Derived units:
The units of all other physical quantities which can be obtained from fundamental units are called derived units.
eg: m/s, Kg/m³ etc.

Conversion of units:-
While doing all the calculations in physics, homogenity of units should be Followed. Many times this requires conversion of units. For this we use conversion factor which can be generated as follows:-
Suppose we wish to convert 5 mi/h into m/s
then 1 mile = 1.6 x 10³ m
factor becomes =
also ln = 3600 s , factor is = f₂
Complete multiplying factor is

[Remember while finding individual factors, the given unit should be in denominator.]

Multiplication Factor:
To mention Physical Quantities Following standard prefixes for certain power of 10 are used:

Power of 10	Prefix	Symbol
+ 12 + 9 + 6 + 3 - 2 - 3 - 6 - 9 - 12	Tera Giga Mega Kilo Centi milli micro nano pico	T G M K C m n P

Dimensional Analysis:
The dimensions of aphysical quantities are the powers to which the base quantities are raised to represent that quantity "
Dimensions of fundamental Quantities are represented as:
Length [l], mass [M], Time[T], Electric current [A], temperature[K], Luminious intensity[Cd], amount of substance [md]
eg: dimensions of volume = [l]³ 0r [m⁰L³T⁰] as its unit is m³
force is mass x acceleration units are Kg m/s²    dimension are [MLT^-2]

Application of Dimensional Analysis:

(1) In conversion of units from one system to another.
This is based on the fact that product of the numerical value and its corresponding unit is constant.
Numerical value x unit = constant
if unit is changed then numerical value changes
Let dimensions of a quantity be [M^a L^b T^c] and numerical value in first units be n₁ [base unit being M₁L₁T₁)
and in seconds unit be n₂ [base unit being M₂L₂T₂]
then n₁[M₁^a L₁^c T₁²] = n₂[M₂^aL₂^bT₂^c

n₂ = n₁
Thus unit conversion can be done.

(ii) Examining dimensional correctness of agiven relation
• For a givenformula to be correct, the dimesion on left hand side and right hand side must be equal.
• Just cdimensional equality doesnot guarentees that formula is correct , but unequal dimensions do state that formula is wrong.

(iii) To establish the relation among various physical quantities
If we know that on what all quantities a particular parameter depends than we can dimensionally work out the correct formula.
• Stokis law has been developed using this principle.

• It is known experimentally that when a ball is in a viscous liquid , then viscous force depends on radius (r) of of the sphere, velocity of the sphere (v) and viscosity of the liquid
F (^x r^y v^z)
or F = K^x r^y v^z
examining dimensionally
[MLT^-2] = [ML^-1T^-1]^x [L]^y[LT^-1]² = [m^x L^{- x + y + z} T ^{-x - z}]
equating powers we get
     x = y = z = 1
F = Krv
K can further be calculated experimentally (K = 6).
In this way, we can easily convert a propwertionality equation in to a formula.

Illustration: The period 'P' of a simple pendlum is the time for one complete swing. How does P depend on the mass m of the bob, the length l of the string and the acceleration due to gravity ?
Solution:- P = Km^x g^z l^y     where k is aconstant
converting the equation into dimensional equation
[T^-1] = [M]^x [LT^-2]² [L]⁴
      = [m^xL^Z+yT^-2z]
on equating the powers of each dimensions on either side of the equation
Thus z = -1/2, y = 1/2, x = 0