Physical quantities and measurements

Class 11 Physics, unit 1, book pages 7–21

Practise 78 MCQs Take the unit quiz Review 38 cards

Physics rests on measurement, so this unit is about how to measure, estimate and report physical quantities honestly. It shows how derived units are built from the SI base units, how dimensions are used to check an equation or to guess a possible formula, and how precision, accuracy and uncertainty describe the quality of a measurement. It ends with the rules for carrying uncertainties through calculations, which are needed in every experiment and numerical problem that follows.

Learning outcomes (7)

Make reasonable estimates of physical quantities (of those quantities that are discussed in the topics of this grade).
Express derived units as products or quotients of the SI base units.
Analyse the homogeneity of physical equations (through dimensional analysis).
Derive formulae in simple cases (through using dimensional analysis).
Analyse and critique the accuracy and precision of data collected by measuring instruments.
Assess the uncertainty in a derived quantity (by simple addition of absolute, fractional or percentage uncertainties).
Justify why all measurements contain some uncertainty.

1.1Estimation of physical quantities

Physics is based on experimental observations. A qualitative observation has no numbers attached: it deals with facts we can notice with our five senses (sight, smell, taste, touch and hearing), such as colours, shapes and textures, or the statement that 'water keeps its level'. A quantitative observation includes numbers and is also called a measurement. We can measure mass, time, distance, speed, pressure, force, torque, momentum and energy, and quantitative observations are the ones a scientist finds most useful.

In daily life we meet questions like these: What is the height of this building? Will this equipment fit in the back of our car or do we need a truck? How long will this download take? How large a current flows in this circuit? How many houses could a proposed power plant power? We usually answer them by making estimations, because often little or no actual data is available.

Estimation: A rough educated guess of the value of a physical quantity by using prior experience and sound physical reasoning.

An estimation includes identifying the correct physical principles and making a good guess about the relevant variables. It is very useful for developing a physical sense. Estimation does not mean guessing a formula or a number at random.

Estimation of length

Anything can be a ruler. To estimate a big length, break it into smaller pieces: estimate the length of one small piece and multiply by the number of pieces. To estimate the height of a building, count its floors, estimate the height of one floor by imagining how many people would have to stand on each other's shoulders to reach the ceiling, and estimate the height of one person.

height of building = (number of floors) \times (height of one floor)

The same idea works in reverse, to find a small length from a bulk of many identical things. To estimate the thickness of a sheet of paper, measure the thickness of a stack of paper and divide by the number of pages in the stack.

thickness of one sheet = \frac{thickness of the stack}{number of pages in the stack}

Breaking big things into small things, or collecting small things into a big thing, can also be used to estimate other quantities such as mass and time. For these, the length, mass and time scales of Table 1.1 may be helpful.

Estimate areas and volumes from lengths

For the area or volume of a complex object, introduce a simple model of it, such as a sphere or a box. Estimate its linear dimensions (the radius of the sphere, or the length, width and height of the box) and then use the standard geometric formulas. If you already have an estimate of the area or volume, you can do the reverse and use the geometric formulas to get an estimate of the linear dimensions.

Table 1.1: The estimation of some physical quantities (orders of magnitude)
Length (m)	Mass (kg)	Time (s)
Diameter of proton: $1 0^{- 15}$	Mass of electron: $1 0^{- 30}$	Mean lifetime of unstable nucleus: $1 0^{- 22}$
Diameter of large nucleus: $1 0^{- 14}$	Mass of proton: $1 0^{- 27}$	Time for a single floating-point operation in a supercomputer: $1 0^{- 17}$
Diameter of hydrogen atom: $1 0^{- 10}$	Mass of bacterium: $1 0^{- 15}$	Time period of visible light: $1 0^{- 15}$
Diameter of typical virus: $1 0^{- 9}$	Mass of mosquito: $1 0^{- 6}$	Time period of an atom in a solid: $1 0^{- 13}$
Width of pinky fingernail: $1 0^{- 2}$	Mass of hummingbird: $1 0^{- 2}$	Time period of nerve impulse: $1 0^{- 3}$
Height of 4-year-old child: $1 0^{0}$	Mass of 1 litre of water: $1 0^{0}$	Time for 1 heartbeat: $1 0^{0}$
Length of football ground: $1 0^{2}$	Mass of a motorcycle: $1 0^{2}$	One day: $1 0^{5}$
Diameter of Earth: $1 0^{7}$	Mass of atmosphere: $1 0^{19}$	One year: $1 0^{7}$
Diameter of solar system: $1 0^{12}$	Mass of Moon: $1 0^{22}$	Human lifetime: $1 0^{9}$
1 light-year: $1 0^{16}$	Mass of Earth: $1 0^{25}$	Recorded human history: $1 0^{11}$
Diameter of Milky Way: $1 0^{21}$	Mass of Sun: $1 0^{30}$	Age of Earth: $1 0^{17}$
Distance between edges of observable universe: $1 0^{26}$	Mass of known universe: $1 0^{53}$	Age of universe: $1 0^{18}$

Estimate mass from volume and density

To estimate the mass of an object, first estimate its volume and then use an estimate of its average density. Density has the dimension of mass divided by volume, so

mass = density \times volume

It helps to remember three values: the density of air is about $1 kg m^{- 3}$ , the density of water is $1 0^{3} kg m^{- 3}$ , and the densest everyday solids have a maximum density of about $1 0^{4} kg m^{- 3}$ . Asking whether an object floats or sinks in air or in water gives a rough idea of its density. You can also do the reverse: from an estimate of the mass and the density you can estimate the volume.

Estimating the energy needed by an adult man to climb the stairs from the ground floor to the first floor: mass about

70 kg

, height gained about

3 m

(Example 1.1).

Example 1.1

Estimate the energy required for an adult man to walk up the stairs from the ground floor to the 1st floor.

Given

$m = 70 kg$ mass of an adult man (estimated)
$h = 3 m$ distance between two floors (estimated)
$g = 10 m s^{- 2}$ rounded

To find

energy required $E$

Walking up the stairs raises the man's body through a height $h$ , so the energy needed is the gain in gravitational potential energy.
$E = m g h$
Substitute the estimated values.
$E = 70 kg \times 10 m s^{- 2} \times 3 m$
Multiply, and recognise that the unit $kg m^{2} s^{- 2}$ is the joule.
$E = 2100 kg m^{2} s^{- 2} = 2100 J$

E \approx 2100 J

Assignment 1.1

Estimate how many floating-point operations a supercomputer can do in 1 day. The time for a single floating-point operation in a supercomputer is

1 0^{- 17} s

Given

$t_{1} = 1 0^{- 17} s$ time for one operation
$t = 1 day \approx 1 0^{5} s$ from Table 1.1

To find

number of operations $N$ in one day

The number of operations is the total time divided by the time taken by one operation.
$N = \frac{t}{t _{1}}$
Use the order of magnitude of one day from Table 1.1 (the exact value $86400 s$ is about $1 0^{5} s$ ).
$N = \frac{1 0 ^{5} s}{1 0 ^{- 17} s}$
Subtract the powers of ten.
$N = 1 0^{5 - (- 17)} = 1 0^{22}$

N \approx 1 0^{22} operations

Worked example: estimating the mass of air in a classroom

Estimate the mass of air in a classroom that measures about

8 m

6 m

with a height of

3 m

. Would your answer change a lot if the room were full of water instead?

Given

$l = 8 m$ length of room
$w = 6 m$ width of room
$h = 3 m$ height of room
$ρ_{ai r} = 1 kg m^{- 3}$ remembered value for air

To find

mass of air $m$
mass if the room held water

The room is modelled as a box. The three accent arrows are the estimated linear dimensions.

Model the room as a box, as the book suggests for a complex shape, and use the volume of a box.
$V = l w h = 8 \times 6 \times 3 = 144 m^{3}$
Mass is density times volume. Use the remembered density of air, about $1 kg m^{- 3}$ .
$m = ρ V = 1 \times 144 = 144 kg \approx 1.4 \times 1 0^{2} kg$
For water use $ρ = 1 0^{3} kg m^{- 3}$ in the same room. The volume is unchanged.
$m_{w a t er} = 1 0^{3} \times 144 = 1.4 \times 1 0^{5} kg$
Compare the two. Water is about a thousand times denser than air, so the mass is about $1000$ times larger.
$\frac{m _{w a t er}}{m _{ai r}} = \frac{1 0 ^{3}}{1} = 1 0^{3}$

m_{ai r} \approx 1.4 \times 1 0^{2} kg, m_{w a t er} \approx 1.4 \times 1 0^{5} kg

1.2Derived units in terms of base units

In Grade 9 we learnt that derived units can be expressed in terms of base units and are obtained by multiplying or dividing base units with each other. Here we express more derived units as products or quotients of the SI base units. The method is to start from the definition of the quantity, replace every quantity on the right by simpler ones until only base quantities remain, and then put in their SI units.

Force

Start from Newton's second law and write acceleration as velocity divided by time, then velocity as displacement divided by time.

Force = mass \times acceleration = mass \times \frac{velocity}{time} = mass \times \frac{displacement}{time ^{2}}

Now put the SI unit for each quantity: N for force, kg for mass, m for displacement and s for time.

N = kg \times m s^{- 2}

Work

Work is force times displacement. Replace force step by step as above.

Work = Force \times displacement = mass \times acceleration \times displacement = mass \times \frac{velocity}{time} \times displacement = mass \times \frac{displacement}{time ^{2}} \times displacement = mass \times \frac{displacement ^{2}}{time ^{2}}

Putting J for work, kg for mass, m for displacement and s for time:

J = kg \times \frac{m ^{2}}{s ^{2}} = kg m^{2} s^{- 2}

In the same way other derived units can be written as products or quotients of the SI base units. Some examples are in Table 1.2.

Table 1.2: Some derived units as products or quotients of the SI base units
Derived quantity	SI unit	Symbol	In terms of base units
Force	newton	N	$kg m s^{- 2}$
Work	joule	J	$N m = kg m^{2} s^{- 2}$
Power	watt	W	$J s^{- 1} = kg m^{2} s^{- 3}$
Pressure	pascal	Pa	$N m^{- 2} = kg m^{- 1} s^{- 2}$
Electric charge	coulomb	C	$A s$

1.3Dimensions of physical quantitiesOften examined

Dimension denotes the qualitative nature of a physical quantity. Length, width, height, distance, displacement and radius are all measured in metres because they have the same nature, so they share the same dimensions. The dimension of a physical quantity is written as a capital letter enclosed in square brackets $[]$ . The dimensions of the base quantities are given in Table 1.3.

Table 1.3: Dimensions of base quantities
Sr. No	Physical quantity	Dimensions
1	mass	$[M]$
2	length	$[L]$
3	time	$[T]$
4	electric current	$[I]$
5	temperature	$[Θ]$
6	intensity of light	$[J]$
7	amount of substance	$[N]$

The dimensions of derived quantities are obtained by multiplying or dividing the dimensions of the base quantities from which they are derived. For example, the dimensions of area, volume, velocity and acceleration are $[L^{2}]$ , $[L^{3}]$ , $[L T^{- 1}]$ and $[L T^{- 2}]$ respectively.

How the dimensions of derived quantities follow

Area is length times length.
$[area] = [L] [L] = [L^{2}]$
Volume is length times length times length.
$[volume] = [L^{3}]$
Velocity is displacement divided by time.
$[v] = \frac{[ L ]}{[ T ]} = [L T^{- 1}]$
Acceleration is velocity divided by time.
$[a] = \frac{[ L T ^{- 1} ]}{[ T ]} = [L T^{- 2}]$

Thus dimensions give the relation of a given physical quantity with the base quantities, that is mass, length, time and so on. Four essential terms are used in dimensional analysis.

Dimensional and dimensionless quantities
	Variable (magnitude changes)	Constant (magnitude fixed)
Has dimensions	Dimensional variables. Length, velocity, acceleration, force, energy etc.	Dimensional constants. Planck's constant $h$ , gravitational constant $G$ , speed of light in vacuum $c$ , ideal gas constant $R$ etc.
No dimensions	Dimensionless variables. Plane angle, solid angle, strain, coefficient of friction etc.	Dimensionless constants. The pure numbers $1, 2, 3, \dots$ , the exponential constant $e = 2.718$ and $π$ .

Dimensional variables: Physical quantities that have dimensions and variable magnitude.
Dimensional constants: Physical quantities that have dimensions and a constant magnitude.
Dimensionless variables: Physical quantities that have no dimensions and have variable magnitudes.
Dimensionless constants: Physical quantities that have no dimensions and have constant magnitude.

1.3.1Advantages of dimensionsOften examined

Using the method of dimensions (called dimensional analysis) we can check the homogeneity of an equation, derive a possible formula and determine the units of physical quantities. Dimensional analysis makes use of the fact that dimensions can be treated as algebraic quantities. That is, quantities can be added or subtracted only if they have the same dimensions, and the terms on both sides of an equation must have the same dimensions.

(i) The homogeneity of an equation

To check the correctness of an equation, we must ensure that both sides have the same dimensions; otherwise the equation cannot be considered a physically correct equation. This is called the principle of homogeneity of dimensions.

Checking

v_{f} = v_{i} + a t

Check whether the equation

v_{f} = v_{i} + a t

is dimensionally correct.

Find the dimensions of the left-hand side, the final velocity.
$[L.H.S.] = [L T^{- 1}]$
Find the dimensions of each term on the right-hand side. The initial velocity has the dimensions of velocity, and $a t$ is acceleration times time.
$[R.H.S.] = [L T^{- 1}] + [L T^{- 2}] [T]$
Simplify the second term: $[L T^{- 2}] [T] = [L T^{- 1}]$ .
$[R.H.S.] = [L T^{- 1}] + [L T^{- 1}] = 2 [L T^{- 1}]$
The number $2$ is a dimensionless constant, so it does not change the dimensions.
$[L.H.S.] = [R.H.S.]$

The equation is dimensionally correct.

(ii) To derive a possible formula

Deriving a relation for a physical quantity depends on correctly guessing the factors on which the quantity depends. As an illustration the book derives the formula for the wavelength of matter waves.

Derivation Wavelength of matter waves by dimensional analysis

Aim. Find a possible formula for the wavelength $λ$ of matter waves from the quantities it may depend on.

Assumptions.

The wavelength depends only on Planck's constant $h$ , the velocity $v$ and the mass $m$ of the particle.
The dependence is a product of powers, with a dimensionless constant.

Write the wavelength as proportional to a product of powers of the quantities it depends on. The powers $a$ , $b$ and $c$ are unknown.
$λ \propto h^{a} m^{b} v^{c}$
Replace the proportionality sign by an equals sign and a constant (the number cannot be found by dimensional analysis).
$λ = (constant) h^{a} m^{b} v^{c} (1)$
Write the dimensions of each quantity. Planck's constant has dimensions $[M L^{2} T^{- 1}]$ , mass $[M]$ , velocity $[L T^{- 1}]$ and wavelength (a length) $[L]$ . Put them on both sides of (1).
$[L] = (constant) [M L^{2} T^{- 1}]^{a} [M]^{b} [L T^{- 1}]^{c}$
Write the left side with all three base dimensions shown, and collect the powers of $M$ , $L$ and $T$ on the right.
$[M^{0} L^{1} T^{0}] = (constant) [M]^{a + b} [L]^{2 a + c} [T]^{- a - c} (2)$
Equate the powers of $M$ on both sides of (2).
$a + b = 0 (3)$
Equate the powers of $L$ on both sides of (2).
$2 a + c = 1 (4)$
Equate the powers of $T$ on both sides of (2).
$- a - c = 0 (5)$
Solve (3), (4) and (5). From (5), $c = - a$ . Put this in (4): $2 a - a = 1$ , so $a = 1$ . Then $c = - 1$ and, from (3), $b = - 1$ .
$a = 1, b = - 1, c = - 1$
Put the values of the powers back in (1).
$λ = (constant) h^{1} m^{- 1} v^{- 1}$

Result

λ = (constant) \times \frac{h}{m v}

The wavelength of a moving particle is inversely proportional to its momentum $m v$ . Dimensional analysis gives the form of the formula but not the value of the constant, which is found by experiment.

Worked example: deriving a speed formula by dimensional analysis

A stone is dropped from rest from a height

h

. Assume that the speed

v

with which it reaches the ground depends only on its mass

m

, the height

h

and the acceleration due to gravity

g

. Use dimensional analysis to find how

v

depends on them.

Write the speed as a product of powers of the variables, with a pure number $k$ in front.
$v = k m^{a} g^{b} h^{c}$
Put in the dimensions of each quantity. The speed is $[L T^{- 1}]$ , the mass is $[M]$ , $g$ is $[L T^{- 2}]$ and the height is $[L]$ .
$[M^{0} L T^{- 1}] = [M]^{a} [L T^{- 2}]^{b} [L]^{c} = [M^{a} L^{b + c} T^{- 2 b}]$
Compare the powers of $[M]$ on both sides. No mass appears on the left.
$a = 0$
Compare the powers of $[T]$ .
$- 2 b = - 1 \Rightarrow b = \frac{1}{2}$
Compare the powers of $[L]$ and use $b = \frac{1}{2}$ .
$b + c = 1 \Rightarrow c = \frac{1}{2}$
Put the powers back into the first equation.
$v = k g^{1/2} h^{1/2} = k g h$

v = k g h

(iii) To determine units of physical quantities

Because dimensions are written with the base quantities, once a quantity has its dimensions the SI unit follows by replacing $[M]$ , $[L]$ and $[T]$ with kg, m and s. For example, a quantity with dimensions $[M L^{2} T^{- 2}]$ has the unit $kg m^{2} s^{- 2}$ , which is the joule.

1.3.2Limitations of dimensional analysis

Dimensional analysis is powerful but has limitations.

Dimensional analysis does not distinguish between physical quantities having the same dimensions. For example, if the dimensional formula of a physical quantity is [ML^{2}T^{-2}] it may represent work, or energy, or torque.
It cannot be used to derive formulas containing trigonometric functions, exponential functions, logarithmic functions, etc.
It cannot determine the dimensionless constant when deriving a possible formula.
It does not always prove that a relation is physically correct although the relation is dimensionally correct. However, a dimensionally wrong equation is always wrong.

Worked example: the simple pendulum

Example 1.2

Derive the formula for the time period of a simple pendulum using dimensional analysis.

Derivation Time period of a simple pendulum by dimensional analysis

Aim. Derive a formula for the time period T of a simple pendulum using dimensional analysis.

Assumptions.

The time period possibly depends on the mass m of the bob, the length l of the pendulum, the angle \theta which the string makes with the vertical, and the acceleration due to gravity g.
The dependence is a product of powers, with a dimensionless constant.

Write the time period as a constant times a product of powers of the possible variables, with unknown powers a, b, c and d.
T=(\text{constant})\,m^{a}\,l^{b}\,\theta^{c}\,g^{d}\qquad(1)
Put the dimensions on both sides. An angle is an arc divided by a radius, so [\theta]=[LL^{-1}]=[M^{0}L^{0}T^{0}]; mass is [M], length [L] and g is [LT^{-2}].
[M^{0}L^{0}T]=(\text{constant})\,[M]^{a}\,[L]^{b}\,[LL^{-1}]^{c}\,[LT^{-2}]^{d}
Collect the powers of each base dimension. The angle contributes nothing.
[M^{0}L^{0}T^{1}]=(\text{constant})\,[M]^{a}\,[L]^{b+d}\,[T]^{-2d}\qquad(2)
Equate the powers of M in (2).
a=0\qquad(3)
Equate the powers of L in (2).
b+d=0\qquad(4)
Equate the powers of T in (2).
-2d=1\ \ \text{or}\ \ d=-\tfrac{1}{2}\qquad(5)
Put d=-\tfrac12 in (4).
b=\tfrac{1}{2}
Put a=0, b=\tfrac12 and d=-\tfrac12 in (1). Since m^{0}=1 the mass drops out. The power c cannot be found because \theta is dimensionless, so it is left in the constant.
T=(\text{constant})\,m^{0}\,l^{1/2}\,g^{-1/2}=(\text{constant})\times\sqrt{\dfrac{l}{g}}

ResultT=2\pi\sqrt{\dfrac{l}{g}}

The time period does not depend on the mass of the bob, and it grows with the square root of the length. Dimensional analysis cannot give the constant; it is found by experiment and is 2\pi.

T=2\pi\sqrt{\dfrac{l}{g}}

Assignment 1.2

Which of the following relationships is dimensionally consistent with an expression yielding a value for acceleration? In these equations x is a distance, t is time and v is velocity. (a) v/t^{2} (b) v/x^{2} (c) v^{2}/t (d) v^{2}/x

Acceleration has dimensions [LT^{-2}]. Find the dimensions of each option, using [v]=[LT^{-1}], [x]=[L] and [t]=[T].
[a]=[LT^{-2}]
Option (a).
\dfrac{[v]}{[t^{2}]}=\dfrac{[LT^{-1}]}{[T^{2}]}=[LT^{-3}]\ne[LT^{-2}]
Option (b).
\dfrac{[v]}{[x^{2}]}=\dfrac{[LT^{-1}]}{[L^{2}]}=[L^{-1}T^{-1}]\ne[LT^{-2}]
Option (c).
\dfrac{[v^{2}]}{[t]}=\dfrac{[L^{2}T^{-2}]}{[T]}=[L^{2}T^{-3}]\ne[LT^{-2}]
Option (d).
\dfrac{[v^{2}]}{[x]}=\dfrac{[L^{2}T^{-2}]}{[L]}=[LT^{-2}]=[a]

\text{(d) }v^{2}/x

1.4Precision and accuracyOften examined

Science is based on observations and experiments, which involve measurements. Precision is how closely repeated measurements of the same physical quantity agree with each other. Accuracy refers to how closely a measurement agrees with a standard or true value.

Precision and accuracy: Precision refers to the closeness of measured values to each other, while accuracy refers to the closeness of a measured value to a standard or true value.

Figure 1.1: Several independent shots at a bullseye target show the difference between accurate and precise. Accuracy is how close an arrow gets to the bull's-eye centre; precision is how close a second arrow is to the first one, regardless of closeness to the target.

A weighing example

A person weighs exactly 160.0 pounds and weighs himself three times on three different scales. The results are:

Scale	Readings (pounds)	True value (pounds)
A	170.1, 169.9, 170.0	160.0
B	161, 162, 158	160.0
C	159.9, 160.0, 160.1	160.0

Scale A is very precise (the readings agree within 0.2 pounds) but not accurate (all are about 10 pounds too high).
Scale B is fairly accurate (the readings are scattered around 160) but not precise (they differ by up to 4 pounds).
Scale C is both precise and accurate.

How precision and accuracy are indicated

The precision of a measurement is associated with the least count of the measuring instrument. The smaller the least count, the greater the precision. Precision is indicated by the absolute uncertainty in the measurement. Accuracy is indicated by the fractional or percentage uncertainty (error) in the measurement: the smaller the fractional or percentage uncertainty, the greater the accuracy.

	Precision	Accuracy
Meaning	Closeness of measured values to each other	Closeness of a measured value to the standard or true value
Indicated by	Absolute uncertainty (set by the least count)	Fractional or percentage uncertainty
Better when	Least count is smaller	Fractional or percentage uncertainty is smaller

Worked example: telling precision from accuracy in three sets of readings

The true length of a steel rod is 50.0\,\mathrm{cm}. Three students each measure it three times with a ruler and record: A: 50.1,\ 50.0,\ 50.2; B: 49.0,\ 51.2,\ 50.4; C: 48.0,\ 48.1,\ 47.9 (all in cm). Find each mean and say which student is precise, which is accurate and which is neither.

Each dot is one reading. The dashed accent line is the true value.

Mean of A.
\bar x_A=\dfrac{50.1+50.0+50.2}{3}=50.1\,\mathrm{cm}
Mean of B.
\bar x_B=\dfrac{49.0+51.2+50.4}{3}=50.2\,\mathrm{cm}
Mean of C.
\bar x_C=\dfrac{48.0+48.1+47.9}{3}=48.0\,\mathrm{cm}
Precision is how closely the readings agree with each other. The spread (largest minus smallest) is 0.2\,\mathrm{cm} for A, 2.2\,\mathrm{cm} for B and 0.2\,\mathrm{cm} for C.
\text{A and C are precise, B is not}
Accuracy is how close the readings are to the true value. A is within 0.1\,\mathrm{cm}. C is about 2\,\mathrm{cm} below the true value in every reading.
\text{A is accurate, C is not}

\text{A: precise and accurate; B: neither; C: precise but not accurate}

1.5UncertaintiesOften examined

In Grade 9 we studied the sources of human error, systematic error and random error in experiments. The difference between the true value and the observed value of a measurement is called error.

\text{Error}=\text{true value}-\text{observed value}

In a measurement the error may occur because of negligence or inexperience of a person, using a faulty apparatus, or an inappropriate method or technique. Errors are divided into three types: personal error, systematic error and random error. Here we study uncertainty.

Uncertainty: The range of possible values within which the true value of the measurement lies. It is a quantitative measurement of the variability in the data.

For example, a measurement of 3.06\,\mathrm{mm}\pm0.02\,\mathrm{mm} means that the experimenter is confident that the actual value of the quantity being measured lies between 3.04\,\mathrm{mm} and 3.08\,\mathrm{mm}.

Why every measurement has uncertainty

All measurements have a degree of uncertainty. It is caused by two factors: the limitations of the measuring instrument (systematic error) and the skill of the experimenter.

Absolute uncertainty

The absolute uncertainty is equal to the least count of a measuring instrument. For example, the length of a glass slab measured with a metre rod is 37.5\,\mathrm{cm}. The least count of the metre rod is 1\,\mathrm{mm}=0.1\,\mathrm{cm}, so the absolute uncertainty in the measured value is \pm0.1\,\mathrm{cm}. An uncertainty of \pm0.05\,\mathrm{cm} develops at each end. If one end of the slab coincides with the 20.5\,\mathrm{cm} mark and the other end with the 58.0\,\mathrm{cm} mark, the length of the slab, with its uncertainty, is

(58.0\pm0.05)\,\mathrm{cm}-(20.5\pm0.05)\,\mathrm{cm}=(37.5\pm0.1)\,\mathrm{cm}

It means that the length of the slab is between 37.4\,\mathrm{cm} and 37.6\,\mathrm{cm}. In this measurement the precision is \pm0.1\,\mathrm{cm}, which is equal to the magnitude of the absolute uncertainty.

Length of a glass slab measured with a metre rod: each end carries an uncertainty of \pm0.05\,\mathrm{cm} (half the least count), so the length carries \pm0.1\,\mathrm{cm}.

Fractional and percentage uncertainty

The accuracy of a measurement is indicated by the magnitude of the fractional error (uncertainty). For the slab above:

\text{Fractional uncertainty}=\dfrac{\text{Absolute uncertainty}}{\text{Measured value}}=\dfrac{\pm0.1}{37.5}=\pm0.003

\text{Percentage uncertainty}=\text{fractional uncertainty}\times100\,\%

The smaller the magnitude of the fractional (relative) uncertainty or error, the greater the accuracy of the measurement.

Worked example: why the same instrument is less accurate for a short length

A metre rod has a least count of 0.1\,\mathrm{cm}, so the absolute uncertainty of every length measured with it is \pm0.1\,\mathrm{cm}. Find the percentage uncertainty when it measures lengths of 2\,\mathrm{cm}, 10\,\mathrm{cm} and 50\,\mathrm{cm}, and draw how the percentage uncertainty changes with the length.

Accuracy is shown by the percentage uncertainty, which is the absolute uncertainty divided by the measured value, times 100\,\%. With \Delta L=0.1\,\mathrm{cm} this is 10/L when L is in cm.
\%\text{ uncertainty}=\dfrac{0.1}{L}\times100=\dfrac{10}{L}
For L=2\,\mathrm{cm}.
\dfrac{10}{2}=5\,\%
For L=10\,\mathrm{cm}.
\dfrac{10}{10}=1\,\%
For L=50\,\mathrm{cm}.
\dfrac{10}{50}=0.2\,\%

The absolute uncertainty is fixed at 0.1\,\mathrm{cm}, but the percentage uncertainty falls as the measured length grows.

5\,\%,\quad1\,\%,\quad0.2\,\%

Kind of uncertainty	Formula	Unit
Absolute	\Delta x (equal to the least count)	same as x
Fractional	\dfrac{\Delta x}{x}	none
Percentage	\dfrac{\Delta x}{x}\times100\,\%	per cent

1.5.1Rules for calculating uncertainties in the final resultOften examined

There are rules for calculating the uncertainty in a final result, but we must be careful whether we use the absolute or the percentage uncertainty in each case. Let x and y be two different physical quantities with uncertainties \Delta x and \Delta y. If z is a physical quantity obtained by operating on x and y, then the propagated uncertainty \Delta z in the result is calculated by the following rules.

(a) Rule for addition and subtraction

If two or more measured quantities are added or subtracted, their absolute uncertainties are added to get the uncertainty in the result.

z=x+y\quad\text{or}\quad z=x-y\qquad\Longrightarrow\qquad\Delta z=\pm(\Delta x+\Delta y)

For example, if x\pm\Delta x=(24.0\pm0.1)\,\mathrm{cm} and y\pm\Delta y=(30.0\pm0.1)\,\mathrm{cm}, then \Delta z=\pm(0.1+0.1)=\pm0.2\,\mathrm{cm}, whether the two lengths are added or subtracted.

(b) Rule for multiplication and division

If two or more quantities are multiplied or divided, their percentage uncertainties are added to get the uncertainty in the result.

\begin{aligned}z&=xy\\ \text{or}\quad z&=\dfrac{x}{y}\\ \Longrightarrow\ \%\text{ uncertainty in }z&=\%\text{ uncertainty in }x+\%\text{ uncertainty in }y\end{aligned}

Worked example: uncertainty in a speed from distance and time

A runner covers a distance s=(120.0\pm0.5)\,\mathrm{m} in a time t=(9.6\pm0.1)\,\mathrm{s}. Find the speed v=s/t with its absolute and percentage uncertainty.

Given

s = (120.0\pm0.5)\,\mathrm{m}
t = (9.6\pm0.1)\,\mathrm{s}

To find

v and \Delta v

Find the speed from the measured values.
v=\dfrac{s}{t}=\dfrac{120.0}{9.6}=12.5\,\mathrm{m\,s^{-1}}
The speed is a quotient, so the percentage uncertainties add. First the percentage uncertainty of s.
\dfrac{0.5}{120.0}\times100=0.42\,\%
Then that of t.
\dfrac{0.1}{9.6}\times100=1.04\,\%
Add them to get the percentage uncertainty of v.
0.42+1.04=1.46\,\%\approx1.5\,\%
Change the percentage back into an absolute uncertainty.
\Delta v=\dfrac{1.46}{100}\times12.5=0.18\,\mathrm{m\,s^{-1}}\approx0.2\,\mathrm{m\,s^{-1}}

v=(12.5\pm0.2)\,\mathrm{m\,s^{-1}}\ \ (\pm1.5\,\%)

(c) Rule for the power of a quantity

The total uncertainty in a power of a quantity is equal to the percentage uncertainty multiplied by that power.

z=x^{3}\qquad\Longrightarrow\qquad\%\text{ uncertainty in }z=\pm3\,(\%\text{ uncertainty in }x)

Operation	Uncertainty to add	Result
z=x\pm y	absolute uncertainties	\Delta z=\pm(\Delta x+\Delta y)
z=xy or z=x/y	percentage uncertainties	\%z=\%x+\%y
z=x^{n}	percentage uncertainty times the power	\%z=n\,(\%x)

(d) Uncertainty in the average of many measurements

The uncertainty in the average value is calculated in three steps.

Find the average of the measured values.
Find the deviation of each value from the average.
The mean deviation is the uncertainty in the average.

\begin{aligned}\bar r&=\dfrac{r_1+r_2+\cdots+r_n}{n}\\ \Delta r_i&=\left|\bar r-r_i\right|\\ \overline{\Delta r}&=\dfrac{\Delta r_1+\Delta r_2+\cdots+\Delta r_n}{n}\end{aligned}

Uncertainty in the average radius of a cylinder

Three readings are recorded for the radius of a small cylinder: r_1=1.50\,\mathrm{cm}, r_2=1.51\,\mathrm{cm} and r_3=1.52\,\mathrm{cm}. Find the uncertainty in the average radius.

Given

r_1 = 1.50\,\mathrm{cm}
r_2 = 1.51\,\mathrm{cm}
r_3 = 1.52\,\mathrm{cm}

To find

average radius \bar r
mean deviation \overline{\Delta r}

Find the average of the three readings.
\bar r=\dfrac{r_1+r_2+r_3}{3}=\dfrac{1.50+1.51+1.52}{3}\,\mathrm{cm}=1.51\,\mathrm{cm}
Find the deviation of each reading from the average (taken as a positive size).
\begin{aligned}\Delta r_1&=1.51-1.50=0.01\,\mathrm{cm}\\ \Delta r_2&=1.51-1.51=0\,\mathrm{cm}\\ \Delta r_3&=\left|1.51-1.52\right|=0.01\,\mathrm{cm}\end{aligned}
Find the mean of the deviations. This is the uncertainty in the average.
\overline{\Delta r}=\dfrac{0.01+0+0.01}{3}\,\mathrm{cm}=0.0067\,\mathrm{cm}\approx0.007\,\mathrm{cm}
State the result with its uncertainty.
r=(1.51\pm0.007)\,\mathrm{cm}

\bar r=1.51\,\mathrm{cm},\qquad\overline{\Delta r}\approx0.007\,\mathrm{cm}

(e) Uncertainty in a timing experiment

The time period T of a vibrating body is found by dividing the time of many vibrations by the number of vibrations.

T=\dfrac{\text{time of multiple vibrations}}{\text{number of vibrations}}

The uncertainty \Delta T in the time period is found by dividing the least count (L.C) of the time-recording device by the number of vibrations.

\Delta T=\dfrac{\text{L.C}}{\text{number of vibrations}}

Time period of a pendulum from 20 vibrations

The time recorded for 20 vibrations of a pendulum is t=35.2\,\mathrm{s}. The least count of the stopwatch is 0.1\,\mathrm{s} (1/10\,\mathrm{s}). Find the time period with its uncertainty.

Given

t = 35.2\,\mathrm{s}
n = 20
\text{L.C} = 0.1\,\mathrm{s}

To find

T
\Delta T

The uncertainty in the measured time is the least count: t=(35.2\pm0.1)\,\mathrm{s}. Divide the time by the number of vibrations.
T=\dfrac{35.2}{20}=1.76\,\mathrm{s}
Divide the least count by the number of vibrations.
\Delta T=\dfrac{0.1}{20}=0.005\,\mathrm{s}
Write the result.
T\pm\Delta T=(1.76\pm0.005)\,\mathrm{s}

T=(1.76\pm0.005)\,\mathrm{s}

Worked examples

Example 1.3

The voltage measured across a conductor is V\pm\Delta V=(7.3\pm0.1) volts and the current is I\pm\Delta I=(2.73\pm0.051) ampere. Find the resistance and the uncertainty in it.

Given

V\pm\Delta V = (7.3\pm0.1)\,\mathrm{V}
I\pm\Delta I = (2.73\pm0.051)\,\mathrm{A}

To find

R\pm\Delta R

Ohm's law gives the resistance.
R=\dfrac{V}{I}=\dfrac{7.3}{2.73}=2.7\,\Omega
Percentage uncertainty in V.
\dfrac{\Delta V}{V}\times100\,\%=\dfrac{0.1}{7.3}\times100\,\%=1.37\,\%\approx1\,\%
Percentage uncertainty in I. The book uses 0.05 for \Delta I here (the value 0.051 rounded).
\dfrac{\Delta I}{I}\times100\,\%=\dfrac{0.05}{2.73}\times100\,\%=1.83\,\%\approx2\,\%
R=V/I is a quotient, so the percentage uncertainties add.
\text{total uncertainty in }R=1\,\%+2\,\%=3\,\%
Convert the percentage back into an absolute uncertainty.
\Delta R=\dfrac{3}{100}\times2.7\,\Omega=0.08\,\Omega
Write the result.
R\pm\Delta R=2.7\,\Omega\pm3\,\%=(2.7\pm0.08)\,\Omega

R=(2.7\pm0.08)\,\Omega

Example 1.4

The radius of a circular disc is measured as 2.25\,\mathrm{cm} with uncertainty \pm0.01\,\mathrm{cm}. Find its surface area with the uncertainty in it.

Given

r = 2.25\,\mathrm{cm}
\Delta r = \pm0.01\,\mathrm{cm}

To find

A\pm\Delta A

Area of a circle.
A=\pi r^{2}=3.14\times(2.25)^{2}=15.90\,\mathrm{cm^{2}}
Percentage uncertainty in the radius.
\dfrac{\Delta r}{r}\times100\,\%=\dfrac{0.01}{2.25}\times100\,\%=0.4\,\%
The area contains r^{2}, so the percentage uncertainty is multiplied by the power 2 (\pi is a constant with no uncertainty).
\dfrac{\Delta A}{A}\times100\,\%=2\times0.4\,\%=0.8\,\%
Absolute uncertainty in the area.
\Delta A=0.8\,\%\times15.90\,\mathrm{cm^{2}}=0.13\,\mathrm{cm^{2}}
Write the result.
A\pm\Delta A=(15.90\pm0.13)\,\mathrm{cm^{2}}

A=(15.90\pm0.13)\,\mathrm{cm^{2}}

Assignment 1.3

The radius of a circle is measured to be (10.5\pm0.2)\,\mathrm{m}. Calculate (a) the area and (b) the circumference of the circle, also give the uncertainty in each value.

Given

r = 10.5\,\mathrm{m}
\Delta r = \pm0.2\,\mathrm{m}

To find

A\pm\Delta A
C\pm\Delta C

Percentage uncertainty in the radius, used for both parts.
\dfrac{\Delta r}{r}\times100\,\%=\dfrac{0.2}{10.5}\times100\,\%=1.9\,\%
(a) Area.
A=\pi r^{2}=3.14\times(10.5)^{2}=346\,\mathrm{m^{2}}
The area has r^{2}, so the percentage uncertainty doubles.
\dfrac{\Delta A}{A}\times100\,\%=2\times1.9\,\%=3.8\,\%
Absolute uncertainty in the area.
\Delta A=0.038\times346\,\mathrm{m^{2}}=13\,\mathrm{m^{2}}
(b) Circumference. It has r to the power 1, so its percentage uncertainty equals that of r.
C=2\pi r=2\times3.14\times10.5=65.9\,\mathrm{m},\qquad\dfrac{\Delta C}{C}\times100\,\%=1.9\,\%
Absolute uncertainty in the circumference.
\Delta C=0.019\times65.9\,\mathrm{m}=1.3\,\mathrm{m}

A=(346\pm13)\,\mathrm{m^{2}},\qquad C=(65.9\pm1.3)\,\mathrm{m}

Formula sheet

Mass from density and volume

m=\rho V

\rho: estimated average density; V: estimated volume

Unit: \mathrm{kg}

Energy to climb a height

E=mgh

m: mass; g: acceleration due to gravity; h: height gained (Example 1.1)

Unit: \mathrm{J}

Newton in base units

\mathrm{N}=\mathrm{kg\,m\,s^{-2}}

force = mass \times displacement / time^{2}

Unit: \mathrm{kg\,m\,s^{-2}}

Joule in base units

\mathrm{J}=\mathrm{N\,m}=\mathrm{kg\,m^{2}\,s^{-2}}

work = force \times displacement

Unit: \mathrm{kg\,m^{2}\,s^{-2}}

Watt in base units

\mathrm{W}=\mathrm{J\,s^{-1}}=\mathrm{kg\,m^{2}\,s^{-3}}

power from Table 1.2

Unit: \mathrm{kg\,m^{2}\,s^{-3}}

Pascal in base units

\mathrm{Pa}=\mathrm{N\,m^{-2}}=\mathrm{kg\,m^{-1}\,s^{-2}}

pressure from Table 1.2

Unit: \mathrm{kg\,m^{-1}\,s^{-2}}

Coulomb in base units

\mathrm{C}=\mathrm{A\,s}

electric charge from Table 1.2

Unit: \mathrm{A\,s}

Dimensions of area, volume, velocity, acceleration

[L^{2}],\ [L^{3}],\ [LT^{-1}],\ [LT^{-2}]

derived from the base dimensions [M],[L],[T],[I],[\Theta],[J],[N]

Unit: none

Principle of homogeneity

[\text{L.H.S.}]=[\text{R.H.S.}]

both sides of a correct equation have the same dimensions

Unit: none

Wavelength of matter waves (dimensional analysis)

\lambda=(\text{constant})\,\dfrac{h}{mv}

h: Planck's constant; m: mass; v: velocity

Unit: \mathrm{m}

Time period of a simple pendulum (dimensional analysis)

T=2\pi\sqrt{\dfrac{l}{g}}

l: length; g: acceleration due to gravity

Unit: \mathrm{s}

Error

\text{Error}=\text{true value}-\text{observed value}

Unit: same as the quantity

Fractional uncertainty

\dfrac{\Delta x}{x}=\dfrac{\text{absolute uncertainty}}{\text{measured value}}

\Delta x: absolute uncertainty (least count); x: measured value

Unit: none

Percentage uncertainty

\dfrac{\Delta x}{x}\times100\,\%

Unit: per cent

Sum or difference

z=x\pm y,\quad\Delta z=\pm(\Delta x+\Delta y)

absolute uncertainties are added

Unit: same as z

Product or quotient

z=xy\ \text{or}\ \dfrac{x}{y},\quad\%z=\%x+\%y

percentage uncertainties are added

Unit: per cent

Power of a quantity

z=x^{n},\quad\%z=n\,(\%x)

percentage uncertainty is multiplied by the power

Unit: per cent

Average and mean deviation

\bar r=\dfrac{r_1+\cdots+r_n}{n},\quad\overline{\Delta r}=\dfrac{\Delta r_1+\cdots+\Delta r_n}{n}

\Delta r_i=|\bar r-r_i|: deviation of each reading; the mean deviation is the uncertainty in the average

Unit: same as r

Time period from many vibrations

T=\dfrac{t}{n},\qquad\Delta T=\dfrac{\text{L.C}}{n}

t: time of n vibrations; L.C: least count of the timer

Unit: \mathrm{s}

Key terms

Qualitative observation: An observation with no numbers, dealing with facts noticed by the five senses, such as colour, shape and texture.
Quantitative observation (measurement): An observation that includes numbers.
Estimation: A rough educated guess of the value of a physical quantity by using prior experience and sound physical reasoning.
Derived unit: A unit obtained by multiplying or dividing base units; it can be expressed in terms of the SI base units.
Dimension: The qualitative nature of a physical quantity, written as a capital letter in square brackets, such as [L].
Dimensional variables: Physical quantities that have dimensions and variable magnitude, such as velocity and force.
Dimensional constants: Physical quantities that have dimensions and a constant magnitude, such as h, G, c and R.
Dimensionless variables: Physical quantities with no dimensions and variable magnitude, such as plane angle, strain and coefficient of friction.
Dimensionless constants: Quantities with no dimensions and constant magnitude, such as pure numbers, e and \pi.
Dimensional analysis: Using dimensions, treated as algebraic quantities, to check an equation, derive a possible formula and determine units.
Principle of homogeneity of dimensions: Both sides of a physically correct equation must have the same dimensions.
Precision: The closeness of measured values to each other; it is indicated by the absolute uncertainty and is greater when the least count is smaller.
Accuracy: The closeness of a measured value to a standard or true value; it is indicated by the fractional or percentage uncertainty.
Error: The difference between the true value and the observed value of a measurement.
Uncertainty: The range of possible values within which the true value of the measurement lies; a quantitative measure of the variability in the data.
Absolute uncertainty: The uncertainty \Delta x in a measurement, equal to the least count of the measuring instrument.
Fractional uncertainty: Absolute uncertainty divided by the measured value, \Delta x/x.
Percentage uncertainty: Fractional uncertainty multiplied by 100\,\%.
Mean deviation: The mean of the deviations of the readings from their average; it is the uncertainty in the average.

Multiple choice questionsp. 19

The mean diameter of a wire is found to be (0.50\pm0.02)\,\mathrm{mm}. The percentage uncertainty in the diameter is:

A2\,\%
B4\,\%Answer
C6\,\%
D8\,\%

Why

The percentage uncertainty is the absolute uncertainty divided by the measured value: \dfrac{0.02}{0.50}\times100\,\%=4\,\%. So B is correct.

A (2\,\%) is the absolute uncertainty 0.02 written as a percentage without dividing by the mean diameter 0.50.

A reaction takes place that is expected to yield 171.9\,\mathrm{g} of product, but it only yields 154.8\,\mathrm{g}. What is the percent error for this experiment?

A17.1\,\%
B90.1\,\%
C111.0\,\%
D9.9\,\%Answer

Why

The percent error compares the error with the expected value: \dfrac{171.9-154.8}{171.9}\times100\,\%=\dfrac{17.1}{171.9}\times100\,\%\approx9.9\,\%. So D is correct.

B (90.1\,\%) is the percentage yield, \dfrac{154.8}{171.9}\times100\,\%, not the error. A is the error in grams, not a percentage.

Three different people weigh a standard mass of 2.00\,\mathrm{g} on the same balance. Each person obtains a reading of exactly 7.32\,\mathrm{g} for the mass of the standard. These results imply that the balance is:

Aboth accurate and precise
Bneither accurate nor precise
Caccurate but not precise
Dprecise but not accurateAnswer

Why

All three readings are the same, 7.32\,\mathrm{g}, so the readings agree closely: the balance is precise. But the true mass is 2.00\,\mathrm{g}, and the readings are 5.32\,\mathrm{g} too high, so the balance is not accurate. So D is correct.

A and C need the readings to be close to 2.00 g, which they are not. B is wrong because the three readings agree exactly, so they are precise.

42026 Annual I, Set B, Q.2 (i)2026 Annual I, Set D, Q.2 (i)

Dimension of universal gravitational constant (G) is:

A[M^{-2}L^{3}T^{-2}]
B[M^{3}L^{-1}T^{-2}]
C[M^{-1}L^{3}T^{-2}]Answer
D[M^{-3}L^{3}T^{-2}]

Why

From F=G\dfrac{m_1m_2}{r^{2}}, G=\dfrac{Fr^{2}}{m_1m_2}. Substituting dimensions: [G]=\dfrac{[MLT^{-2}][L^{2}]}{[M][M]}=[M^{-1}L^{3}T^{-2}]. So C is correct.

A has M^{-2}, which misses the one mass already contained in F. B and D give the wrong sign to the power of M.

A measurement which, on repetition, gives the same or nearly the same result is called:

Aaccurate
Baverage
CpreciseAnswer
Destimated

Why

Repeated readings that agree with each other show precision, which is the closeness of measured values to each other. So C is correct.

Accuracy needs comparison with the true or standard value, which repetition alone cannot show, so A is wrong. An average and an estimate describe how a value is obtained, not how well repeated readings agree, so B and D are wrong.

A student is measuring the time of an event by using a stopwatch. He takes 5 measurements as: 3.0\,\mathrm{s}, 3.2\,\mathrm{s}, 3.4\,\mathrm{s}, 2.8\,\mathrm{s}, 3.1\,\mathrm{s}. What is the uncertainty in the results?

A\pm0.3\,\mathrm{s}Answer
B\pm0.6\,\mathrm{s}
C\pm3.1\,\mathrm{s}
D\pm7.75\,\mathrm{s}

Why

For a small set of readings the uncertainty is taken as half the spread of the readings: \Delta t=\dfrac{t_{\max}-t_{\min}}{2}=\dfrac{3.4-2.8}{2}=0.3\,\mathrm{s}. This is also the largest deviation from the average, \bar t=\dfrac{15.5}{5}=3.1\,\mathrm{s}, so the result is (3.1\pm0.3)\,\mathrm{s}.

B (\pm0.6\,\mathrm{s}) is the whole spread, not half of it. C is the average itself, not an uncertainty, and D has no meaning here.

Note. The book prints no answer. The mean deviation of these readings is 0.16\,\mathrm{s}, which is not an option, so the intended method is half the spread (equal here to the largest deviation from the mean), giving A.

Which of the following quantity has a different dimension?

Aforce
Bweight
Cmodulus of elasticityAnswer
Dtension

Why

Force, weight and tension are all forces, with dimension [MLT^{-2}]. Weight is mg, which has the dimensions of force, so B is not the odd one out.

Modulus of elasticity is a stress, a force per unit area: \dfrac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}]. This differs from [MLT^{-2}], so C is correct.

If the dimensions of a physical quantity are given by [L^{a}M^{b}T^{c}], then the physical quantity will be:

Aforce, if a=-1, b=0, c=-2
Bpressure, if a=-1, b=1, c=-2Answer
Cvelocity, if a=1, b=0, c=1
Dacceleration, if a=1, b=1, c=-2

Why

Pressure is \dfrac{\text{force}}{\text{area}}=\dfrac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}], which matches a=-1, b=1, c=-2. So B is correct.

A is wrong because force is [MLT^{-2}], so a=1 and b=1. C is wrong because velocity [LT^{-1}] needs c=-1. D is wrong because acceleration [LT^{-2}] needs b=0.

Order of magnitude of (10^{6}+10^{3}) is:

A10^{18}
B10^{9}
C10^{6}Answer
D10^{3}

Why

10^{6}+10^{3}=1\,000\,000+1\,000=1.001\times10^{6}. Its order of magnitude is 10^{6}, so C is correct.

B (10^{9}) comes from multiplying the powers, 10^{6}\times10^{3}, instead of adding the numbers. The term 10^{3} is too small to change the order of magnitude, so D is wrong.

Which of the following may be used as a valid formula to calculate the speed of ocean waves? [v = speed, g = acceleration due to gravity, \lambda = wavelength, \rho = density, h = depth].

Av=\sqrt{\lambda g}Answer
Bv=\rho g h
Cv=gh/\lambda
Dv=\lambda g h

Why

Check the dimensions. [\lambda g]=L\cdot LT^{-2}=L^{2}T^{-2}, so \sqrt{\lambda g} has dimension LT^{-1}, the dimension of speed. So A is dimensionally correct.

[\rho gh]=ML^{-3}\cdot LT^{-2}\cdot L=ML^{-1}T^{-2} (B is not a speed). [gh/\lambda]=LT^{-2} (C is not a speed). [\lambda gh]=L^{3}T^{-2} (D is not a speed).

Note. The book labels both the second and the third option C; the second option is relabelled B here. Dimensional analysis checks only dimensions, so it cannot fix a dimensionless constant (unit 1.3.2, limitation 3).

Short questionsp. 20

1.1

Draw a table to show a reasonable estimate of some physical quantities.

Solution

Each estimate is given to the nearest power of ten, using the values of Table 1.1 of this unit.

Reasonable estimates of some physical quantities.
Physical quantity	Estimate	Unit
Diameter of a hydrogen atom	10^{-10}	m
Width of a pinky fingernail	10^{-2}	m
Height of a 4-year-old child	1	m
Length of a football ground	10^{2}	m
Diameter of the Earth	10^{7}	m
Mass of a hummingbird	10^{-2}	kg
Mass of 1 litre of water	1	kg
Mass of a motorcycle	10^{2}	kg
Time for one heartbeat	1	s
One day	10^{5}	s

1.2

Express the units of the following derived quantities in terms of base units. (a) Force (b) Work (c) Power (d) Pressure (e) Electric charge.

Solution

(a) Force.

Force is mass times acceleration, and acceleration has the unit \mathrm{m\,s^{-2}}.

F=m\,a=\mathrm{kg}\times\mathrm{m\,s^{-2}}=\mathrm{kg\,m\,s^{-2}}

\mathrm{kg\,m\,s^{-2}}

(b) Work.

Work is force times displacement.

W=F\,d=\mathrm{kg\,m\,s^{-2}}\times\mathrm{m}=\mathrm{kg\,m^{2}\,s^{-2}}

\mathrm{kg\,m^{2}\,s^{-2}}

Power is work divided by time.

P=\dfrac{W}{t}=\dfrac{\mathrm{kg\,m^{2}\,s^{-2}}}{\mathrm{s}}=\mathrm{kg\,m^{2}\,s^{-3}}

\mathrm{kg\,m^{2}\,s^{-3}}

(d) Pressure.

Pressure is force divided by area.

P=\dfrac{F}{A}=\dfrac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m^{2}}}=\mathrm{kg\,m^{-1}\,s^{-2}}

\mathrm{kg\,m^{-1}\,s^{-2}}

(e) Electric charge.

Charge is current times time, and the ampere is a base unit.

Q=I\,t=\mathrm{A}\times\mathrm{s}=\mathrm{A\,s}

\mathrm{A\,s}

1.3

Why is it important to use an instrument with the smallest resolution?

Solution

The resolution (least count) of an instrument is the smallest change in a quantity that it can show. The absolute uncertainty of a reading equals the least count, so a smaller least count gives a smaller uncertainty and a more precise measurement.

For example, the length of a slab measured with a meter rule of least count 0.1\,\mathrm{cm} is (37.5\pm0.1)\,\mathrm{cm}, a fractional uncertainty of \dfrac{0.1}{37.5}\approx0.3\,\%. A ruler with divisions of 1\,\mathrm{cm} could only give (38\pm1)\,\mathrm{cm}, a fractional uncertainty of about 3\,\%, roughly ten times larger.

So the instrument with the smallest resolution gives the smallest uncertainty, which means more precise readings and a result with a smaller fractional uncertainty, which is more accurate.

1.42026 Annual I, Set B, Q.2 (viii)

What is the importance of increasing the number of readings in an experiment?

Solution

Every reading contains some random error, so single readings scatter on either side of the true value. Taking more readings and averaging them lets these random errors partly cancel, so the average is more reliable.

More readings also show the spread of the data. The mean deviation of the readings from their average is the uncertainty in the average, so a larger set gives a clearer picture of how well the readings agree (precision). Extra readings also help to spot a reading that lies far from the others, so that it can be checked.

1.52026 Annual I, Set D, Q.2 (v) OR2025 Annual I, Set B, Q.2 (i)

What is the difference between precision and accuracy?

Solution

Precision is the closeness of measured values to each other. It is indicated by the absolute uncertainty, which equals the least count of the instrument.

Accuracy is the closeness of a measured value to the standard or true value. It is indicated by the fractional or percentage uncertainty: the smaller it is, the more accurate the measurement.

Example from the book: a person whose true weight is 160.0 lb weighs on three scales. Scale A reads 170.1, 169.9 and 170.0 lb, so it is precise (the readings agree) but not accurate (about 10 lb too high). Scale B reads 161, 162 and 158 lb, so it is fairly accurate but not precise. Scale C reads 159.9, 160.0 and 160.1 lb, so it is both precise and accurate.

1.6

What is the principle of homogeneity of dimensions?

Solution

The principle states that both sides of a physically correct equation must have the same dimensions. Quantities can be added or subtracted only if they have the same dimensions, so every term of an equation must have the dimensions of the left side.

For example, in v_f=v_i+at the left side has dimension [LT^{-1}]. The right side has [LT^{-1}] from v_i and [LT^{-2}][T]=[LT^{-1}] from at. Both sides agree, so the equation is homogeneous.

1.7

A ball is thrown in the air and 5 different students are individually measuring the time it takes to fall back down using stopwatches. The times obtained by each student are the following: 6.2\,\mathrm{s}, 6.0\,\mathrm{s}, 6.4\,\mathrm{s}, 6.1\,\mathrm{s}, 5.8\,\mathrm{s}. (i) What is the uncertainty of the results? (ii) How should the resulting time be expressed?

Solution

(i) Uncertainty of the results.

Find the average of the five readings.
\bar t=\dfrac{6.2+6.0+6.4+6.1+5.8}{5}=\dfrac{30.5}{5}=6.1\,\mathrm{s}
Find the deviation of each reading from the average, ignoring the sign: 0.1, 0.1, 0.3, 0.0 and 0.3 s.
|\Delta t|=0.1,\ 0.1,\ 0.3,\ 0.0,\ 0.3\ \mathrm{s}
The uncertainty in the average is the mean deviation.
\Delta t=\dfrac{0.1+0.1+0.3+0.0+0.3}{5}=\dfrac{0.8}{5}=0.16\,\mathrm{s}\approx0.2\,\mathrm{s}

\Delta t\approx\pm0.2\,\mathrm{s}

(ii) How the resulting time should be expressed.

The result is written as the average plus or minus its uncertainty, with the average given to the same decimal place as the uncertainty.

t=(6.1\pm0.2)\,\mathrm{s}

1.8

The energy of a photon is given by E=hf, where f is frequency. Find the dimensions of Planck's constant h.

Solution

Rearrange the equation as h=\dfrac{E}{f}. Energy is work, with dimension [ML^{2}T^{-2}] (unit \mathrm{kg\,m^{2}\,s^{-2}}), and frequency has dimension [T^{-1}].

[h]=\dfrac{[ML^{2}T^{-2}]}{[T^{-1}]}=[ML^{2}T^{-1}]

[h]=[ML^{2}T^{-1}]

1.9

Justify why all measurements contain some uncertainty.

Solution

No measurement can be exact, for two reasons. First, every instrument has a limited least count, so a reading can only be made to within that value, and the absolute uncertainty is at least equal to it. Second, the experimenter's skill limits the result: personal errors (negligence or inexperience), systematic errors (a faulty apparatus or an inappropriate method) and random errors that change unpredictably from one reading to the next.

Because the true value is never known exactly, a measurement is always given as a range of possible values, for example (37.5\pm0.1)\,\mathrm{cm}.

Comprehensive questionsp. 20

1.1

Define and explain the term uncertainty.

Solution

Definition

Uncertainty is the range of possible values within which the true value of a measurement lies. It is a quantitative measure of the variability in the data. For example, 3.06\,\mathrm{mm}\pm0.02\,\mathrm{mm} means that the experimenter is confident that the true value lies between 3.04\,\mathrm{mm} and 3.08\,\mathrm{mm}.

The uncertainty range of 3.06\,\mathrm{mm}\pm0.02\,\mathrm{mm}: the true value lies somewhere between 3.04 mm and 3.08 mm.

Why every measurement has an uncertainty

All measurements have a degree of uncertainty. It is caused by the limitations of the measuring instrument, which fixes the least count, and by the skill of the experimenter. Errors are of three kinds: personal error (negligence or inexperience), systematic error (from a faulty apparatus or an inappropriate method) and random error (unpredictable changes from one reading to the next).

Absolute and fractional uncertainty

The absolute uncertainty is equal to the least count of the instrument. A meter rule with a least count of 0.1\,\mathrm{cm} gives an absolute uncertainty of \pm0.1\,\mathrm{cm} for each end reading.

\begin{aligned}\text{Fractional uncertainty}&=\dfrac{\text{absolute uncertainty}}{\text{measured value}}=\dfrac{\Delta x}{x}\\ \text{Percentage uncertainty}&=\dfrac{\Delta x}{x}\times100\,\%\end{aligned}

The absolute uncertainty indicates precision, and the fractional or percentage uncertainty indicates accuracy. For a slab of length (37.5\pm0.1)\,\mathrm{cm}, the fractional uncertainty is \dfrac{0.1}{37.5}\approx\pm0.003, which is \pm0.3\,\%.

1.2

Discuss the rules for calculating uncertainty propagation in the final results in different cases.

Solution

Let x and y be two measured quantities with uncertainties \Delta x and \Delta y. If z is obtained by operating on x and y, the propagated uncertainty \Delta z is found by the rules below. Absolute and percentage uncertainties must not be mixed up.

Rules for propagated uncertainty.
Case	Relation	Rule
Addition or subtraction	z=x\pm y	\Delta z=\Delta x+\Delta y (absolute uncertainties add)
Multiplication or division	z=xy or z=x/y	\%\Delta z=\%\Delta x+\%\Delta y
Power	z=x^{n}	\%\Delta z=n\times\%\Delta x
Average of many readings	\bar x	\Delta\bar x is the mean deviation of the readings
Timing a number of vibrations	T=t/n	\Delta T=\mathrm{LC}/n, where LC is the least count

Addition and subtraction

If two or more measured quantities are added or subtracted, their absolute uncertainties are added. Take x=(24.0\pm0.1)\,\mathrm{cm} and y=(30.0\pm0.1)\,\mathrm{cm}, and z=x+y.

Add the values.
z=24.0+30.0=54.0\,\mathrm{cm}
Add the absolute uncertainties.
\Delta z=0.1+0.1=0.2\,\mathrm{cm}

z=(54.0\pm0.2)\,\mathrm{cm}

Multiplication and division

If quantities are multiplied or divided, their percentage uncertainties are added. Take Example 1.3 of this unit: V=(7.3\pm0.1)\,\mathrm{V} and I=(2.73\pm0.051)\,\mathrm{A}, with R=V/I.

Percentage uncertainty in V.
\dfrac{0.1}{7.3}\times100\,\%=1.37\,\%\approx1\,\%
Percentage uncertainty in I.
\dfrac{0.051}{2.73}\times100\,\%=1.87\,\%\approx2\,\%
Find R and add the percentage uncertainties.
R=\dfrac{7.3}{2.73}=2.7\,\Omega,\qquad \%\Delta R=1\,\%+2\,\%=3\,\%
Find the absolute uncertainty in R.
\Delta R=\dfrac{3}{100}\times2.7\,\Omega=0.08\,\Omega

R=(2.7\pm0.08)\,\Omega

Powers

If a quantity is raised to a power n, its percentage uncertainty is multiplied by n. For the area of a circular disc, A=\pi r^{2}, with r=(2.25\pm0.01)\,\mathrm{cm} (Example 1.4 of this unit):

Percentage uncertainty in r.
\dfrac{0.01}{2.25}\times100\,\%\approx0.4\,\%
The power 2 doubles it.
\%\Delta A=2\times0.4\,\%=0.8\,\%
Area and its absolute uncertainty.
A=\pi(2.25)^{2}=15.90\,\mathrm{cm^{2}},\qquad \Delta A=0.008\times15.90=0.13\,\mathrm{cm^{2}}

A=(15.90\pm0.13)\,\mathrm{cm^{2}}

Average of many measurements

The uncertainty in an average is the mean deviation of the readings from their average. Take three readings of a radius: r_1=1.50\,\mathrm{cm}, r_2=1.51\,\mathrm{cm} and r_3=1.52\,\mathrm{cm}.

Average.
\bar r=\dfrac{1.50+1.51+1.52}{3}=1.51\,\mathrm{cm}
Deviation of each reading from the average.
|\Delta r_1|=0.01\,\mathrm{cm},\quad |\Delta r_2|=0\,\mathrm{cm},\quad |\Delta r_3|=0.01\,\mathrm{cm}
Mean deviation, the uncertainty in the average.
\Delta\bar r=\dfrac{0.01+0+0.01}{3}=0.0067\,\mathrm{cm}\approx0.007\,\mathrm{cm}

\bar r=(1.51\pm0.007)\,\mathrm{cm}

Timing experiments

A period is found by dividing the time for several vibrations by their number, T=t/n. Its uncertainty is the least count of the timer divided by n. Take t=(35.2\pm0.1)\,\mathrm{s} for n=20 vibrations, timed with a stopwatch of least count 0.1\,\mathrm{s}.

Period.
T=\dfrac{35.2}{20}=1.76\,\mathrm{s}
Uncertainty in the period.
\Delta T=\dfrac{0.1}{20}=0.005\,\mathrm{s}

T=(1.76\pm0.005)\,\mathrm{s}

1.3

What does the dimension of a physical quantity mean? What are its advantages? Explain with examples.

Solution

Meaning of dimension

The dimension of a physical quantity denotes its qualitative nature, such as length or time, and does not depend on the units used. It is written with a capital letter in square brackets: [L] for length. Length, width, height, distance and radius all have the same dimension, [L], because they have the same nature.

Dimensions of the base quantities.
Base quantity	Dimension
Mass	[M]
Length	[L]
Time	[T]
Electric current	[I]
Temperature	[\theta]
Intensity of light	[J]
Amount of substance	[N]

Dimensions of derived quantities are products and quotients of the base dimensions. For example, area is [L^{2}], volume is [L^{3}], velocity is [LT^{-1}] and acceleration is [LT^{-2}]. Pure numbers and plane angles have no dimensions.

Advantages of dimensions

Dimensions can be treated as algebraic quantities, so the method of dimensions (dimensional analysis) can be used in three ways: to check the homogeneity of an equation, to derive a possible formula, and to find the units of a quantity.

Checking homogeneity

Take the equation v_f=v_i+at. Quantities can be added only if they have the same dimensions, so the check is:

Dimensions of the left side.
[v_f]=[LT^{-1}]
Dimensions of the right side.
[v_i]+[at]=[LT^{-1}]+[LT^{-2}][T]=[LT^{-1}]+[LT^{-1}]=2[LT^{-1}]

The number 2 is a dimensionless constant, so both sides have dimension [LT^{-1}] and the equation is dimensionally correct.

Deriving a formula: the simple pendulum

Derivation Time period of a simple pendulum

Aim. Find how the time period T depends on the mass m of the bob, the length l, the angle \theta and the acceleration due to gravity g.

Assumptions.

The period may depend on m, l, \theta and g, and on no other quantity.
The angle \theta is a ratio of two lengths, so it has no dimensions.

A simple pendulum: the bob of mass m swings through the angle \theta from the vertical, and the string has length l.

Write the relation with unknown powers, where k is a constant.
T=k\,m^{a}\,l^{b}\,\theta^{c}\,g^{d}
Replace each quantity by its dimensions. \theta has none.
[T]=[M]^{a}[L]^{b}[1]^{c}[LT^{-2}]^{d}=[M^{a}][L^{b+d}][T^{-2d}]
Equate the powers of M.
a=0
Equate the powers of L.
b+d=0
Equate the powers of T.
-2d=1\quad\Rightarrow\quad d=-\dfrac12
Find b from the power of L.
b=-d=\dfrac12
Substitute the powers into the relation. The power c cannot be found, because heta has no dimensions.
T=k\,l^{1/2}g^{-1/2}=k\sqrt{\dfrac{l}{g}}

ResultT=k\sqrt{\dfrac{l}{g}}

The period does not depend on the mass of the bob. The constant k is found by experiment to be 2\pi, so T=2\pi\sqrt{l/g}.

Deriving a formula: matter waves

Derivation Wavelength of matter waves

Aim. Find how the wavelength \lambda of matter waves depends on Planck's constant h, the mass m of the particle and its speed v.

Assumptions.

\lambda may depend on h, m and v only.

Write the relation with unknown powers.
\lambda=k\,h^{a}m^{b}v^{c}
Replace each quantity by its dimensions. Planck's constant has dimension [ML^{2}T^{-1}].
[L]=[ML^{2}T^{-1}]^{a}[M]^{b}[LT^{-1}]^{c}=[M^{a+b}L^{2a+c}T^{-a-c}]
Equate the powers of M, L and T.
a+b=0,\qquad 2a+c=1,\qquad -a-c=0
Solve the three equations.
a=1,\qquad b=-1,\qquad c=-1
Substitute the powers into the relation.
\lambda=k\,h\,m^{-1}v^{-1}=k\,\dfrac{h}{mv}

Result\lambda=k\,\dfrac{h}{mv}

The constant k is dimensionless, so dimensional analysis cannot find it. Rearranging gives [h]=[\lambda][m][v]=[ML^{2}T^{-1}], so the unit of h is \mathrm{kg\,m^{2}\,s^{-1}}.

Limitations of dimensional analysis

It does not distinguish quantities that have the same dimensions, for example work, energy and torque all have [ML^{2}T^{-2}].
It cannot derive formulas containing trigonometric, exponential or logarithmic functions.
It cannot determine dimensionless constants such as k above.
A dimensionally correct equation is not always physically correct, but a dimensionally wrong equation is always wrong.

1.4

What is meant by estimation of a physical quantity? Explain with examples.

Solution

Meaning of estimation

An estimation is a rough educated guess of the value of a physical quantity, made by using prior experience and sound physical reasoning. It is not a random guess of a formula or a number. An estimation identifies the correct physical principles and then makes a good guess about the relevant variables.

Strategies

Length: break a big thing into smaller things. Estimate the length of one smaller part and multiply by the number of parts. For example, the height of a building is about the number of floors times the height of one floor.
Reverse: estimate the length of a small thing that makes up a big one. For example, the thickness of one sheet of paper is the thickness of a stack divided by the number of pages.
Area and volume: model a complicated object as a sphere or a box, estimate its linear dimensions, and use a standard geometric formula. The reverse also works.
Mass from volume and density: estimate the volume, then use density, \rho=\dfrac{m}{V}. Air has a density of about 1\,\mathrm{kg\,m^{-3}}, water about 10^{3}\,\mathrm{kg\,m^{-3}}, and the densest everyday solids about 10^{4}\,\mathrm{kg\,m^{-3}}.

Example 1: energy to climb stairs

Estimate the energy an adult man needs to walk up the stairs from the ground floor to the first floor. The energy needed is E=mgh. An adult has a mass of about 70\,\mathrm{kg} and the distance between two floors is about 3\,\mathrm{m}.

Substitute the estimates, with g=10\,\mathrm{m\,s^{-2}}.
E=70\,\mathrm{kg}\times10\,\mathrm{m\,s^{-2}}\times3\,\mathrm{m}
Evaluate.
E=2100\,\mathrm{kg\,m^{2}\,s^{-2}}=2100\,\mathrm{J}

E\approx2.1\times10^{3}\,\mathrm{J}

Example 2: heartbeats in a lifetime

Table 1.1 gives a human lifetime of about 10^{9}\,\mathrm{s} and the time for one heartbeat as about 1\,\mathrm{s}. The number of heartbeats is therefore about

N=\dfrac{10^{9}\,\mathrm{s}}{1\,\mathrm{s}}=10^{9}\ \text{heartbeats}

Both examples use the same method: identify the physics, pick sensible values from experience or a table, and keep only one significant figure.

Numerical problemsp. 20–21

1.1

Estimate number of heartbeats in a lifetime of 60 years?

Solution

Given

t = 60\,\mathrm{years}
T_{\text{beat}} = 1\,\mathrm{s} time for one heartbeat, from Table 1.1

To find

N, the number of heartbeats

Convert the lifetime to seconds, using 1\,\mathrm{yr}\approx3.15\times10^{7}\,\mathrm{s}.
t=60\times3.15\times10^{7}\,\mathrm{s}\approx1.9\times10^{9}\,\mathrm{s}
Divide by the time for one heartbeat.
N=\dfrac{t}{T_{\text{beat}}}=\dfrac{1.9\times10^{9}}{1}\approx10^{9}

N\approx10^{9}\ \text{heartbeats}

Book answer. 10^{9}

1.2

Determine the dimensions of each of the following quantities. Here v is speed, a is acceleration, x is distance and t is time.

Solution

(a) \dfrac{v^{2}}{ax}

Dimensions of the numerator.
[v^{2}]=[LT^{-1}]^{2}=[L^{2}T^{-2}]
Dimensions of the denominator.
[ax]=[LT^{-2}][L]=[L^{2}T^{-2}]
Divide. The quotient has no dimensions.
\left[\dfrac{v^{2}}{ax}\right]=\dfrac{[L^{2}T^{-2}]}{[L^{2}T^{-2}]}=[M^{0}L^{0}T^{0}]

[M^{0}L^{0}T^{0}]\ \text{(dimensionless)}

(b) \dfrac{at^{2}}{2}

Dimensions of at^{2}.
[at^{2}]=[LT^{-2}][T]^{2}=[L]
The 2 is a pure number, so it has no dimensions.
\left[\dfrac{at^{2}}{2}\right]=[L]

[L]

Book answer. (a) No, (b) [L]

Note. The book's 'No' in (a) means the quantity has no dimensions, so the answer is [M^{0}L^{0}T^{0}], that is, dimensionless.

1.3

If A=\dfrac{X^{2}}{Y^{2}Z}, then find the percentage uncertainty in A. The percentage uncertainties in X, Y and Z are 1\,\%, 1\,\% and 2\,\% respectively.

Solution

Given

\%\Delta X = 1\,\%
\%\Delta Y = 1\,\%
\%\Delta Z = 2\,\%

To find

\%\Delta A

Power rule for X^{2}: the power 2 doubles the percentage uncertainty.
\%\Delta(X^{2})=2\times1\,\%=2\,\%
Power rule for Y^{2} in the denominator.
\%\Delta(Y^{2})=2\times1\,\%=2\,\%
Add all the percentage uncertainties, because the quantities are multiplied and divided.
\%\Delta A=2\,\%+2\,\%+2\,\%=6\,\%

\%\Delta A=6\,\%

Book answer. 6\,\%

1.4

A spherical ball of radius r experiences a resistive force F due to the air as it moves through it at speed v. The resistive force is given by the expression F=c\,r\,v, where c is a constant. By using dimensions, derive the SI base unit of the constant c.

Solution

Rearrange for c.
c=\dfrac{F}{r\,v}
Dimensions of force, radius and speed.
[F]=[MLT^{-2}],\qquad [r]=[L],\qquad [v]=[LT^{-1}]
Substitute the dimensions.
[c]=\dfrac{[MLT^{-2}]}{[L][LT^{-1}]}=[ML^{-1}T^{-1}]
Replace each dimension by its SI base unit.
[ML^{-1}T^{-1}]=\mathrm{kg\,m^{-1}\,s^{-1}}

\mathrm{kg\,m^{-1}\,s^{-1}}

Book answer. \mathrm{kg\,m^{-1}\,s^{-1}}

1.52025 Annual I, Set B, Q.2 (xiii)2025 Annual I, Set D, Q.2 (i)

The pressure P at a depth h in an incompressible fluid of density \rho is given by P=\rho g h, where g is acceleration due to gravity. Check the homogeneity of this equation.

Solution

Dimensions of the left side. Pressure is force per unit area.
[P]=\dfrac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}]
Dimensions of the right side. Density is mass per unit volume.
[\rho][g][h]=[ML^{-3}]\,[LT^{-2}]\,[L]=[ML^{-1}T^{-2}]
Compare the two sides.
[P]=[\rho g h]=[ML^{-1}T^{-2}]

Both sides have the same dimensions, so the equation is homogeneous and dimensionally correct.

[P]=[\rho g h]=[ML^{-1}T^{-2}]\quad\text{(homogeneous)}

1.6

Estimate how many protons are there in a bacterium. (Take mass of bacterium as 10^{-15}\,\mathrm{kg} and mass of proton 10^{-27}\,\mathrm{kg}.)

Solution

Given

m_{\text{bact}} = 10^{-15}\,\mathrm{kg}
m_p = 10^{-27}\,\mathrm{kg}

To find

N, the number of protons

Divide the mass of the bacterium by the mass of one proton.
N=\dfrac{m_{\text{bact}}}{m_p}=\dfrac{10^{-15}\,\mathrm{kg}}{10^{-27}\,\mathrm{kg}}=10^{12}

N\approx10^{12}\ \text{protons}

Book answer. 10^{12} protons

Note. This is a rough estimate that takes the whole mass of the bacterium to be protons, as the question intends.

1.7

Estimate how many hydrogen atoms does it take to stretch across the diameter of the Sun? (Take diameter of the Sun as 10^{5}\,\mathrm{km} and diameter of proton 10^{-14}\,\mathrm{km}.)

Solution

Given

D_{\text{Sun}} = 10^{5}\,\mathrm{km}
d = 10^{-14}\,\mathrm{km} diameter of one atom, as given

To find

N, the number of hydrogen atoms

Convert both diameters to metres, using 1\,\mathrm{km}=10^{3}\,\mathrm{m}.
D_{\text{Sun}}=10^{5}\times10^{3}\,\mathrm{m}=10^{8}\,\mathrm{m},\qquad d=10^{-14}\times10^{3}\,\mathrm{m}=10^{-11}\,\mathrm{m}
Divide the diameter of the Sun by the diameter of one atom.
N=\dfrac{D_{\text{Sun}}}{d}=\dfrac{10^{8}\,\mathrm{m}}{10^{-11}\,\mathrm{m}}=10^{19}

N\approx10^{19}\ \text{hydrogen atoms}

Book answer. 10^{19} hydrogen atoms

Note. The answer follows the data given in the question. Table 1.1 gives 10^{-10}\,\mathrm{m} for a hydrogen atom, which would give 10^{18} atoms.

1.8

The current passing through a resistor R=(13\pm0.5)\,\Omega is I=(3\pm0.1)\,\mathrm{A}.

Solution

Given

R = (13\pm0.5)\,\Omega
I = (3\pm0.1)\,\mathrm{A}

To find

(a) P, correct to one significant figure
(b) \%\Delta I
(c) \%\Delta R
(d) \Delta P

The power consumed in a resistor is P=I^{2}R.

(a) Calculate the power consumed, correct to one significant figure.

Substitute the central values.
P=(3)^{2}\times13=9\times13=117\,\mathrm{W}

P\approx1\times10^{2}\,\mathrm{W}

(b) Find the percentage uncertainty of the current passing through the resistor.

Divide the uncertainty by the value and convert to a percentage.
\%\Delta I=\dfrac{0.1}{3}\times100\,\%=3.3\,\%\approx3\,\%

3\,\%

Divide the uncertainty by the value and convert to a percentage.
\%\Delta R=\dfrac{0.5}{13}\times100\,\%=3.85\,\%

3.85\,\%

(d) Find the absolute uncertainty of the electrical power.

Power is I^{2}R, so the percentage uncertainty in P is the percentage uncertainty in R plus twice that in I.
\%\Delta P=\%\Delta R+2\,\%\Delta I=3.85\,\%+2\times3.3\,\%=10.5\,\%
Take this percentage of P=117\,\mathrm{W}.
\Delta P=\dfrac{10.5}{100}\times117\,\mathrm{W}=12.3\,\mathrm{W}\approx12\,\mathrm{W}

\Delta P\approx12\,\mathrm{W}

Book answer. 117 W, 3 %, 3.84 %, 11.7 W

Note. (a) The book prints 117\,\mathrm{W}, the unrounded value; to one significant figure the answer is 1\times10^{2}\,\mathrm{W}. (c) 0.5/13 is 3.85\,\%; the book prints 3.84\,\%, a rounding difference. (d) The rules of this unit give \Delta P\approx12\,\mathrm{W}. The book's 11.7\,\mathrm{W} is exactly 10\,\% of 117\,\mathrm{W} and does not follow from its own percentage uncertainties.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 11 mark

Which of the following equations is dimensionally correct?

Av_f = v_i + atAnswer
Bv_f^2 = v_i^2 + 2at
Cs = v_i t + \dfrac{1}{2}at
DF = ma^2

Why

An equation is dimensionally correct only if every term has the same dimensions. In option A, [v_f]=[v_i]=[LT^{-1}] and [at]=[LT^{-2}][T]=[LT^{-1}], so every term has dimensions [LT^{-1}]. So A is correct.

The tempting options look like equations of motion but are wrong. In B the term 2at has dimensions [LT^{-1}], while v^{2} has [L^{2}T^{-2}] (the correct equation has 2aS, with [LT^{-2}][L]=[L^{2}T^{-2}]). In C the term \tfrac12at has dimensions [LT^{-1}] but s has [L] (the correct term is \tfrac12at^{2}). In D, [ma^{2}]=[M][LT^{-2}]^{2}=[ML^{2}T^{-4}], which is not the dimensions of force [MLT^{-2}].

2025 Annual I, Set B, MCQ 11 mark

The dimensions \left[ML^2T^{-2}\right] show:

APower
BWorkAnswer
CVelocity
DAcceleration

Why

Work is force times displacement, so [W]=[F][s]=[MLT^{-2}][L]=[ML^{2}T^{-2}]. These are the dimensions given, so the quantity is work (energy has the same dimensions). So B is correct.

Option A (power) is work per unit time, [ML^{2}T^{-2}]/[T]=[ML^{2}T^{-3}], which has an extra T^{-1}. Velocity is [LT^{-1}] and acceleration is [LT^{-2}], and neither contains mass.

2026 Annual I, Set B, MCQ 21 mark

Pascal (Pa) is equal to:

AN/m
BN/m^2Answer
CN^2/m^2
Dkg/m

Why

The pascal is the SI unit of pressure. Pressure is force per unit area, so 1\,\mathrm{Pa}=1\,\mathrm{N\,m^{-2}}. In base units, \mathrm{N\,m^{-2}}=\mathrm{kg\,m\,s^{-2}}\,\mathrm{m^{-2}}=\mathrm{kg\,m^{-1}\,s^{-2}}. So B is correct.

Option A, N/m, is force per unit length (the unit of surface tension), not force per unit area. Option C squares both force and area, which does not give pressure. Option D, kg/m, has the wrong base units: pressure is \mathrm{kg\,m^{-1}\,s^{-2}}, so a \mathrm{s^{-2}} is missing.

2025 Annual I, Set B, MCQ 21 mark

If r = 2.25 \pm 0.01\,\mathrm{cm} then percentage uncertainty in 'r' will be:

A0.2%
B0.3%
C0.4%Answer
D0.5%

Why

Percentage uncertainty is the absolute uncertainty divided by the measured value, times 100: \dfrac{0.01}{2.25}\times100\,\%=0.444\,\%\approx0.4\,\%. An uncertainty is quoted to one significant figure, so C is correct.

Option A (0.2\,\%) would need an absolute uncertainty of about 0.005\,\mathrm{cm}, and option D (0.5\,\%) comes from rounding 0.444 upwards, which is not how 0.444 rounds to one figure.

2026 Annual I, Set B, Q.2 (i)3 marks

Deduce the dimensions of k from Coulomb's law \left(F = k\dfrac{q_1q_2}{r^2}\right).

Solution

Coulomb's law is F=k\dfrac{q_1q_2}{r^{2}}. Make k the subject, then put the dimensions of each quantity on the right (1 mark).

Rearrange (1 mark).
k=\dfrac{F\,r^{2}}{q_1q_2}
Dimensions of the quantities. Force [F]=[MLT^{-2}], distance [r]=[L]. Charge is current times time, q=It, so [q]=[AT].
[F]=[MLT^{-2}],\qquad [r^{2}]=[L^{2}],\qquad [q_1q_2]=[AT]^{2}=[A^{2}T^{2}]
Substitute (1 mark). Collect the powers of each base dimension.
[k]=\dfrac{[MLT^{-2}]\,[L^{2}]}{[A^{2}T^{2}]}=[ML^{3}T^{-4}A^{-2}]

As a check, the unit of k is \mathrm{N\,m^{2}\,C^{-2}}=\mathrm{kg\,m^{3}\,s^{-4}\,A^{-2}}, which has the same dimensions.

[k]=[ML^{3}T^{-4}A^{-2}]

2025 Annual I, Set B, Q.2 (i)1+1+1 = 3 marks

Briefly explain 'precision' and 'accuracy' with a diagram.

Solution

Precision (1 mark). Precision is the closeness of repeated measurements of the same quantity to each other. It depends on the least count of the instrument and is indicated by the absolute uncertainty: the smaller the least count, the greater the precision.

Accuracy (1 mark). Accuracy is the closeness of a measured value to the standard or true value. It is indicated by the fractional or percentage uncertainty: the smaller it is, the more accurate the measurement.

Diagram (1 mark). Shots at a target show the idea. The centre is the true value, and each dot is a reading.

Figure 1.1: Several independent shots at a bullseye target show the difference between accurate and precise. (a) tight group at the centre; (b) tight group away from the centre; (c) scattered shots around the centre; (d) scattered shots away from the centre.

In (a) the readings agree with each other and lie on the true value (precise and accurate). In (b) they agree with each other but are all shifted from the true value (precise, not accurate). In (c) they scatter but average out around the true value (accurate, not precise). In (d) they scatter and miss (neither).

2026 Annual I, Set B, Q.2 (v) OR3 marks

A body of mass (27.2\pm0.1)\,\mathrm{kg} has a volume of (2.0\pm0.1)\,\mathrm{m^3}. Calculate the density of the body? Find uncertainty in the density measurement.

Solution

Given

m = (27.2\pm0.1)\,\mathrm{kg}
V = (2.0\pm0.1)\,\mathrm{m^{3}}

To find

density \rho
uncertainty in \rho

Density (1 mark). Density is mass per unit volume.
\rho=\dfrac{m}{V}=\dfrac{27.2}{2.0}=13.6\,\mathrm{kg\,m^{-3}}
Percentage uncertainties of the data. Percentage uncertainty is absolute uncertainty divided by the measured value, times 100.
\dfrac{\Delta m}{m}\times100=\dfrac{0.1}{27.2}\times100=0.37\,\%,\qquad \dfrac{\Delta V}{V}\times100=\dfrac{0.1}{2.0}\times100=5\,\%
Quotient rule (1 mark). When two quantities are divided, their percentage uncertainties add.
\%\,\text{uncertainty in }\rho=0.37\,\%+5\,\%=5.37\,\%\approx5.4\,\%
Absolute uncertainty (1 mark). Multiply the density by the fractional uncertainty. Quote it to one significant figure.
\Delta\rho=\dfrac{5.37}{100}\times13.6=0.73\approx0.7\,\mathrm{kg\,m^{-3}}

\rho=(13.6\pm0.7)\,\mathrm{kg\,m^{-3}}\ \ (\pm5.4\,\%)

2026 Annual I, Set B, Q.2 (viii)1+2 = 3 marks

What is meant by precision? Also write the importance of taking multiple readings in an experiment.

Solution

Precision (1 mark). Precision is the closeness of repeated measurements of the same quantity to each other. It depends on the least count (resolution) of the instrument and is indicated by the absolute uncertainty: the smaller the uncertainty, the greater the precision.

Importance of multiple readings (2 marks).

Every reading contains some random error, so single readings fall on either side of the true value. Taking many readings and averaging them lets these random errors partly cancel, so the mean is closer to the true value and more reliable.
The spread of the readings shows how well they agree. The mean deviation of the readings from their average gives the uncertainty in the mean, so more readings give a better estimate of the precision.
A reading that lies far from the others (a mistake) is easy to spot among many readings and can be checked or rejected.

2025 Annual I, Set B, Q.2 (xiii)3 marks

Use dimensional analysis to prove the equation P = \rho g h is dimensionally homogeneous.

Solution

Dimensions of the left side (1 mark). Pressure is force per unit area, and force has the dimensions of mass times acceleration.
[P]=\dfrac{[F]}{[A]}=\dfrac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}]
Dimensions of the right side (1 mark). Density is mass per unit volume, g is an acceleration and h is a length.
[\rho gh]=[\rho]\,[g]\,[h]=[ML^{-3}]\,[LT^{-2}]\,[L]=[ML^{-1}T^{-2}]
Compare (1 mark). Both sides have the same dimensions.
[P]=[\rho gh]=[ML^{-1}T^{-2}]

By the principle of homogeneity of dimensions, an equation can be correct only if every term has the same dimensions. Here both sides have the dimensions [ML^{-1}T^{-2}], so the equation P=\rho gh is dimensionally homogeneous. (This shows the dimensions are consistent; it cannot fix a numerical constant, which here is 1.)

[P]=[\rho gh]=[ML^{-1}T^{-2}]\quad\text{(homogeneous)}

2025 Annual I, Set D, MCQ 11 mark

In a simple pendulum experiment, percentage errors (uncertainty) in 'L' and 'T' are 0.1% and 2% respectively. Maximum percentage error (uncertainty) in 'LT^2' is:

A4.1%Answer
B3%
C2.1%
D3.2%

Why

For a product, the percentage uncertainties add, and a power multiplies the percentage uncertainty of that quantity by the power. In LT^{2}: \%\Delta(LT^{2})=\%\Delta L+2\,\%\Delta T=0.1\,\%+2\times2\,\%=4.1\,\%. So A is correct.

Option C (2.1\,\%) is what you get from 0.1\,\%+2\,\%, forgetting that T is squared, so its percentage uncertainty is doubled.

2026 Annual I, Set D, MCQ 21 mark

The dimensional formula of velocity is:

A[M^0L^0T^0]
B[LT^{-1}]Answer
C[LT^{-2}]
D[L^2T^{-1}]

Why

Velocity is displacement divided by time, so [v]=\dfrac{[L]}{[T]}=[LT^{-1}]. So B is correct.

Option C, [LT^{-2}], is the dimensions of acceleration (velocity per unit time). Option A, [M^{0}L^{0}T^{0}], belongs to a dimensionless quantity such as an angle, and D, [L^{2}T^{-1}], is area per unit time.

2026 Annual I, Set D, MCQ 31 mark

Joule can be expressed as:

A\mathrm{kg\,m^{-2}\,s^{-2}}
B\mathrm{kg\,m\,s^{-1}}
C\mathrm{kg\,m^{2}\,s^{-1}}
D\mathrm{kg\,m^{2}\,s^{-2}}Answer

Why

A joule is the unit of work, which is force times distance, so \mathrm{J}=\mathrm{N\,m}=(\mathrm{kg\,m\,s^{-2}})\,\mathrm{m}=\mathrm{kg\,m^{2}\,s^{-2}}. So D is correct.

Option B, \mathrm{kg\,m\,s^{-1}}, is the unit of momentum, and C, \mathrm{kg\,m^{2}\,s^{-1}}, is the unit of angular momentum (and of Planck's constant). They differ from the joule by a factor of \mathrm{s^{-1}}. Option A has \mathrm{m^{-2}} instead of \mathrm{m^{2}}, which would come from dividing by a distance instead of multiplying.

2026 Annual I, Set D, Q.2 (i)3 marks

Deduce the dimensions of universal gravitational constant (G). \left(F = G\dfrac{m_1m_2}{r^2}\right)

Solution

Newton's law of gravitation is F=G\dfrac{m_1m_2}{r^{2}}. Make G the subject and put in the dimensions (1 mark for the rearrangement).

Rearrange (1 mark).
G=\dfrac{F\,r^{2}}{m_1m_2}
Dimensions of the quantities. Force [F]=[MLT^{-2}], distance [r]=[L], mass [m]=[M].
[F]=[MLT^{-2}],\qquad [r^{2}]=[L^{2}],\qquad [m_1m_2]=[M^{2}]
Substitute (1 mark).
[G]=\dfrac{[MLT^{-2}]\,[L^{2}]}{[M^{2}]}=[M^{-1}L^{3}T^{-2}]

The unit of G is \mathrm{N\,m^{2}\,kg^{-2}}=\mathrm{m^{3}\,kg^{-1}\,s^{-2}}, which agrees.

[G]=[M^{-1}L^{3}T^{-2}]

2025 Annual I, Set D, Q.2 (i)3 marks

Show that the equation P = \rho g h is dimensionally homogeneous.

Solution

Dimensions of the left side (1 mark). Pressure is force per unit area, and force has the dimensions of mass times acceleration.
[P]=\dfrac{[F]}{[A]}=\dfrac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}]
Dimensions of the right side (1 mark). Density is mass per unit volume, g is an acceleration and h is a length.
[\rho gh]=[\rho]\,[g]\,[h]=[ML^{-3}]\,[LT^{-2}]\,[L]=[ML^{-1}T^{-2}]
Compare (1 mark). Both sides have the same dimensions.
[P]=[\rho gh]=[ML^{-1}T^{-2}]

[P]=[\rho gh]=[ML^{-1}T^{-2}]\quad\text{(homogeneous)}

2025 Annual I, Set D, Q.2 (ii)3 marks

Time for 10 vibrations of a simple pendulum is 17.6\,\mathrm{s}. Least count of stop watch is 0.1\,\mathrm{s}. Find time period of simple pendulum. Also determine uncertainty in it.

Solution

Given

t = 17.6\,\mathrm{s}
n = 10\ \text{vibrations}
\text{L.C} = 0.1\,\mathrm{s}

To find

time period T
uncertainty \Delta T

Time period (1 mark). The time period is the time of many vibrations divided by their number.
T=\dfrac{t}{n}=\dfrac{17.6}{10}=1.76\,\mathrm{s}
Uncertainty (1 mark). The uncertainty in the time period is the least count of the stopwatch divided by the number of vibrations.
\Delta T=\dfrac{\text{L.C}}{n}=\dfrac{0.1}{10}=0.01\,\mathrm{s}
Result (1 mark). Write the period with its uncertainty, with the same number of decimal places in both.
T=(1.76\pm0.01)\,\mathrm{s}

Timing 10 vibrations together shares the stopwatch's single uncertainty of 0.1\,\mathrm{s} among all of them, so the uncertainty in one period is 10 times smaller than in timing one vibration. The percentage uncertainty is \dfrac{0.01}{1.76}\times100\approx0.57\,\%.

T=(1.76\pm0.01)\,\mathrm{s}

2026 Annual I, Set D, Q.2 (v) OR3 marks

Differentiate between accuracy and precision of measurements. (Give a graphical example)

Solution

Precision and accuracy compared (2 marks)
	Precision	Accuracy
Meaning	Closeness of repeated measurements of the same quantity to each other	Closeness of a measured value to the standard or true value
Depends on	The least count (resolution) of the instrument	How well the instrument is calibrated and used (systematic errors)
Indicated by	The absolute uncertainty: smaller means more precise	The fractional or percentage uncertainty: smaller means more accurate
Example	Readings 2.51, 2.52, 2.51 cm: close to each other	Readings centred on the true value 2.50 cm

Graphical example (1 mark). Shots at a target show the idea. The centre is the true value and each dot is a reading.

2026 Annual I, Set D, Q.2 (viii)3 marks

Write any three limitations of dimensional analysis.

Solution

Dimensional analysis checks equations and suggests possible formulas, but it has these limitations (any three, 1 mark each).

It cannot distinguish quantities with the same dimensions. For example [ML^{2}T^{-2}] may be work, energy or torque, so the dimensions alone do not say which one it is.
It cannot handle trigonometric, exponential or logarithmic functions. Such functions have dimensionless arguments and cannot be written as a product of powers of M, L and T.
It cannot find dimensionless constants. In the time period of a pendulum, T=(\text{constant})\sqrt{l/g}, the constant 2\pi has to be found by experiment.
A dimensionally correct equation need not be physically correct. Dimensions can only reject a wrong equation, never prove a right one. For example s=\tfrac12at is wrong, but s=\tfrac12 at^{2} and s=at^{2} are both dimensionally correct and only one of them is true.

2025 Annual I, Set D, Q.2 (viii)3 marks

Show that: \dfrac{\mathrm{Volt}}{\mathrm{meter}} = \dfrac{\mathrm{Newton}}{\mathrm{Coulomb}}

Solution

The volt is the unit of potential difference and the newton per coulomb is a force per unit charge. Show that both have the same base units (or dimensions).

Dimensions of volt per metre (1.5 marks). Potential difference is work done per unit charge, and charge is current times time.
\begin{aligned}{[V]}&=\dfrac{[W]}{[q]}=\dfrac{[ML^{2}T^{-2}]}{[AT]}=[ML^{2}T^{-3}A^{-1}]\\ \Rightarrow\ \left[\dfrac{V}{m}\right]&=\dfrac{[ML^{2}T^{-3}A^{-1}]}{[L]}=[MLT^{-3}A^{-1}]\end{aligned}
Dimensions of newton per coulomb (1.5 marks). Force has dimensions [MLT^{-2}].
\left[\dfrac{N}{C}\right]=\dfrac{[MLT^{-2}]}{[AT]}=[MLT^{-3}A^{-1}]
Compare. The dimensions are equal.
\left[\dfrac{V}{m}\right]=\left[\dfrac{N}{C}\right]=[MLT^{-3}A^{-1}]

In base units, \dfrac{\mathrm{V}}{\mathrm{m}}=\dfrac{\mathrm{J/C}}{\mathrm{m}}=\dfrac{\mathrm{N\,m}}{\mathrm{C\,m}}=\dfrac{\mathrm{N}}{\mathrm{C}}, because a joule is a newton metre. Both are \mathrm{kg\,m\,s^{-3}\,A^{-1}}.

\dfrac{\mathrm{Volt}}{\mathrm{meter}}=\dfrac{\mathrm{Newton}}{\mathrm{Coulomb}}=\mathrm{kg\,m\,s^{-3}\,A^{-1}}

Exam focus

Deduce the dimensions of a constant from a law, e.g. k in Coulomb's law or G in Newton's law of gravitation. (short)
Set B Q.2 (i) (k) and Set D Q.2 (i) (G), 3 marks each in the 2026 paper. Same as MCQ 1.4. Method: rearrange for the constant, put in [F]=[MLT^{-2}] and [q]=[AT].
Go to answer
Check the homogeneity of an equation such as P=\rho gh, or decide which equation is dimensionally correct. (short)
2025: Set B Q.2 (xiii) and Set D Q.2 (i) (P=\rho gh, 3 marks each) and Set D Q.2 (viii) (V/m=N/C). 2026: Set B MCQ 1 (which equation is dimensionally correct). Book numerical 1.5.
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Explain precision and accuracy, with a target diagram, and say how each is indicated. (short)
2025 Set B Q.2 (i); 2026 Set B Q.2 (viii) (precision and why take many readings) and Set D Q.2 (v) OR (accuracy against precision with a graphical example). Short questions 1.4 and 1.5.
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Percentage uncertainty and absolute uncertainty in a result, e.g. \rho=m/V or LT^{2}. (numerical)
2026 Set B Q.2 (v) OR (density from m and V, 3 marks). 2025 Set B MCQ 2 and Set D MCQ 1, and 2025 Set D Q.2 (ii) (period and its uncertainty from 10 vibrations, T=(1.76\pm0.01)\,\mathrm s). Book exercise MCQ 1 and numerical 1.3 use the same rules.
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Write the limitations of dimensional analysis, with an example for each. (short)
2026 Set D Q.2 (viii), 3 marks (any three limitations). Section 1.3.2 lists them.
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Identify a quantity from its dimensions or base units: velocity [LT^{-1}], joule \mathrm{kg\,m^{2}\,s^{-2}}, work [ML^{2}T^{-2}]. (mcq)
2026 Set B MCQ 2 (pascal =\mathrm{N\,m^{-2}}), Set D MCQ 2 (velocity) and MCQ 3 (joule). 2025 Set B MCQ 1 (work) and Set D Q.2 (viii). Book exercise MCQs 7 and 8 and short question 1.2.
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Define and explain uncertainty, and discuss the rules for combining uncertainties (sum, difference, product, quotient, power). (long)
Comprehensive questions 1.1 and 1.2. Not asked as a long question in 2025 or 2026, but the numericals and MCQs of both years rest on these rules.
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Use dimensions to find the base unit of a constant (for example F=c\,r\,v) or to derive a possible formula. (numerical)
Book numerical 1.4 and the pendulum example in the notes. Not in the 2025 or 2026 papers, but it is the usual method question for this unit.
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Estimation: powers-of-ten estimates such as heartbeats in a lifetime, or protons in a bacterium. (numerical)
Book numericals 1.1, 1.6 and 1.7 and short question 1.1. No estimation question appeared in the 2025 or 2026 papers.
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