**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.

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