The post CIE A Level Physics Past Papers Solutions (Paper 1) appeared first on ServantBoy.

]]>I have decided to provide explanatory answers to CIE a level physics past papers especially paper 1. And the reason is because CIE has already provided answers to a level physics paper 2 and paper 4 which a degree of explanation. But this is not so for paper 1 (the correct options are only supplied without detailed explanation on how they arrived at the answers).

I want to be uploading solutions to CIE A level physics past papers 1. This will help you to understand some CIE A level physics questions better.

Note: you will need to have the CIE past papers with you for you to relate it with the solutions

To download past papers for physics click here

The solution below is Cambridge A level physics past question for Oct/Nov 2017. Click and download

I will upload for other years as soon as possible.

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]]>The post Physics Formulas Kinematics With Graphs appeared first on ServantBoy.

]]>Kinematics in physics is about the motion of a body or an obeject. And the terms associated with kinematics are displacement, speed, velocity, and acceleration. These are the variables physicists play around in this topic.

Kinematics formulas

The starting point is from the definition of acceleration which is the rate of change of velocity. This is like a car travelling at a certain speed then the driver decided to move faster or slower by pressing the accelerator. The speedometer in the car will change and at that instance the car has either accelerate or decelerate.

Acceleration = change of velocity / time = (v – u) / t

u is the initial velocity, v is the final velocity, t is the time taken, and a is the acceleration

cross multiply,

Make v the subject of the equation

The first formula above is the Newton’s first equation of motion

The next step,

From average velocity,

**Average velocity = s / t **

**Average velocity = (v +u)/2 **

Equate the two equations together, you get

**s / t = (v +u)/2 **

Cross multiply and rearrange

*s = [(v + u)t]/2 *

substitute ** v = u + at **into

you have

This is the Newton’s second equation of motion

Recommended: Short notes on motion

Final step:

Make t the subject of the Newton’s first equation of motion

t = (v – u) / a

substitute t = (v – u) / a into Newton’s second equation of motion

Cross multiply,

This is the Newton’s third equation of motion

The next thing is to look at Kinematics graphs

The post Physics Formulas Kinematics With Graphs appeared first on ServantBoy.

]]>The post View And Download JAMB Syllabus For Physics pdf appeared first on ServantBoy.

]]>I have also made an E-book for all candidates writing physics in JAMB. Click here Preparatory Physics Guide to read about the book and how to buy the book. The name of the book is Preparatory Physics Guide.

**Subject Topics/ Sub Topics (JAMB Syllabus For Physics)**

CAPACITORS

– CAPACITORS

CHANGE OF STATE

– CHANGE OF STATE

CHARACTERISTICS OF SOUND WAVES

– CHARACTERISTICS OF SOUND WAVES

CONDUCTION OF ELECTRICITY

– CONDUCTION OF ELECTRICITY THROUGH GASES

– CONDUCTION OF ELECTRICITY THROUGH LIQUIDS

CURRENT ELECTRICITY

– CURRENT ELECTRICITY

DAMS AND ENERGY PRODUCTION

– DAMS AND ENERGY PRODUCTION

DISPERSION OF LIGHT AND COLOURS

– DISPERSION OF LIGHT AND COLOURS

EDDY CURRENT

– EDDY CURRENT

ELASTICITY

– ELASTICITY

ELECTRIC CELLS

– ELECTRIC CELLS

ELECTRICAL ENERGY AND POWER

– ELECTRICAL ENERGY AND POWER

ELECTROMAGNETIC INDUCTION

– ELECTROMAGNETIC INDUCTION

ELECTROMAGNETIC SPECTRUM

– ELECTROMAGNETIC SPECTRUM

ELECTROSTATICS

– ELECTROSTATICS

ELEMENTARY MODERN PHYSICS

– ELEMENTARY MODERN PHYSICS

ENERGY AND SOCIETY

– ENERGY AND SOCIETY

EQUILIBRIUM OF FORCES

– CENTRE OF GRAVITY AND STABILITY

– CONDITIONS FOR EQUILIBRIUM OF RIGID BODIES UNDER THE ACTION OF PARALLEL AND NON-PARALLEL

FORCES

– EQUILIBRIUM OF PARTICLES

– PRINCIPLES OF MOMENTS

FORCE ON A CURRENT-CARRYING CONDUCTOR IN A MAGNETIC FIELD

– FORCE ON A CURRENT-CARRYING CONDUCTOR IN A MAGNETIC FIELD

FRICTION

– FRICTION

GAS LAWS

– GAS LAWS

GRAVITATIONAL FIELD

– GRAVITATIONAL FIELD

HEAT TRANSFER

– HEAT TRANSFER

INDUCTANCE

– INDUCTANCE

INTRODUCTORY TO ELECTRONICS

– INTRODUCTORY TO ELECTRONICS

LIGHT ENERGY

– PROPAGATION OF LIGHT

– SOURCE OF LIGHT

LIQUIDS AT REST

– LIQUIDS AT REST

MAGNETS AND MAGNETIC FIELDS

– MAGNETS AND MAGNETIC FIELDS

MEASUREMENTS AND UNITS

– DERIVED PHYSICAL QUANTITIES AND THEIR UNITS

– DIMENSIONS

– FUNDAMENTAL PHYSICAL QUANTITIES

– LENGTH, AREA AND VOLUME

– LIMITATIONS OF EXPERIMENTAL MEASUREMENTS

– MASS

– MEASUREMENT, POSITION, DISTANCE AND DISPLACEMENT

– TIME

MOTION

– LINEAR MOTION

– MOTION

– MOTION IN A CIRCLE

– NEWTON’S LAWS OF MOTON

– PROJECTILES

– SIMPLE HARMONIC MOTION (S.H.M.)

NUCLEAR ENERGY

– NUCLEAR ENERGY

OPTICAL INSTRUMENTS

– OPTICAL INSTRUMENTS

PRESSURE

– ATMOSPHERIC PRESSURE

– PRESSURE IN LIQUIDS

PROPAGATION OF SOUND WAVES

– PROPAGATION OF SOUND WAVES

QUANTITY OF HEAT

– QUANTITY OF HEAT

REFLECTION OF LIGHT AT PLANE AND CURVED SURFACES

– REFLECTION OF LIGHT AT PLANE AND CURVED SURFACES

REFRACTION OF LIGHT THROUGH

– GLASS PRISM

REFRACTION OF LIGHT THROUGH A PLANE AND CURVED SURFACES

– REFRACTION OF LIGHT THROUGH A PLANE AND CURVED SURFACES

SCALARS AND VECTORS

– SCALARS AND VECTORS

SIMPLE A.C. CIRCUITS

– SIMPLE A.C. CIRCUITS

SIMPLE MACHINES

– SIMPLE MACHINES

SOLAR ENERGY

– SOLAR ENERGY

STRUCTURE OF MATTER AND KINETIC THEORY

– KINETIC THEORY

– MOLECULAR NATURE OF MATTER

TEMPERATURE AND ITS MEASUREMENT

– TEMPERATURE AND ITS MEASUREMENT

THERMAL EXPANSION

– LIQUIDS

– SOLIDS

VAPOURS

– VAPOURS

WAVES

– CHARACTERISTICS/PROPERTIES

– CLASSIFICATION

– PRODUCTION AND PROPAGATION

WORK, ENERGY AND POWER

– WORK, ENERGY AND POWER

The post View And Download JAMB Syllabus For Physics pdf appeared first on ServantBoy.

]]>The post Magnetic Resonance Imaging (MRI operation and uses) appeared first on ServantBoy.

]]>Principle of operation of MRI

Nuclei of hydrogen (and certain other nuclides) have spin/have a magnetic axis. In a strong magnetic field, the nuclei precess about the direction of the field with a frequency known as the Larmor frequency. Radio wave pulse at Larmor frequency causes resonance and nuclei precess in high-energy state. After pulse has ceased, nuclei relax, emitting an RF signal.

How MRI works

MRI scanning involves radio frequency electromagnetic radiation. The patient lies on a bed in strong magnetic field. RF pulse at Larmor frequency transmitted to patient. RF emissions from patient are detected and processed. Hydrogen nuclei within patient have Larmor frequency dependent on magnetic field strength so that location (and concentration) of hydrogen nuclei can be detected. Total image built up by varying the non-uniform field to give specific field strength at different positions within the patient. The receiving coils pick up the relaxation signal and pass it to the computer. Therefore, a picture of the patient’s insides is built up by a computer.

Recommended: Uses and operation of Ultrasound

**Component of magnetic resonance imaging**

Superconducting magnet

RF transmission coil

RF receiver coil

Gradient coils

Computer

MRI uses

It is use in medical diagnosis (spinal/back problems and fine detail provided of soft and hard tissues)

Advantages of MRI

i) It does not use ionizing radiation which causes a hazard to patients and staff

ii) There are no moving mechanisms, just changing currents and magnetic fields

iii) The patient feels nothing during a scan (although the gradient coils are noisy as they are switched), and there are no after-effects.

iv) MRI gives better soft-tissue contrast than a CT scan, although it does not show bone as clearly

v) Computer images can be generated showing any section through the volume scanned, or as a three-dimensional image

Disadvantage of magnetic resonance imaging

Every metallic objects in the patient, such as surgical pins, can become heated. Loose steel objects must not be felt in the room as these will be attracted to the magnet, and the room must be shielded from external radio fields.

Reference

David Sang et al, Cambridge International AS and A level Physics Coursebook

The post Magnetic Resonance Imaging (MRI operation and uses) appeared first on ServantBoy.

]]>The post Ultrasound: Uses, Operation and how it is generated appeared first on ServantBoy.

]]>**How Ultrasound is generated**

i) Like audible sound, ultrasound is produced by a vibrating source. The frequency of the source is the same as the frequency of the waves it produces.

ii) Ultrasound waves may be generated and detected using a piezo-electric transducer.

iii) A transducer converts energy from one form to another.

iv) Electrical energy is converted into ultrasound energy by means of a piezo-electric crystal such as quartz(it is widely used because of its mechanical strength, small dielectric loss, resistance to moisture and the stability of its properties in the face of temperature change).

v) The structure of quartz is made up of a large number of tetrahedral silicate.

vi) An external electric field is applied to the crystal, the ions in each unit cell are displaced by electrostatic forces, resulting in the mechanical deformation of the whole crystal i.e. expand or contract

vii) An alternating voltage with frequency f causes the crystal to contract and expand at the same frequency f. The alternating voltage applied across the crystal is then acts as the vibrating source of the ultrasound waves.

**Uses of Ultrasound**

- To detect objects and measure distance
- Sonography(ultrasonic imaging) used both in veterinary and human medicine
- For nondestructive testing of products and structures
- To detect invisible flaws
- For cleaning and for mixing, and to accelerate chemical processes

**Principle of Ultrasound scan**

There are several different types of ultrasound scan which are used in practice, but their principle are the same(i.e. echo sounding)

- Direct the ultrasound waves into the body
- These waves pass through various tissues and are partially reflected at each other boundary where the wave speed changes
- The reflected waves are then detected and used to construct internal image of the body

Based on the principle of an ultrasound scan a diagnostic information can be obtained.

A-scan and B-scan are two techniques commonly use for the display of an ultrasound scan.

In an A-scan a pulse of ultrasound is transmitted into the body through the coupling medium. At each boundary between, some of the energy of the pulse is reflected and some transmitted. The transducer detects the reflected pulses. The signal is amplified and displayed on a cathode ray oscilloscope.

Recommended: Short note on electronic sensor

The post Ultrasound: Uses, Operation and how it is generated appeared first on ServantBoy.

]]>The post Solutions to Cambridge A level Physics Oct/Nov 2016 P1 appeared first on ServantBoy.

]]>The force *F *between two point charges *q*1 and *q*2, a distance *r *apart, is given by the equation

where *k *is a constant.

What are the SI base units of *k *? {Cambridge A level oct/nov 2016, ques 2, p11}

Solution

The SI base unit of force is Kgms-2

Charge* q*1 is As

Charge q2 is As

distance r is m

SI base unit of k = (Kgms-2 x m) / (As x As) = Kgm2s-4A-2 (B is the correct option)

Question 2

A student uses a cathode-ray oscilloscope (c.r.o.) to measure the period of a signal. She sets the time-base of the c.r.o. to 5 ms cm–1 and observes the trace illustrated below. The trace has a length of 10.0 cm.

What is the period of the signal? {Cambridge A level oct/nov 2016, ques 5, p11}

Solution

From the graph there are 3.5 oscillations

The distance to cover one oscillations = 10/3.5 cm

Since the time-base of the c.r.o is 5 ms cm–1

The period of the signal = (10/3.5) x 5 = 14.3 ms = 1.4 x 10-2 s (D is the correct option)

Question 3

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into the air and onto a track that is vertically 2.0 m lower.

The cyclist travels a horizontal distance of 6.0 m in the air. Air resistance is negligible.

What is the horizontal velocity *v *of the cyclist at the end of the higher track? {Cambridge A level oct/nov 2016, ques 6, p11}

Solution

There are important points to note in this question:

The horizontal velocity v is used to calculate the horizontal distance

The time to reach the maximum height is the time to travel the horizontal distance

At maximum height u = 0

Using H = ut + 1/2gt2

2 = 0 + 1/2×9.81xt2

(t=0.6395s)

Horizontal distance = horizontal velocity(v) x time(t)

6 = 0.6395v

V = 9.4ms-2 (B is the correct option)

Question 4

A car is travelling at constant velocity. At time *t *= 0, the driver of the car sees an obstacle in the

road and then brakes to a halt. The graph shows the variation with *t *of the velocity of the car.

How far does the car travel in the 5.0 s after the driver sees the obstacle? {Cambridge A level oct/nov 2016, ques 8, p11}

Solution

The distance travelled by the car = 20 x 0.8 + ½ x 20 x(5 – 0.8) =16 +42 = 58m (C is the correct option)

Question 5

A car has mass *m*. A person needs to push the car with force *F *in order to give the car acceleration *a*. The person needs to push the car with force 2*F *in order to give the car acceleration 3*a*.

Which expression gives the constant resistive force opposing the motion of the car? {Cambridge A level oct/nov 2016, ques 11, p11}

Solution

Resultant force = applied force – resistive force

Ma = F- R

R = F- ma —-i

3ma = 2F – R

R = 2F – 3ma —–ii

Substitute for R in eq i

2F – 3ma = F – ma

F = 2ma

Therefore, R = 2ma – ma = ma

Resistive force = ma (A is the correct option)

Recommended: Solutions to Cambridge A level May/June Paper 1 Physics questions 2015 and 2016

Question 6

A car travels at a constant speed of 25 m s–1 up a slope. The wheels driven by the engine exert a forward force of 3000 N. There is a drag force due to air resistance and friction of 2100 N. The

weight of the car has a component down the slope of 900 N.

What is the rate at which thermal energy is dissipated? {Cambridge A level oct/nov 2016, ques 20, p12}

Solution

Rate at which thermal energy is dissipated = power loss

Power = force x velocity

Rate at which thermal energy is dissipated = drag force x velocity = 25 x 2100 = 5.3 x 104 W (C is the correct option)

Question 7

Two parallel circular metal plates X and Y, each of diameter 18 cm, have a separation of 9.0 cm. A potential difference of 9.0 V is applied between them.

Point P is 6.0 cm from the surface of plate X and 3.0 cm from the surface of plate Y.

What is the electric field strength at P? {Cambridge A level oct/nov 2016, ques 30, p12}

Solution

The kind of field in this this is a uniform electric field. Therefore, at any point in the field the electric field strength is constant.

Electric field strength = potential difference / distance between the plate = 9 / 0.09

Electric field strength at P = 100 Nc-1 (B is the correct option)

The post Solutions to Cambridge A level Physics Oct/Nov 2016 P1 appeared first on ServantBoy.

]]>The post Free download CIE Physics 9702 past questions appeared first on ServantBoy.

]]>Tips to excel

(1) Be inquisitive – Ask the right questions in class. It is when you ask that you get answers, clarity, and understanding

(2) Relate with your colleagues that are intelligent – Get around people that are ahead of you academically. Whenever important topics are been discussed by your colleagues, always find your place around them. From my personal experience, sometimes studying alone may be boring or not interesting but when you get around friends discussing a particular topic, a level of interest rise up in you. There is a level of challenge you get that will propel you to read. It is that fuel that will make that particular topic interesting. In some cases, the reason why some of us read at times is that we heard some people discussing a topic that we know little or nothing about. This leads to a hunger to know more about the topic. In summary being close to people that are ahead of you academically can be of help to your personal growth.

(3) Get enough materials on physics expecially the ones recommended by CIE – Different materials have a way in which they discuss different topics. And the more the available materials you have the better your understanding in that area. It is possible that you may not get a satisfactory explanation from a textbook but having more than one textbook will give you an option to check other’s views.

(4) Determination and consistency – People who are successful aren’t people who have only motivation but are consistent and determined in their motivation.

The free download of Cambridge A level CIE Physics 9702 past questions and markscheme are below:

Oct/Nov 2017

CIE Physics may/june 2017 free download

Markschemes

Questions

May/June 2017

Cambridge International Examination Physics may/june 2017 free download

May/June 2016

Markscheme CIE Physics may/june 2016

9702_s16_ms_11

9702_s16_ms_12

9702_s16_ms_13

9702_s16_ms_21

9702_s16_ms_22

9702_s16_ms_23

9702_s16_ms_31

9702_s16_ms_41

9702_s16_ms_42

9702_s16_ms_43

9702_s16_ms_51

9702_s16_ms_52

9702_s16_ms_53

Past physics question papers Cambridge A level may/june 2016

9702_s16_qp_11

9702_s16_qp_12

9702_s16_qp_13

9702_s16_qp_21

9702_s16_qp_22

9702_s16_qp_23

9702_s16_qp_31

9702_s16_qp_41

9702_s16_qp_42

9702_s16_qp_43

9702_s16_qp_51

9702_s16_qp_52

9702_s16_qp_53

Oct/Nov 2015

Past physics markscheme Cambridge A level oct/nov 2015

9702_w15_ms_11

9702_w15_ms_12

9702_w15_ms_13

9702_w15_ms_21

9702_w15_ms_22

9702_w15_ms_33

9702_w15_ms_41

9702_w15_ms_42

9702_w15_ms_43

9702_w15_ms_51

9702_w15_ms_52

9702_w15_ms_53

Past physics question papers Cambridge A level oct/nov 2015

9702_w15_qp_11

9702_w15_qp_12

9702_w15_qp_13

9702_w15_qp_21

9702_w15_qp_22

9702_w15_qp_23

9702_w15_qp_31

9702_w15_qp_33

9702_w15_qp_41

9702_w15_qp_42

9702_w15_qp_43

9702_w15_qp_51

9702_w15_qp_52

9702_w15_qp_53

The post Free download CIE Physics 9702 past questions appeared first on ServantBoy.

]]>The post How to solve some UTME physics past questions appeared first on ServantBoy.

]]>Question 1

Which of the following consists entirely vector quantities? {UTME 2001}

A. Work, pressure and moment B. Velocity, magnetic flux and reaction.

C. Displacement, impulse and power. D. Tension, magnetic flux and mass.

Solution

Vector quantities has both magnitude and direction

Option A – pressure and work are scalar quantities

Option B – they are all vector quantities

Option C- power is a scalar quantity

Option D – mass is a scalar quantity

B is the correct option

Question 2

A plane sound wave of frequency 85.5Hz and velocity 342ms-1 is reflected from a vertical wall. At what distance

from the wall does the wave have an antinode?{UTME 2001}

A. 0. 1m B 1m C. 2m D. 3m

Solution

V =fℷ

ℷ = V/f = 342/85.5 = 4 m

distance to have an antinode = ℷ/4 = 4/4 = 1m

B is the correct option

Question 3

A string is fastened tightly between two walls 24cm apart. The wavelength of the second overtone is{UTME 2001}

A. 12cm B. 24cm C. 8cm D. 16cm

Solution

third harmonic is the second overtone

second overtone = ℷ/2 + ℷ/2 +ℷ/2 = 3ℷ/2

3ℷ/2 = 24

ℷ = 16 cm

D is the correct option

Question 4

Find the frequencies of the first three harmonics of a piano string of length 1.5m, if the velocity of the waves on the

string is 120ms-1.{UTME 2001}

A. 180Hz, 360Hz, 540Hz. B. 360Hz, 180Hz, 90Hz.

C. 40Hz, 80Hz, 120Hz. D. 80Hz, 160Hz, 240Hz.

Solution

first harmonic = F0 = v/2l = 120/2*1.5 = 120/3 = 40Hz

second harmonic = 2f0 = 2*40 = 80Hz

third harmonic = 3f0 = 3*40 = 120Hz

C is the correct option

Question 5

A gas with initial volume 2 x 10-6m3 is allowed to expand to six times its initial volume at constant pressure of

2 x 105Nm-2. The work done is{UTME 2001}

A. 4.0J B. 12.0J C. 2.0J D. 1.2J

Solution

workdone = pdv

dv =V2 – V1 = (6*2 x 10-6) – 2 x 10-6

dv = 12 x 10-6 – 2 x 10-6 = 10 x 10-6

workdone = 2 x 10^5 * 10 x 10-6 = 20 x 10^-1 = 2.0J

C is the correct answer

Question 6

The process of energy production in the sun is{UTME 2001}

A. radioactive decay B. electron collision.

C. Nuclear fission. D. Nuclear fusion

Solution

The answer is Nuclear Fusion

D is the correct option

Question 7

A student is at a height 4m above the ground during a thunderstorm. Given that the potential difference between

the thunderstorm and the ground is 107V, the electric field created by the storm is{UTME 2001}

A. 2.0 x 106NC-1. B. 4.0 x 107NC-1.

C. 1.0 x 107NC-1. D. 2.5 x 106NC-1

Solution

E = V/d = 10^7 / 4 = 2.5 x 10^6

D is the correct option

Question 8

An object is weighed at different locations on the earth. What will be the right observation?{UTME 2010}

A. Both the mass and weight vary B. The weight is constant while the mass varies

C. The mass is constant while the weight varies D. Both the mass and weight are constant.

Solution

The mass of an object doesnt change but weight changes because the force of gravity varies from place to place on the surface of earth.There are two reasons behind this variation:

The shape of earth and the rotation of the earth.

C is the correct option

Question 9

In a hydraulic press, the pump piston exerts a pressure of 100 Pa on the liquid. What force is exerted in the second piston of cross-sectional area 3m2?{UTME 2010}

A. 200 N B. 100 N C. 150 N D. 300 N

Solution

Pressure = Force /Area

100 = force/3

force exerted = 300 N

D is the correct answer

Question 10

If the angle between two vectors P and Q is 0 degree, the vectors are said to

A. be perpendicular B. be parallel C. interest at angle 60o. D. intersect at angle 45o. (UTME 2004)

Solution

The angle between two parallel lines is zero, therefore, B is the correct option

Question 11

What happens to the rays in a parallel beam of light?

A. They diverge as they travel. B. They meet at infinity. C. They intersect D. They converge as they travel. (UTME 2004)

Solution

Parallel beam of light meet at infinity

B is the correct option

Question 12

The process whereby a liquid turns spontaneously into vapour is called

A. boiling B. evaporation C. sublimation D. relegation. (UTME 2005)

Solution

Solid to gas is sublimation

liquid to vapour at all temperature is evaporation

liquid to vapour at a fixed temperature is boiling

Since the word spontaneous is used in the question which means occur without having been planned, therefore, evaporation is the best answer.

B is the correct option

The post How to solve some UTME physics past questions appeared first on ServantBoy.

]]>The post How to solve Questions on Nuclear physics for UTME appeared first on ServantBoy.

]]>Question 1

A piece of radioactive material contains 1000 atoms. If its half-life is 20 seconds, the time taken for 125

atoms to remain is {UTME 2012}

A. 20 seconds B. 40 seconds C. 60 seconds D. 80 seconds

Solution

N = N0 exp(-ℷt)

ℷ = 0.693/T

ℷ = 0.693/20

125 = 1000 exp(-0.03465t)

125/1000 = exp(-0.03465t)

1000/125 = exp(0.03465t)

8 = exp(0.03465t)

0.03465t = In 8

0.03465t = 2.0794415

t = 2.0794415 /0.03465 = 60 seconds

C is the correct option

Question 2

A radioactive isotope has a decay constant of 10-5s-1. Calculate its half life. {UTME 2011}

A. 6.93 x 10-6s B. 6.93 x 10-5s C. 6.93 x 105s D. 6.93 x 104s

Solution

T = 0.693/ℷ

T = 0.693 / 0.00001

T = 6.93 x 10^4 s

D is the correct option

Question 3

The radioisotope 235U92 decays by emitting two alpha particles, three beta particles and a gamma ray. What is the mass and atomic numbers of the resulting daughter element? {UTME 2010}

A. 91 and 227 B. 92 and 238 C. 227 and 91 D. 215 and 88.

Solution

You should note that in nuclear reaction, proton number and nucleon number must be conserved

235U92 – 2(4He2) + 3(0e-1) +ϒ + 227X91

227X91 means nucleon number(mass number) is 227 and proton number(atomic number) is 91.

C is the correct option

Question 4

A piece of radioactive material contains 10^20 atoms. If the half life of the material is 20 seconds, the number of

disintegrations in the first second is {UTME 2009}

A. 3.47 x 1018 B. 6.93 x 1020 C. 3.47 x 1020 D. 6.93 x 1018

Solution

N = N0 exp(-ℷt)

ℷ = 0.693/T

ℷ = 0.693/20

N = 10^20 exp(-1*0.03465)

N = 10^20/1.03536 = 9.658 x 10^19

the number of disintegrations in the first second is = N0 – N = 10^20 – 9.658 x 10^19

the number = 0.342 x 10^19 = 3.42 x 10^18

A is the correct answer

Question 5

If the decay constant of a radioactive substance is 0.231s-1, the half-life is {UTME 2009}

A. 3.00s B. 0.12s C. 0.33s D. 1.50s

Solution

T = 0.693/ℷ

T = 0.693 / 0.23

T = 3.01 s

A is the correct option

Question 6

In the equation above, the particle X is {UTME 2008}

A. a proton B. a neutron C. an α-particle D. a β-particle

Solution

In nuclear reaction, proton number and nucleon number must be conserved

On the reactant side, the total number of proton = 7 + 2 = 9, the total nucleon number = 14 + 4 = 18

For the product side to be conserved, the nucleon number on X must be 1 and proton number must be 1

That is, 1X1 = proton

A is the correct option.

Question 7

A radioactive substance has a half-life of 20 days. What fraction of the original radioactive nuclei will remain after 80

days? {UTME 2007}

A. 1/16 B. 1/8 C. 1/4 D. 1/32

Solution

let the original value be N

N – N/2 …20 days

N/2 – N/4 … 40days

N/4 – N/8 ….60 days

N/8 – N/16 …. 80 days

after 80 days 1/16 0f the original radioactive nuclei will remain

A is the correct option

Question 8

The time it will take a certain radioactive material with a half-life of 50 days to reduce to 1/32 of its original

number is {UTME 2005}

A. 150 days B. 200 days C. 250 days D. 300 days.

Solution

let the original value be N

N – N/2…50 days

N/2 – N/4 …100 days

N/4 – N/8 …150 days

N/8 – N/16 … 200 days

N/16 – N/32 … 250 days

It will take 250 days for a certain radioactive material with a half-life of 50 days to reduce to 1/32 of its original

number

C is the correct option

Question 9

In the reaction above, X is {UTME 2005}

A. proton B. neutron C. electron D. neutrino

Solution

In nuclear reaction, proton number and nucleon number must be conserved

On the reactant side, the total number of proton = 92 + 0 = 92, the total nucleon number = 235 + 1 = 236

On the product side, the total number of proton = 56 + 36 + A = 92, A = 0, the total nucleon number = 144 + 90 + 2Z = 236, Z =(236-234)/2 = 2/2 = 1

For the product side to be conserved, the nucleon number on X is 1 and proton number is 0

1X0 = neutron

B is the correct option

Short notes on Nuclear physics

Question and answer on Nuclear for A level

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]]>Question 1

The pressure of a given mass of a gas changes from 300Nm-2 to 120Nm-2 while the temperature drops from 127oC to –73oC. The ratio of the final volume to the initial volume is{UTME 2001}

A. 2 : 5 B. 4 : 5 C. 5 : 2 D. 5 : 4

Solution

P1V1/T1 = P2V2/T2

127 degree = 273+127 = 400 k

-73 degree = 273-73 = 200 k

300*V1/400 = 120*V2/200

300*200*V1 = 120*400*V2

V1/V2 = 120*400/300*200 = 24/30 = 4:5

V2/V1 = 5:4

D is the correct option

Question 2

The pressure of 3 moles of an ideal gas at a temperature of 27oC having a volume of 10-3m3 is{UTME 2002}

A. 2.49 x 105Nm-2 B. 7.47 x 105Nm-2 C. 2.49 x 106Nm-2 D. 7.47 x 106Nm-2

[R = 8.3J mol-1K-1]

Solution

PV = nRT

P = nRT/V

P = 3*8.3*(273+27)/10-3

P = 3*8.3*300/0.001

P = 7470000Nm-2 = 7.47 x 106Nm-2

D is correct option

Question 3

Which of the following gas laws is equivalent to the work done?{UTME 2007}

A. Pressure Law B. Van der Waal’s Law C. Boyle’s Law D. Charles’ Law

Solution

PV = K(k=constant), this is Boyle’s Law and is equivalent to workdone

C is the correct option

Question 4

A sealed flask contains 600cm3 of air at 27oC and is heated to 35oC at constant pressure. The new volume is{UTME 2008}

A. 508cm3 B. 516cm3 C. 608cm3 D. 616cm3

Solution

V1/T1 = V2/T2

600/(273+27) = V2/(273+35)

600/300 = V2/308

2/1 = V2/308

V2 = 308*2 = 616cm3

D is the correct option

Question 5

At 40C, the volume of a fixed mass of water is{UTME 2009}

A. constant B. minimum C. maximum D. zero.

Solution

Anomalous behaviour of water is between 0 – 4 degree. At this temperature water contract i.e volume of water decreases. So the volume will be minimum

B is the correct option

Question 6

The pressure of two moles of an ideal gas at a temperature of 270C and volume 10-2m3 is{UTME 2009}

A. 4.99 x 105 Nm-2 B. 9.80 x 103 Nm-2 C. 4.98 x 103 Nm-2 D. 9.80 x 105 Nm-2

[R = 8.313 J mol-1 K-1]

Solution

PV = nRT

P = nRT/V

P = 2*8.313*(273+27)/10-2

P = 2*8.313*300/0.01

P = 498780Nm-2 = 4.99 x 105 Nm-2

A is correct option

Question 7

The pressure of one mole of an ideal gas of volume 10-2m3 at a temperature of 270C is {UTME 2010}

A. 2.24 x 104 Nm-2 B. 2.24 x 105 Nm-2 C. 2.49 x 105 Nm-2 D. 2.49 x 104 Nm-2.

[Molar gas constant = 8.3 Jmol-1K-1]

Solution

PV = nRT

P = nRT/V

P = 1*8.3*(273+27)/10-2

P = 1*8.313*300/0.01

P = 249000Nm-2 = 2.49 x 105 Nm-2

C is correct option

Question 8

2000cm3 of a gas is collected at 27oC and 700mmHg. What is the volume of the gas at standard temperature and

pressure?{UTME 2012}

A. 1896.5cm3 B. 1767.3cm3 C. 1676.3cm3 D. 1456.5cm3

Solution

At s.t.p pressure = 760mmHg, temperature = 273K

P1V1/T1 = P2V2/T2

700*200/300 = 760*V2/273

V2 = 700*2000*273/(300*760) =1676.3cm3

C is the correct option

Question 9

A gas at a pressure of 105Nm-2 expands from 0.6m3 to 1.2m3 at constant temperature, the work done is{UTME 2013}

A. 6.0 x 104J B. 7.0 x 107J C. 6.0 x 106J D. 6.0 x 105J

Solution

workdone = P(V2 – V1)

workdone = 100000(1.2 – 0.6) = 100000(0.6)

workdone = 60000 J = 6.0 x 104J

A is the correct option

short notes on gas law for UTME

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