## Fundamental Physical Quantities, Unit, Symbols

**UNITS AND DIMENSIONS**

**Units **“ To measure or represent any physical quantity we need units”

**Exception: **Although there are few physical quantities which do not

need any unit like refractive index, Relative density or specific gravity etc.

Physical quantity

nu = Constant

**Example:** In a particular system, the unit of length,

mass and time are chosen to be 10cm, 10gm and 0.1 sec

respectively. The unit of force in this system will be

equivalen to:

- 0.1 N
- 1 N
- 10 N
- 100 N

Sol. Unit of force Þ F = ma = Kg metre /sec^{2}

Here

F = 0.1 N **So the Correct option is (a)**

**Example: **The density of a material in CGS System of unit

is 4gm/cm^{3}. In a system of units in which unit

of length is 10cm and unit of mass is 100 gm the

value of density of material will be

(a) 0.04

(b) 0.4

(c) 40

(d) 400

**Solution**. Given n_{1} u_{1} = 4 gm/cm^{3}

n_{2} u_{2} = n_{1} u_{1}

n_{2} = 40 ** Answer (a)**

**FUNDAMENTAL PHYSICAL QUANTITIES, UNITS, SYMBOLS**

**SUPPLEMENTARY UNITS**

**Derived Units:** The units which may derive by

fundamental or Supplementry units

Ex. Force= ma = Kg mt/sec^{2}

Conventions adopted while writing a unit

1. Even if a unit is named after a person it should

not be Capital initial letter.

Newton (no) Joule (no) Ampere (no)

or or etc.

newton (yes) joule (yes) ampere (yes)

2. In symbol for a unit named after a person

Newton ® N

Ampere ® A

Joule ® J

Watt ® w

3. The symbol or units are not expresse

d in plural form : 50 m (yes) , 10 joule (yes)

50 ms (no) , 10 joules (no)

4. Not more than one Solidus is used

1 poise = 1g/s cm or 1 gs^{-1}

## Homogeneity

**Law of homogeneity **

Here we can jump through +, - , = (signs)

and a,b, c- are Constant then find dimensional formula of a, b, c-

**Apply Law of homoginity**

Dimensional formula of v = dim-formula of at = v

Similarly we can say t = c Þ C = [T]

b = [L]

**Example: **The equation of state of some gases can be expressed as here p is pressure, V is the volume, T is temp. then the dimensional formula of Constants a and b will be ?

Apply Law of homoginity Dimensional formula of p = Dim-formula of

a = PV^{2} = ML ^{-1}T^{-2}L^{6} =[ML^{5}T^{-2}]

Similarly dimensional formula of v = Dim. Formula of b b = v

= [L^{3}]

**ANS** a = [ML^{5} T^{-2}]

b = [L^{3}]

**Application No-3 **

**Dimension less quantities or functions**

(i) Quantities having units but dimensionless

So all trigonometric functions will be dimension less i.e. sin *x*, cos *x*, tan *x* – etc.

(ii) Quantities neither having units nor dimension

Refraction index

Reynolds number, Relative density etc.

(iii) Dimension less mathematical functions

(a) Logarithmic functions: *log _{e}*

*x*or

*log*

_{10}

*x*

(b) Exponential functions: e^{x}, a^{x}

(a) P

(b) R

(c) T

(d) V

**Correct answer is (a)**

**Example** Nuclear force between nucleons is given by Here F is force, r is distance, and c, k are Constants then find dimensional formula of C and k = ?

Here e^{-kr} is dimension less

So C = Fr^{2} = M L T^{-2} L^{2} = [M L^{3} T^{-2}]

No because e^{-kr} is dimension less

So k = 1/r

= [L^{-1}]

**Ans:** dimensional formula of

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

K = [L^{-1}]

## Systems of Units, Dimensional Formluas

**SYSTEMS OF UNITS**

**(1) MKS AND MKSA SYSTEM**

Here in this system M® meter, K® Kilogram, S ®Second and A® ampere

Physical quantity Unit in MKS or MKSA System

…….. and so on

**(2) SI SYSTEM (INTERNATIONAL SYSTEM)**

Here in this system unit of physical quantities are use to be named after different great Scientists.

Physical quantity

Unit in SI System

- Force ® newton or N
- Energy ® joule or J
- Power ® watt or W

**(3) CGS SYSTEM**

Here C® Centimeter, G ® gram, S ® Second

Physical quantity Unit in CGS System

….and so on

**(4) FPS SYSTEM **

Here F ® foot, P ® pound , S ® second

Force ® pd ft/ sec^{2}, Similarly other physical quantities

**Dimensional and dimensional formula **

Dimensional formula

Almost all physical quantities can be expressed in terms of the seven fundamental units in symbolic form.

**Dimensions **

Dimensions of a physical quantity are the powers to which we must be raised to fundamental quantity represent the given physical quantity.

……. And so on

**APPLICATIONS OF DIMENSIONAL FORMULAS **

Application No. (1)

**To establish results **

Suppose any mass m is moving along a circular path of radius R with uniform speed of V. If the force required to do so is F then establish the result or relation between F, m, v and R.

Given ,

F = k m^{x }v^{Y} R^{Z}

F a m^{x }v^{Y} R^{Z}

[*M L T ^{-2}*] =

*K*[

*M*]

*[*

^{x}*LT*

^{-1}]

*[*

^{y}*L*]

^{z}[*M LT ^{-2}*] =

*K*[

*M*]

^{x }L^{y+z }T^{-y}Comparing both sides

*X *=1, *y *+ *z* = 1, - *y* = - 2 Þ *y* = 2

2 + *z* = 1

*Z* = 1 - 2 = - 1

By experiment Constant *K* = 1

So we have result F = m^{1} v^{2} R^{-1} Þ

**Example 1** Given a simple pendulum of very small bob of mass connected with a light String of length *l*. Also given acceleration due to gravity is g and time period *T* of simple pendulum depends on mass of the bomb, length of the String *l* and acceleration due to gravity g then establish result for T.

**Given**

*T* a *m ^{x} l ^{y} g ^{z}*

*T* = *k m ^{x} l ^{y} g ^{z}*

[*M° L° T*^{1}] = *K* [*M*]* ^{x}* [

*L*]

*[*

^{y}*LT*

^{-2}]

^{z}Þ [*M*° *L*° *T*^{1}] = *k* [*M ^{x}*

*L*

^{y}^{ + z }

*T*

^{-2z}]

Compare both Side

*x* = 0, *i.e*. Time period will not depend on mass of the bob *Y* + *z* = 0

Compare both Side

x = 0,i.e. Time period will not depend on mass of the bob Y+ z = 0

By experiment k = 2p

**Example: **In a system of units , if force F, acceleration A and time T are taken as fundamental units then the dimensional formula of energy is

(a) [F A^{2} T]

(b) [F A T^{2}]

(c) [F^{2} AT]

(d) [FAT]

**Solution:** Let energy denote by E

Then E a F^{x} A^{y} T^{z}

E = F^{x} A^{y} T^{z}

[M L^{2} T^{-2}] = [M L T^{-2}]^{x} [ LT^{-2}]^{y} [T]^{Z}

[M L^{2} T^{-2}] = [M^{x} L^{x+y} T^{-2x-2y +z}]

Compare both Sides

x =1 , - 2x – 2y + z = -2

x + y = 2 - 2x1 -2 x1 + z = -2

Þ Y = 1 - 4 + z = -2

Z = 4 – 2

Þ Z = 2

Therefore dimensional formula of energy E in Terms of FAT is [ F^{1} A^{1} T^{2}] Þ [ FAT^{2}] **Ans B**

**Application No.2: **To check the validity of any result.