverse, (a)
shortest wavelength is____ I,II,III, and IV (b) I,II and IV (c) I,III and IV only
ANSWER- VIOLET. The spectrum ROYGBIV (d) II,III and IV only.
is in the order of decreasing wavelength ANSWER- B . The E-M spectrum in order
By Kaymath (call 08068552755) 27

of increasing frequency is RMIVUXY. Blue

light is a part of visible light (white light). In
the order, infrared came before visible light ALWAYS KNOW THAT IT IS GOD
and same applies to blue light III is wrong. WHO GIVES GOOD SUCCESS,
102. Radio carrier waves are generated by
causing electrons to oscillate rapidly in__ ALWAYS TRUST HIM AND BE
ANSWER- A TRANSMITTER. Radio waves CLOSE TO HIM. GLORY BE TO
are generated by alternating electric
current within transmitters.
GOD.
103. Which of the following is not part of
the electromagnetic spectrum? JOIN KAYMATH AND OTHER STUDENTS OF TOTAL
ANSWER- ALPHA RAYS. Alpha particles are SOLUTIONS
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