Category: Physic Tutorials

The idea of temperature is rooted in qualitative ideas of “hot” and “cold” based on the sense of touch. As we know an object that feels hot usually has a higher temperature than when it feels cold.

We are used to the idea that a thermometer shows the temperature of an object with which is in contact.

Temperature is a measure of the degree of hotness or coldness of an object on some scale. You can also say it tells us the direction in which energy flows (energy flows from a region of higher temperature to a region of lower temperature).

When two objects in contact have the same temperature, there will be no transfer of thermal energy between them. This is referred to as thermal equilibrium.

The unit of temperature includes Celsius, Kelvin, and Fahrenheit. However, the S.I base unit of temperature is Kelvin. Measuring temperature in Kelvin scale is better than that of Celsius scale in that one of its fixed points, absolute zero has much significance than the higher or lower fixed points of Celsius scale.

Note: it is impossible to have a temperature lower than zero Kelvin (0 K is absolute zero).

Read: 50, 30, 60, 40, 45, 100 Celsius to Fahrenheit

Conversion of different temperature scales

To convert from Celsius to Kelvin = temperature in Celsius + 273.15

To convert from Kelvin to Celsius = temperature in Kelvin – 273.15

Example

Convert -23 degree Celsius to Kelvin

Solution

-23 + 273.15 = 250.15 K

Note: the triple point of water is the temperature at which ice, water, and water vapour can co-exist. And it is defined as 273.16 K (273.16 – 273.15 = 0.01 C)

To convert from Celsius to Fahrenheit or vice-versa

C/5 = (F-32)/9

C is Celsius scale while F is Fahrenheit scale

Example

Convert 50 degree Celsius to Fahrenheit

50/5 = (F-32)/9

10 = (F-32)/9

10 X 9 = F-32

90 = F -32

F = 90 + 32

F = 122

How to determine the centre of gravity of an object and concept of centre of gravity is what students of physics ought to understand especially when answering questions on moment of forces or rotational motion in classical mechanics.

Every particle or object has weight because of the force of gravity on them, said differently; every object is attracted towards the centre of the earth by the gravitational force. The point where all the weight of the object may appear to act is referred to as centre of gravity.

Stability of an object (stable, unstable, and neutral) depends on the CG of that object. There are factors that affect the C.G. of an object, they are:

• The distribution of the masses
• The gravitational free strength which is the same as acceleration of free fall g

 

How to find the centre of gravity

It is easier to find the centre of gravity of a regular object (uniform object) than irregular object. The C.G of a uniform object is at its mid-point.

• The C.G for a person standing upright is in the middle of the body behind the navel
• The C.G of a lamina or thin sheet or triangular plate is two-thirds of the distance of the median from the corresponding point. And this can be found by suspending any of the objects freely from two to three points.

For instance, the C.G of a rigid body in the figure below is the single point through which the weight of the body is considered to act.

The center of mass of an object is sometimes called its center of gravity. However, the center of mass is defined independently of any gravitational effect.

Centre of mass of an object is the point where the total mass appears to act.

If the distribution of masses and gravitational field strength (g) remains constant on all part of the object, the CG will coincide with the CM of the object.

Example on centre of gravity

A man of weight 600 N stands at the end of a uniform wooden plank, which is pivoted as shown in the diagram. What is the weight of the wooden plank?

Solution

The CG of the plank is roughly at the midle since it is a uniform wooden plank. Therefore, the distance of the weight of the plank is 1.5m from the man.

Sum of clockwise momnet = sum of anticlockwise moment

w x 1 = 600 x 0.5 (the moment is taking from the pivot)

w = 300N

To view more example, click here

Before I explain why ammeter is connected in series and voltmeter is connected in parallel in a circuit, there is a need to explain the function of the two measuring instruments. This will help you to understand what I am about to explain.

Function of Ammeter

It is used to measure the value of flowing current in a circuit.

Function of Voltmeter

It is used to measure the value of the potential difference or voltage across the load (resistor). The reading on the voltmeter indicates the energy transferred to the component by each unit of the charge.

Now you need to understand series and parallel connection;

In series connection,

• Current are the same through the loads
• In a closed circuit only ONE path for current to flow
• Total potential difference = sum of individual potential difference across the loads
• Total resistance = sum of individual resistances

In parallel connection,

• Potential difference across each resistor
• Total current = sum of individual currents
• Effective resistance less than individual resistance

Why ammeter is connected in series

The ammeter is connected in series with the load so that there is the same current in both. This can only be possible if the resistance on the ammeter is very low so as to ensure correct measurement of current in the circuit.

For an ideal ammeter, the value of the resistance should be zero. This means that all the voltage should appear across the load, so that current can be measured accurately.

From, V =IR

Voltage is directly proportional to current flow. If there is no voltage on the ammeter, then current won’t flow. And this can only be possible if the resistance on the ammeter is zero.

Why voltmeter is connected in parallel

A voltmeter is designed to have a high resistance and should be connected in parallel in order to measure accurately.

By this, no current will flow through the voltmeter i.e. it won’t disrupt the current that should flow across a resistor or load.

For current not to flow through a voltmeter, the resistance of an ideal voltmeter should be infinity.

Kindly note current in a circuit will choose to flow through a less resistive path. This is also a reason why the resistance of a voltmeter must be high to avoid current flow across it.

Mathematically,

If the load is r and the voltmeter resistance is infinity connected in parallel to each other, the effective resistance will be

1/r + 1/infinity = 1/R

1/infinity = 0

1/r + 0 = 1/R

R = r (i.e. the effective resistance is the same as the resistance on the load)

This shows clearly that volt-meter will not alter the flow of current in the circuit.

Recommended: Note on Direct Current

This article is intended to help students who finds it difficult to solve questions on dynamics. Below are some questions on dynamics and steps on how to answer them.

Question 1

A brick weighing 20 N rests on an inclined plane. The weight of the brick has a component of 10 N
parallel with the plane. The brick also experiences a frictional force of 4 N.

What is the acceleration of the brick down the plane? Assume that the acceleration of free fall g is
equal to 10 m s–2.

Question 2

The diagram shows a barrel suspended from a frictionless pulley on a building. The rope
supporting the barrel goes over the pulley and is secured to a stake at the bottom of the building

A man stands close to the stake. The bottom of the barrel is 18 m above the man’s head. The
mass of the barrel is 120 kg and the mass of the man is 80 kg.
The man keeps hold of the rope after untying it from the stake and is lifted upwards as the barrel
falls.
What is the man’s upward speed when his head is level with the bottom of the barrel? (Use
g = 10 m s–2.)

Question 3

A box of mass 8.0 kg rests on a horizontal, rough surface. A string attached to the box passes
over a smooth pulley and supports a 2.0 kg mass at its other end

When the box is released, a frictional force of 6.0 N acts on it.
What is the acceleration of the box?

Question 4

The momentum of an object changes from 160 kg m s–1 to 240 kg m s–1 in 2 s.
What is the mean resultant force on the object during the change?

Question 5

Two spheres approach each other along the same straight line. Their speeds are u1 and u2
before collision. After the collision, the spheres separate with speeds v1 and v2 in the directions
shown below.

Which equation must be correct if the collision is perfectly elastic?
A u1 – u2 = v2 + v1
B u1 – u2 = v2 – v1
C u1 + u2 = v2 + v1
D u1 + u2 = v2 – v1

Solutions

Solution Q1

weight of the box is 20N

weight = mg

20 = m * 10

m = 2kg

since the box slide down the plane, it means there is a resultant force or net force

F – Fr = ma

F = force parallel to the plane = 10N

10 – 4 = 2 * a

6 = 2a

a = 3 ms-2

Solution Q2

The force acting in this questions are the weight of the man, the weight of the barrel and the tension on the rope.

As the man moves upward, this is what happens. The tension on the rope acting upward is greater than the weight of the man acting downward. So this gives a resultant force

mathematically,

T – wm = m1a1……..iing eqn i

wm represents the weight of the man

m1 is the mass of the man

a1 is the acceleration of the man

As the barrel moves downward. The tension on the rope acting upward is less than the weight on the barrel acting downwards. So this gives a resultant force

dont forget that weight(gravitational force) is always acting downwards to the centre of the earth.

mathematically,

wb – T = m2a2……..ii

wb = weight of the barrel

m2 = mass of the barrel

a2 = acceleration of the barrel

combining eqn i and ii and eliminate T

you will have

wb – wm =m2a2 + m1a1

note the two objects will have the same acceleration

a2 = a1

wb – wm = (m1 + m2) a1

wb = 120*10 = 1200N

wm = 80 *10 = 800N

1200 – 800 = (120 + 80) a1

400 = 200*a1

a1 = 2 ms-2

recall,

v²=u²+2as

the man and the barrel would have covered half of the distance when his head is level with the bottom of the barrel

i.e    s = 9m

v² = 0 + 2 * 2 * 9

v²= 36

v = 6 ms-1

Solution Q3

The same approach we use in Q2 is applicable here

wb – fr = (m1 + m2)a1

wb = is the weight of the box

m1 is the mass of the box, m2 is the mass at the other end

20 – 6 = 10 * a

14 = 10a

a = 1.4 ms-2

Solution to Q4

change in momentum = p1 – p2

change in momentum = 240 – 160 = 80kgms-1

resultant force = change in momentum / time  = 80 / 2 = 40N

Recommended: Short note on dynamics

Solution to Q5

Before collision, the balls move in opposite direction to each other

After collision, the balls move in the same direction

recall that,

relative speed of approach = relative speed of separation

u1 – u2 = v2 – v1 ( when they are moving in the same direction both before and after collision)

But in this case, they move in opposite direction before collision

in this case

u1 + u2 = v2 – v1

The answer is D

OR

If you look closely, u1 and v1 are in the same direction,  that means

u1 + v1

u2 and v2 are in opposite direction, that means

v2 – u2

using conservation of linear momentum

u1 + v1 = v2 – u2

rearrange

u1 + u2 = v2 – v1

D is the correct answer

 

More worked examples on CIE Physics Dynamics

Example 1

A ball of mass m. It is dropped onto a horizontal plate as shown in the figure below

Just has the ball makes contact with the plate, it has a velocity v, momentum p and kinetic energy E_k.

1. Write down the momentum p in terms of mass m and velocity v
2. Show that E_k =p2/2m

iii. Just before the impact with the plate, the ball of mas 35g has speed 4.5ms-1. It bounces from the plate so that its speed immediately after losing contact with the plate is 3.5ms-1. The ball is in contact with the plate for 0.14s. Calculate for the time that the ball is in contact with the plate, the average force, in the addition to the weight of the ball, that the plate exert on the ball.

Solution

i   p = mv

ii   E_{k = mv2/2    ………..   1}

       _{p = mv     ………………    2}

_{eqn1 can be written as}

_{E_k=(m^2 v^2)/2m}

_therefore

_{Ek =p2/2m}

iii.

Change in momentum ∆p= m(u+v)

The reason for positive sign is because the ball moves in opposite direction after rebound i.e ∆p = 0.035(3.5+4.5)

∆p=0.28〖kgms〗^(-1)

F= ∆p/∆t

F is the force of impact

F=0.28/0.14

F = 2N

Example 2

A ball falls vertically onto horizontal ground and rebounds, as shown

The ball has momentum p1 downwards just before hitting the ground. After rebounding, the ball

leaves the ground with momentum p2 upwards. The ball is in contact with the ground for 0.020 s.

During this time interval, an average resultant force of 25 N acts on the ball.

What is a possible combination of values for p1 and p2?

Solution

F= (p2- p1)/∆t

Since the ball is in opposite direction, it implies that

F= (p2+p1)/∆t

p2+p1 = 25* 0.02 = 0.5

Since the ball has the same mass the velocity before hitting the plate will be higher than the velocity after it bounces.

So therefore p1 > p2, from the option C is the correct answer.

In this article, i will solve questions on linear expansivity, define area and cubic expansion. Linear expansivity is defined as the increase in length, of the unit length of the material for one degree temperature rise.

To solve any question on L.E:

• Write down the formula
• Write down the data given
• Understand what you are asked to find
• Make proper substitution
• Evaluate

Advantages and disadvantages

Advantages

1. Used for making bimetallic strips, which are used in thermostats
2. Removal of tight glass stopper
3. Red-hot rivets in shipbuilding
4. Expansion of metals is used in bimetallic thermometer

Disadvantages

1. Expansion of railway lines
2. Sagging of overhead wire
3. Expansion of balance wheel or wrist-watch
4. Cracking of glass cup when hot water is poured into the glass cup
5. Expansion of metals or concrete bridges

Area Expansivity: The increase in the area of an object with temperature change is called superficial expansivity or area expansivity.

The superficial expansivity of a material is defined as,

Its value is twice that of L.E.

Cubic Expansivity: Increasing temperature usually causes an increase in volume for both solid and liquid materials.

The cubic expansivity of a solid is defined as

Its value is thrice that of L.E.

Steps on how to solve questions on linear expansivity

Question 1

What is meant by the statement: The linear expansivity of a solid is 1.0 x 10-5 k-1

Solution

The statement means that a unit length of the solid will expand in length by a fraction of 1.0 x 10-5 of the original per Kelvin rise in temperature.

Question 2

A wire of length 100.0m at 300C has a L.E of 2 x 10-5K-1. Calculate the length of the wire at a temperature of -100C {UTME 2013}
A. 99.92m                    B. 100.08m                 C. 100.04m                         D. 99.96m

Solution

l1 = new length

l0 = original length

Q = temperature change

0.00002 = change in length / 100(30-(-10))

0.00002 = change in length / 100*40

0.00002*4000 = change in length

change in length = 0.08

change in length = 100 – x

x = 100 – 0.08 = 99.92m

A is the correct option

Question 3

Two metals P and Q are heated through the same temperature difference. If the ratio of the L.E of P to Q is 2: 3 and the ratio of their lengths is 3:4 respectively, the ratio of the increase in lengths of P to Q is {UTME 2012}
A. 1 : 2                           B. 2 : 1                      C. 8 : 9                                D. 9 : 8

Solution

Temperature difference =( L.E * length)/increase in length

For metal P,

Qp = (L.E * length)p/increase in length p

For metal Q,

Qq = ( L.E * length q)/increase in length q

( linear expansivity * length)p/increase in length p = ( L.E * length q)/increase in length q

increase in length of p:increase length q = ( L.E * length)p / ( L.E * length q)

increase in length of p:increase length q = 2/3 *3 /4 = 6 / 12 = 1:2

A is the correct option

Question 4

A metal of volume 40cm3 is heated from 300C to 900C, the increase in volume is {UTME 2011}
A. 0.40cm3                     B. 0.14cm3                 C 4.00cm3                             D. 1.20cm3.
[L.E of the metal= 2.0 x 10-5K-1]

Solution

cubic expansivity = 3* L.E = 3 * 0.00002 = 0.00006

0.00006 = increase in volume / 40*(90 – 30)

0.00006 = increase in volume / 40*60

increase in volume = 0.00006*2400 = 0.144cm3

B is the correct answer.

Question 5

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10-6K-1. The area expansivity is {UTME 2007}
A. 6.3 x 10-6K-1                B. 4.2 x 10-6K-1                    C. 2.1 x 10-6K-1             D. 2.0 x 10-6K-1

Solution

cubic expansivity = 3* L.E

0.0000063 = 3* L.E

L.E = 0.0000063/3 = 0.0000021

Area expansivity = 2*0.0000021 = 0.0000042

B is the correct option

Question 6

Calculate the length which corresponds to a temperature of 20^0C if the ice and steam points of an ungraduated thermometer are 400 mm apart
A. 80mm
B. 20mm
C. 30mm
D. 60mm

Solution

ice point = lower fixed point = 0^oC

steam point = upper fixed point = 100^oC

(100 -0)/(20 -0) = 400/(x – 0)

100/20 = 400 / x

20*400 = 100x

x = 8000/100

x = 80mm

A is the correct option

Question 7

A steel bridge is built in the summer when its temperature is 35.0°C. At the time of construction, its length is 80.00m. What is the length of the bridge on a cold winter day when its temperature is -12.0°C? (L.E of steel is 1.2 x10^-5)

Solution

initial length = 80 m

initial temperature = 35.0°C

final length = ?

final temperature = -12.0°C

Change in temperature = final – initial = -12-35 = -47.0°C

let the change in length = x

o.oooo12 = x/80(-47)

x = 0.000012*80*(-47) = -0.04512 m

note

x = final length – initial length

-0.04512 = final length – 80

final length = 80 -0.04512 = 79.95488 m

The length of the bridge on a cold winter day = 79.95488 m

Question 8

A copper rod whose linear expansivity =1.70×10^-5°c is 20cm longer than an aluminum rod whose L.E =2.20×10^-5°c. How long should the copper rod be if the difference in their length is to be independent of temperature?

Solution

For copper,

L.E of copper = x

change in length of cooper = y

the original length of copper = c

temperature change = t

x = y/(c*t)

For aluminum,

linear expansivity of aluminum = a

change in length of aluminum = b

the original length of aluminum = d

temperature change = r

a = b/(d*r)

According to the question, copper is 20cm longer than an aluminum rod, i.e., c = d + 0.2

y/t = x*c = x*(d+0.2) = 1.70×10^-5*(d+0.2)

b/r = a*d = 2.20×10^-5*d

note, y/t = b/r since the difference in length is independent on temperature

2.20×10^-5*d = 1.70×10^-5*(d+0.2)

d = 1.7/2.2 * (d+0.2)

d = 0.7727d + 0.1545

d-0.7727d = 0.15

0.2273d = 0.1545

d = 0.68m

c = d + 0.2 = 0.68 +0.2 = 0.88m

Question 9

A metal of volume 40cm³ is heated from 30⁰C to 90⁰C; the increase in volume is

A. 0.40cm³ B. 0.14cm³ C 4.00cm³ D. 1.20cm³.

[L.E of the metal = 2.0 x 10^-5K^-1]

Answer

Volume expansivity = 3 x Linear expansivity = 3 x 2.0 x 10^-5K^-1 = 6.0 x 10^-5K^-1

Increase in volume = x

V.E = (x) / (initial volume x temp rise)

6.0 x 10^-5 = x / (40 x (90 – 30))

X = 6.0 x 10^-5 x 40 x 60 = 14400 x 10^-5 = 0.144 cm³

Question 10

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10^-6K^-1. The area expansivity is

A. 6.3 x 10^-6K^-1 B. 4.2 x 10^-6K^-1 C. 2.1 x 10^-6K^-1 D. 2.0 x 10^-6K^-1

Answer

The cubic expansivity = 6.3 x 10^-6K-1

Recall that

cubic expansivity = 3 x L.E

Therefore,

L.E = (6.3 x 10-6)/3 = 2.1 x 10^-6

Area expansivity = 2 x L.E = 2 x 2.1 x 10^-6 = 4.2 x 10^-6 K^-1

Recommended: Solved questions on other topics in physics

This article provides steps on how to solve questions on Work, Energy and Power. Therefore, the solutions below gives an easy step to how to solve questions on work, Energy, and Power for students writing Cambridge A level and other exams like UTME or Post UTME.

Power: It is the rate of energy expended per unit time. Its unit is Watt

Energy: It is the ability to do work .Its unit is Joule

 

How to solve questions on Work, Energy and Power

Question 1

A hammer with 10 J of kinetic energy hits a nail and pushes it 5.0 mm into a plank. Both the hammer and nail come to rest after the collision. What is the approximate average force that acts on the nail while it moves through 5.0 mm?{ Cambridge A level may/june 2016 p11}
A 0.050 N                   B 2.0 N                       C 50 N                         D 2000 N

Solution

workdone by a net force = change in kinetic energy of a body

F x s = Ek

F x s = 10

F x 0.005 = 10

F = 10 / 0.005

F = 2000N

D is the correct option

Q2, 3 and 4 are from cambridge A level may/june 2016 p13

Question 2

An object of mass 0.30 kg is thrown vertically upwards from the ground with an initial velocity of 8.0 m s–1. The object reaches a maximum height of 1.9 m. How much work is done against air resistance as the object rises to its maximum height?
A 4.0 J                            B 5.6 J                                 C 9.6 J                                          D 15 J

Solution

workdone by a net force = change in kinetic energy of a body

workdone = 1/2 m v2

v2 = u2 – 2as

v2 = 64 – 2*9.81*1.9

v2 = 64 – 37.278

v2 = 26.722

workdone = 1/2 * 0.3 * 26.722

workdone = 4.0 J

A is the correct option

Recommended: Short note on work, energy and power

Question 3

A racing car has an output power of 300 kW when travelling at a constant speed of 60 m s–1. What is the total resistive force acting on the car?
A 5 kN                                 B 10 kN                             C 50 kN                                     D 100 kN

solution

power = force x velocity

300000 = force x 60

force = 300000 / 60

force = 5000 = 5KN

A is the correct option

Question 4

The diagram shows the design of a water wheel which drives a generator to produce electrical power. The flow rate of the water is 200 kg s–1. The generator supplies a current of 32 A at a voltage of 230 V.

Ignoring any changes in kinetic energy of the water, what is the efficiency of the system?
A 14%                     B 16%                            C 22%                                             D 47%

Solution

efficiency = power output / power input

power output = IV

power output = 32*230 = 7360

power input = flow rate * a * h

a is the acceleration due to gravity

power input = 200*8*9.81 = 15696

efficiency = (7360 / 15696)*100%

efficiency = 47%

Q 5 and 6 are from cambridge A level may/june 2016 p12

Question 5

A boy on a bicycle starts from rest and rolls down a hill inclined at 30° to the horizontal. The boy and bicycle have a combined mass of 25 kg. There is a frictional force of 30 N, which is independent of the velocity of the bicycle.
What is the kinetic energy of the boy and the bicycle after rolling 20 m down the slope?
A 1850 J                     B 2450 J                              C 3050 J                                  D 3640 J

Solution

mgsinθ – fr = ma

25 * 9.81*sin30 – 30 = ma

122.625 – 30 = ma

92.625 = ma

ma is the net force

the kinetic energy = net force X distance

kinetic energy = 92.625 * 20 = 1852 J = 1850J

A is the correct answer

Question 6

An escalator in an underground station has 250 people standing on it and is moving with a velocity of 4.3 m s–1. The average mass of a person is 78 kg and the angle of the escalator to the horizontal is 40°.
What is the minimum power required to lift these people?
A 54 kW                          B 64 kW                                  C 530 kW                            D 630 kW

Solution

the vertical force = mgsinθ = 250*78*9.81*sin40 = 122962 N

minimum power = vertical force x velocity = 122962 x 4.3 = 530 Kw

C is the correct option

Question 7

A man has a mass of 80 kg. He ties himself to one end of a rope which passes over a single fixed pulley. He pulls on the other end of the rope to lift himself up at an average speed of 50 cm s–1.

What is the average useful power at which he is working? (Cambridge A level May/June 2017 p13 q17)

A 40 W    B 0.39 kW    C 4.0 kW    D 39 kW

Solution

Power is the rate of energy expended per unit time i.e power = energy/time = force x velocity

force = mass x acceleration due to gravity. note g in CIE is always 9.81ms-2

f = 80 x 9.81 = 784.8N

velocity in ms-1 = 50/100 = 0.5ms-1

power = 784.8 x 0.5 = 392.4 watt = 0.39KW (B is the correct answer)

Question 8

Calculate the apparent weight loss of a man weighing 70kg in an elevator moving downwards with an acceleration of 1.5ms-2.{2013 UTME Physics – Type U}
A. 105N                       B. 686N                              C. 595N                            D. 581N

solution

when an elevator is moving down

net force = ma

net force = 70*1.5

net force = 105 N

the weight loss = net force

the weight loss = 105 N

A is the correct option

Click here to download for free Preparatory guide Physics for Cambridge A level, Post UTME, and UTME

Plotting of graph in physics practical class is not something strange to students performing experiments in physics laboratory. However, some students do encounter some problems when trying to plot graph or to choose appropriate scale for a graph.

I will try to put you through the following

1. How to plot graph in physics practical
2. How to plot graph in physics practical
3. How to choose an appropriate scale for a graph

Before I start the explanation, it is important to note that you do your physics practical making sure you avoided all possible errors for your readings to be accurate. If errors are not avoided, it may make the graph a little bit difficult for you to plot.

Now, going to the subject matter “How to plot graph”

Read: Guide on WAEC physics practical questions and answers

How to Plot Graph And Choose Scale in Physics Practical

• The first thing you need to get is a graph sheet. I will recommend that you buy a standard graph sheet.
• The vertical axis is the Y-axis while the horizontal axis is the x-axis
• Count the number of y- axis boxes and x-axis boxes of your graph sheet

In the above graph, there are 24 Y-axis small boxes  while there are 20 X-axis small boxes. This information is what you will use to choose an appropriate scale.

(4) For Y-axis, divide the number of boxes by the highest value y in your table. For X-axis, divide the number of boxes with the highest value of x in your table.  For instance, if the highest value for y is 12, it means 24boxes/12 = 2 boxes. This means 2boxes i.e 2cm on your graph will represent I unit for y-axis. You can repeat the same for x-axis.

Note: Your plot must be 2/3 of your graph.

For better understanding, I will give you a table of value which i will choose an appropriate scale for and then plot.

Example how plotting Graph in Physics

Plot the graph of Y= 4, 6, 8, 10, 12 against X= 1, 2, 3, 4, 5

The graph I plotted is shown below

In this graph, i decided to leave six boxes (6cm) down the graph and six boxes (6cm) to the left hand side of the graph. This is not a law, it depends on how neat you want your graph to be.

This means I have 18 boxes left for y-axis and 14 boxes left for x-axis. The scale I used was 2cm on the graph represents 2 unit on the y-axis and 2cm on the graph represent 1 unit on the x-axis.

This is how i choose the scale, for y-axis (18 boxes/12 = 1.5 boxes; i rounded it up and i make it 2 boxes to represent 2 unit), for x-axis (14 boxes/5 = 2.8 boxes; i made this 2 cm to represent 1 unit)

The next thing is that Located the values on the graph and mark it with a (+) for both y and x axis. Then I joined all the points using a transparent ruler and ensure a line of best fit. The line of best fit is drawn so that the points are evenly distributed on either side of the line.

Note: Make sure you indicate each axis with the symbols you are given ( V against I) with their units. Also indicate your choosen scale at the top of the graph. This information attracts make in physics practical exams.

General note on ploting graphs in physics:

• For scales to be reasonable, graph must occupy at least 1/3 of the page
• Scales using multiples or sub-multiples of prime numbers such as 3,7,9,13 etc are not good enough
• Points should be plotted correctly to nearest half square on both axes
• To obtain the suitable line of best fit mark, at least three points must be correctly plotted
• Starting from origin is part of graph if intercept is required
• To find slope, large right-angled triangle implies that it occupies at least 1/3 of graph page.

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Primary and Secondary Colours: Red, green and blue colours are primary colours. When they are mixed equally, white light is produced. Others such as yellow, magenta, and cyan are called secondary colours.

The adding of primary colours to produce other colours is known as additive colour mixing.

Red and blue produce magenta

Red and green produce yellow

Blue and green produce cyan

Secondary colours are magenta, yellow and cyan while complimentary colours are colours which can be added to get white.

Red and cyan produce white

Blue and yellow produce white

Green and magenta produce white

Colour triangle

Colours of objects

Colour of an opaque object is due to selective reflection, when white light falls on an object, it absorbs all colours except its characteristics colour which it reflects. A red object appears red because it absorbs all other colours except red which it reflects. If a red object is viewed in green or blue light it will appear dark because this light do not contain red. But in yellow light it will look red.

Questions on primary and secondary colours

(1) The angle of deviation of light of various colours passing through a triangular prism increases in the order

A. blue → green → red

B. red → green → blue

C. green → violet → blue

D. blue → red → green

Red is the least deviated while violet is the most deviated. From ROYGBIV, Red – Orange -Yellow – Green- Blue – Indigo – Violet. B is the correct answer

(2) Disperson of white light is the ability of white light to

A. Penetrate air, water and glass

B. Move in a straight line

C. Move around corners

D. Separate to its component Colours

The separation of colours by prism is referred to as dispersion. D is the right answer

(3) When a red rose flower is observed in blue light, what colour does the observer see?

A. Yellow   B. Red   C. Blue     D. Magenta.

Red and blue produce magenta. D is the right answer

(4) Which of the following is a secondary colour?

A.Blue       B. Orange     C. Red         D. Green.

Blue, red, and green are primary color except orange. B is the right answer

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