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

1. Force ® newton or N
2. Energy ® joule or J
3. 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/ sec2, 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  mvY RZ

F a mvY RZ

[M L T -2] = K [M] x [LT -1] y [L] z

[M LT-2] = K [Mx Ly+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 = m1 v2 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 mx l y g z [M° L° T1] = K [M]x [L]y [LT-2]z

Þ [M° L° T1] = 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 A2 T]

(b) [F A T2]

(c) [F2 AT]

(d) [FAT]

Solution:  Let energy denote by E

Then E a Fx Ay Tz

E = Fx Ay Tz

[M L2 T-2] = [M L T-2]x [ LT-2]y [T]Z

[M L2 T-2] = [Mx Lx+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  [ F1 A1 T2] Þ [ FAT2Ans B

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

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