PHYSICS CLASS: Unit And Dimension

QUANTITIES AND UNIT

 The term has to do with measurement. It is the property of things that can be measured in measurement, the question being asked is how much, how big, how long, how fast etc. The two types of quantities are

Fundamental quantities

Derived quantities

Fundamental quantities are the basic quantities upon which other quantities are built. The units of the basic quantities are called fundamental units.

FUNDAMENTAL QUANTITIES	SYMBOL	FUNDAMENTAL UNIT
Mass	M	Kilogram       Kg
Length	L	Meter             M
Time	t	Seconds         S
Angle	ᶿ	Radian           Rad
Amount of substance	N	Mole               Mol
Temperature	T	Kelvin             K
Electric Current	I	Ampere           A
Luminous Intensity		Candela          Cd

(B) DERIVED QUANTITIES

They are quantities which can be derived from the fundamental quantities. They include quantities such as; Area, volume, velocity, acceleration, momentum etc. The units of these derived quantities are called derived units.

EXAMPLES

Area =   length (m) x length (m) = Meter squared = M²

Velocity = distance (m) = meter per seconds = MS^-1

                                 Time (s) 

 Accelerator = Velocity (MS^-1) = MS^1-1 = MS^-2

                          Time (S) 

DIMENSION

The dimensions of physical quantities tell us how those quantities relate to the fundamental quantities. The three fundamental quantities in mechanics and their dimensions and unit are

Quantity	Dimensions	Unit
Mass	M	Kg
Length	L	M
Time	T	S

Below is the dimension of some physical quantities.

S/No	Quantity	Unit	Dimensions
1	Velocity	MS-1	LT-1
2	Acceleration	MS-2	LT-2
3	Force	Kgm-2	MLT-2
4	Momentum	KgmS-1	MLT-1
5	Area	M2	L2
6	Volume	M3	L3
7	Density	Kgm-3	ML-3
8	Pressure	Kgm-1S-2	ML-1T-2
9	Energy	Kgm2S-2	ML2T-2
10	Power	KgM2S-3	ML2T-3
11	Surface tension	KgS-2	MT-2
12	Young modulus	Kgm-1S-2	ML-1S-2
13	Impulse	KgmS-1	MLS-1
14	moment	Kgm2S-2	ML2S-2

SIMPLIFIED 1: If the formula for the period of Acceleration of a simple pendulum bulb is T = L^x g^y find the value of x and y

Solution:

 L^x g^y

 And L are already in dimension but g = Acceleration due to gravity thus SI unit is the SI unit of acceleration

=MS^{- 2} = LT^-2

 = L^x (LT^-2)^y

 = L^x L^yT^-2y

 = L^x+y  T^-2y

()      1 = -2Y                  

Y= - 1/2

(L) = 0 = x +y

But y = - 1/2

O = x + (-1/2)

O = x – 1/2

1/2 = x

Y = -1/2, x = 1/2

SIMPLIFIED 2: at what respective values of x, y and z would the unit of force be dimensionally equivalent to M^XL^YT^Z

Solution

The dimension of force is

MLT ^-2

 MLT^-2 = M^xL^yT^z

M^xL^yT^z = M¹ L¹ T^-2

X = 1, y = 1, z = -2