Chapter 1 Physical Quantities and Units Checked

1 PHYSICAL QUANTITIES AND UNITS

TOPICS LEARNING OUTCOMES

1.1. Basic quantities and SI Candidates should be able to:
units a) List base quantities like mass (kg), length (m), time (s), current (A),

1.2. Dimensions of physical temperature (K), and quantity of matter (mol),and write their SI units.
quantities b) Deduce units for derived quantities if the definitions are given.
c) Use dimensional analysis to determine the dimensions of derived
1.3. Scalars and Vector
1.4. Uncertainties in quantities.
d) Check the homogeneity of equations using dimensional analysis
measurements e) Construct empirical equations using dimension analysis.
f) Determine the sum , the scalar product and vector product of coplanar

vector ;
g) Resolve a vector to two perpendicular components.
h) Calculate the uncertainty in a derived quantity (a rigorous statistical

treatment is not required).
i) Write a derived quantity to an appropriate number of significant figures.

1.1 Basic quantities and SI units

Physical quantity is a term used to include many different measurable features, with each comprises a
numerical magnitude and a unit. For example, the speed of a car is 5 m s-1. ‘Speed’ is a physical quantity, ‘5’ is
the numerical magnitude and ‘m s-1’ is the unit. Units of all the different physical quantities can be related to the
base or fundamental units. The SI system of measurement uses 7 base units for 7 base quantities.

1.2. Dimensions of physical quantities

Dimensions are used to refer to the physical nature of a quantity and the type of unit used to specify it. The
seven base quantities with their respective units and dimensions are shown below:

QUANTITY UNIT DIMENSION
length m L
mass kg M
time s T
A A
electric current K 
thermodynamic temperature mol
cd N
amount of substance I
luminous intensity

Dimensional analysis is used to check whether the mathematical relations of physical quantities have
consistency in terms of their dimensions.
A physically correct equation must have the same dimensions or units on both the right-hand (RHS) and the
left-hand side (LHS) of the equation.

Supposedly a car starts from rest and accelerates to a speed v in a time t and the distance x travelled by the car is
assumed to be either x  1 vt2 or x  1 vt .

22

Checking the dimensions on both sides of the equations

. x  1 vt2 x  1 vt
2 2

Dimensions [L] = [ L ][T]2 = [L][T] L
T [L] = [ ][T] = [L]

T

For the right hand side equation, dimension on the left of the equal sign matches that on the right, so the relation
is dimensionally correct. The equation is homogeneous and is said to be dimensionally consistent.

Example 1
1. (a) Determine the dimension of Young’s modulus.

(b) The Young modulus can be determined by propagating a wave of wavelength  with velocity v into a
solid material of density  . Using dimensional analysis, derive a formula for Young modulus.

Solution:

(a) Y  F/A Note: Correct dimension for F: MLT-2
e /

[Y ]  [F][]
[ A]  [e]

M LT2L
= L2L
= ML-1T-2

(b) Let Y = (k) x vyz or Y  x vyz

[] = L
[v] = LT-1
[] = ML-3

Then ML-1T-2 = (L)x(LT-1)y ( ML-3)z

M : 1 = z or z =1

T : -2 = -y or y = 2
L : -1 = x +y – 3z

-1 = -3 + x + 2

x=0

Y = k v2

1.3. Scalars and Vector

Scalars and vectors

A scalar quantity is one which has magnitude but no direction, e.g. time, wavelength of light, energy, and
density.
A vector quantity is one which has direction and magnitude. For representing vectors, common notations used
are ⃗⃗⃗ or .

VECTOR ADDITION

(A) Parallelogram Method


Supposedly two vectors A and B are represented in magnitude and direction by the adjacent sides of a
parallelogram. Then, the diagonal represents the vector sum A  B in magnitude and direction.

The parallelogram method is equivalent to the vector triangle shown here.

Parallelogram Method Vector Triangle


If two vectors A and B are at an angle  to each other, then the magnitude of the vector sum or the resultant

R is given by R = A2  B2  2AB cos 

VECTOR SUBTRACTION

   
If a vector Q is subtracted from vector P , then the resultant vector R is P - Q . If vector P is 10 units to the
right, then vector - P is 10 units to the left.


P -P

10 units 10 units

  
P - Q can be rewritten as P  (-Q) . We can use vector addition to find the resultant.

RESOLUTION OF VECTORS


Consider a vector R , which is represented by an arrow
from origin (0, 0) to the point (x, y), making an angle 
with the x-axis. Then the vector R is said to beresolvable
into 2 mutually perpendicular components R x and R y .

By usingtrigonometry, ; 
R x = R cos  R y = R sin 

 R 2  R 2
The magnitude of vector R is R = x y

The direction of vector  is  = tan-1 Ry
R
Rx

Example 2
Consider a block resting on a frictionless inclined plane. Given F is the gravitational force, find Fx and Fy.

Solution:

 Using trigonometry,

Fy = F cos  ; Fx = F sin 

Fy is the force that presses the block against the plane; whereas, Fx is the force that accelerates the block down
the inclined plane.

Example 3

:
Solution:

   
PQRQ PR.

Answer: B

1.4. Uncertainties in measurements

A measurement is a process by which a physical
quantity is compared to a standard unit. For example,
the length of the pencil is obtained as follows:

The smallest division on the meter rule = 1 mm or 0.1 cm
The absolute uncertainty for 1 reading = ± 0.5 mm or ± 0.05 cm
Since the length of the pencil is obtained from the subtraction of two boundary readings,
The uncertainty of the measurement = ± (0.5 + 0.5) mm = ±1 mm or ±0.1 cm
The measured length = (100 ± 1) mm or (10.0 ± 0.1) cm
The measured value has the same number of decimal places as the maximum uncertainty while the latter is
generally expressed as to one significant figure only.

ADDITION AND SUBTRACTION OF UNCERTAINTIES

When two physical quantities are added or subtracted, their absolute uncertainties must be added together.

Example 4

External diameter of test tube, D = (1.23 ± 0.01) cm
Internal diameter of test tube, d = (1.13 ± 0.01) cm

Calculate the thickness of a wall of the test tube.

Solution:

Thickness of a wall of the test tube, t = − = 1.23−1.13 = 0.05 cm
2 2
Absolute error of the thickness of the test tube wall, ∆t = |12 ∆D| + |12 ∆d| = 0.01 cm

Thickness of the test tube wall = (0.05 ± 0.01) cm

MULTIPLICATION AND DIVISION OF UNCERTAINTIES
When a physical quantity is derived from multiplication or division of other physical quantities, its absolute
uncertainty is the sum of the latter’s fractional uncertainties.

Example 5
The dimensions of a piece of A4 paper are as follow: length, x = (297 ± 1) mm and width, y = (209 ± 1) mm.
Calculate (a) the fractional uncertainty of its length, (b) the area with its uncertainty.

Solution:

(a) Fractional uncertainty of the length, x = x  1 = 0.00337
x 297

(b) Area = length × width = 297 × 209 = 62073 mm2

Fractional uncertainty of area: A  x  y
Axy

The uncertainty of the area, A  ( 1  1 )(62073)  506 mm2
297 209

= 5 × 102mm2 (correct to one significant figure)
Area, A = (6.21± 0.05) × 104 mm2