**Unit 1- Mechanics **

**Chapter 1 – Physical Quantities **

__Measurement__

The process of comparing an unknown physical quantity with known fixed unit quantity is called measurement.

__Physical Quantity__

Those quantity which can be measured and can be expressed in numerical value are called physical quantity. Example : Mass, Length, Time etc

A physical quantity is expressed as; Physical quantity = Numerical value (N) x Unit (U)

__The physical quantity can be divided into two parts__

- Fundamental Quantity
- Derived Quantity

__Fundamental Quantity :__

The basic physical quantity which is taken as standard to measure other physical quantities known as fundamental quantity .

For example: Length, Mass, Time, Temperature, Electric current, Luminous intensity , Amount of chemical substance.

__Derived Quantity :__

A quantity obtained from fundamental quantities is called a derived quantity.

For example: Area, Volume, Density, Speed , Electric intensity

__Unit :__

The standard quantity in terms of which the given physical quantities can be compared is called unit. For example: Second, Newton etc.

__There are two types of unit :__

- Fundamental Unit
- Derived Unit

__Fundamental Unit:__

The units of fundamental quantities are called fundamental units.

For example: m, s, k etc

__Derived unit__

The unit of derived quantities are called derived unit.

For example: m/s, m/s^{2}, etc

__System Units:__

**F.P.S System**

In this system ** Length is measure in foot**,

**,**

*mass in pound***.**

*time in second***M.K.S System**

In this system ** Length is measure in meter**,

**,**

*mass in kilogram***.**

*time in second***C.G.S System**

In this system ** Length is measure in centimeter**,

**,**

*mass in gram***.**

*time in second*__Characteristics of standard Unit__

- It should be well defined and of a suitable so.
- It should be easily available, so that easily reproduce in laboratories.
- It should not change with time and place.
- It should not change with change in physical condition.
- It should be universally agreed, so that result of measurement at different places can be compared .

__S I System :__

The internationally Standardize unit of measurement is called SI system.

In SI system of unit 7 fundamental units and 2 supplementary units were proposed. These units are written below :

Fundamental Quantities | S.I. Units | Symbol |

Length | meter | m |

Mass | kilogram | kg |

Time | second | S |

Temperature | Kelvin | K |

Luminous Intensity | candela | cd |

Electric Current | ampere | A |

Amount of Substance | mole | mol |

**Supplementary Quantities**

Plane angle | radian | rad |

Solid Angle | steradian | sr |

** Dimensions of physical quantity:** The power raised to the fundamental quantities which are involved in derived physical quantities is called dimensions of physical quantity

__Dimensions formula of different physical quantities:__

$speed=\frac{distance}{time}=\frac{\left[L\right]}{\left[T\right]}=\left[L{T}^{\u20131}\right]$

The dimensional formula of speed = $[{M}^{0}L{T}^{\u20131}]$

$speed=\frac{distance}{time}=\frac{\left[L\right]}{\left[T\right]}=\left[L{T}^{\u20131}\right]$

The dimensional formula of velocity = $[{M}^{0}L{T}^{\u20131}]$

From the above Both speed and velocity has same dimensional formula, But speed is scalar quantity and velocity is vector quantity. Hence, from dimensional formula we cannot say physical quantity is scalar or vector.

__Dimensional Equation:__

An equation containing physical quantities, each quantity is represented by its dimensional formula the resulting equation is known as dimensional equation.

__Alternative Definition:__

When the dimension of a quantity is found and expressed in the form of an equation, the equation is called dimensional equation.

__Principle of homogeneity of dimension:__

The physical relations must obey the principle of homogeneity. According to this concept: *“every term in physical relation must have the same dimension.”*

$Supposeinphysicalrelations,S=ut+\frac{1}{2}a{t}^{2}intheequationtherearethreeterms:$

$S,ut,and\frac{1}{2}a{t}^{2}Allthetermsmusthavethesamedimensionsi.e$

$\left[S\right]=\left[ut\right]=[\frac{1}{2}a{t}^{2}]toobeytheprincipleofthehomogeneity.$

__Table of Units and Dimensions of Physical Quantities:__

S.N | Fundamental Physical Quantity | Dimensional formula | SI unit of physical Quantity |

1 | Mass | [M] | kilogram |

2 | Length | [L] | meter |

3 | Time | [T] | second |

4 | Temperature | [K] or [θ] | kelvin |

5 | Electric current | [Ɪ] or [A] | ampere |

6 | Amount of substance | [N] | Mole(mol) |

7 | Luminous intensity | [J] | Candela(cd) |

**A + B = C + D / or A + B – C – D = 0**

**According to homogeneity**

**Dimension of A = Dimension of B = Dimension C = Dimension of D**

__Derived Physical Quantities:__

S.N | Physical quantity | Formula Relation | Dimension formula | SI unit |

1 | Density | mass/volume | [M^{1} L^{-3} T^{0}] | Kg m^{-3} |

2 | Speed or Velocity | distance/time | [M^{0} L^{1} T^{-1}] | m/s |

3 | Acceleration | velocity/time | [M^{0} L^{1} T^{-2}] | m/s^{2} |

4 | momentum | mass x velocity | [M^{1} L^{1} T^{-1}] | Kg m s^{-1} |

5 | Force | mass x acceleration | [M^{1} L^{1} T^{-2}] | Newton (N) |

6 | Pressure | Force/area | [M^{1} L^{-1} T^{2}] | Nm^{-2 }or Pa |

7 | Work | Force x displacement | [M^{1} L^{2} T^{-2}] | J (joule) |

8 | Energy | Work(E=mc^{2}) | [M^{1} L^{2} T^{-2}] | J |

9 | Gravitational constant (G) | Force Xd^{2}(mass)^{2} | [M^{-1} L^{3} T^{-2}] | Nm^{2}kg^{-2} |

10 | Surface Tension | Force/length | [M^{1} L^{0} T^{-2}] | Nm^{-1} |

11 | Moment of inertia | Mass x (distance)^{2} | [M^{1} L^{2} T^{0}] | Kg m^{2} |

12 | Angular Momentum | Moment of inertia x angular velocity | [M^{1} L^{2} T^{-1}] | Kg m^{2 }s^{-1} |

13 | Torque or couple | Force x 1^{r} distance | [M^{1} L^{2} T^{-2}] | Nm |

14 | Frequency | 1/second | [T^{-1}] | Hz |

15 | Angular velocity (w) | Velocity/radius | [T^{-1}] | S^{-1} |

16 | Specific Heat | Energy/mass x temp^{r} | [M^{0} L^{2} T^{-2 }k^{-1}] | J kg^{-1 0}C^{-1} |

17 | Stress | Force/Area | [M^{1} L^{-1} T^{-2}] | N/m^{2} |

18 | Strain | Δl/l | [L^{0}]dimensionless | |

19 | Refractive index | V1/v2 | Dimensionless | |

20 | Mechanical Advantage | Load/Effort | Dimensionless | |

21 | Electric Charge | I x t | [M^{0} L^{2} T^{1 }I^{1}] | Coulomb |

22 | Electric resistance | V / I | [M^{1} L^{2} T^{-3 }I^{-2}] | Ohm(Ω) or volt |

23 | Boltzmann’s constant (K) | Energy/temperature | [M^{1} L^{2} T^{-2 }K^{-1}] | J/K |

24 | Plank’s Constant (h) | E = hf | [M^{1} L^{2} T^{-1}] | J.S or eV.S. |

25 | Power of lens (P) | P = 1/f | [L^{-1}] | dioptre |

__Uses of Dimensional formula:__

**Check whether the physical equation v ^{2} = u^{2} + 2as is dimensionally correct or not.**

Solution here,

Given formula

** **v^{2} = u^{2} + 2as

[L.H.S] = [v^{2}] = [LT^{-1 }]^{2} = [L^{2}T^{-2}]…………………… (i)

[R.H.S] = [ u^{2} + 2as] = [L T^{-1}]^{2} + [L T^{-2 }.L] = [L^{2 }T^{-2}] + [L^{2 }T^{-2}]

Since, the number ‘2’ is dimensionless,

[R.H.S] = [L^{2} T^{-2}] + [L^{2} T^{-2}] ………………………. (ii)

From equation (i) and (ii), we have

[L.H.S] = [R.H.S]

Hence, the given equation is dimensionally correct.

**Example :**

**Derive the dimensional relation of time period of pendulum with length and acceleration due to gravity.**

Solution,

$T\propto {l}^{a}{g}^{b}$

$T=K{l}^{a}{g}^{b}.....\left(i\right)$

Where k is dimensionless constant.

$\left[T\right]=[L{]}^{a}[L{T}^{\u20132}{]}^{b}$

$[{M}^{0}{L}^{0}{T}^{1}]=[{M}^{0}{L}^{a+b}{T}^{\u20132b}]$

Equating dimension on both sides, we get

$\left(i\right)1=\u20132b$

$b=\u2013\frac{1}{2}$

putting the value of b in equation

$\left(\mathrm{ii}\right)\mathrm{a}+\mathrm{b}=0$

$\mathrm{or},\mathrm{a}=\frac{1}{2}$

putting the value of a & b in equation (i)

$\mathrm{T}=\mathrm{k}{\mathrm{l}}^{\frac{1}{2}}{\mathrm{g}}^{\frac{\u20131}{2}}$

$\mathrm{T}=\mathrm{k}\frac{{\mathrm{l}}^{\frac{1}{2}}}{{\mathrm{g}}^{\frac{1}{2}}}$

$\mathrm{T}=\mathrm{k}\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$

__# To convert a unit from one system into another:__

**Convert 10 dyne into Newton.**

Solution

Let 10 dyne **=** N_{2} Newton.

Where, dyne is the unit of force in CGS, and Newton in SI.

Dimension formula of force as

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

: a = 1, b = 1, and c = -2 in mass, length and time respectively,

CGS system | SI system |

N_{1 }= 10 dyne | N_{2 }= ? |

M_{1} = 1 g | M_{2} = 1 kg |

L_{1} = 1cm | L_{2} = 1m |

T_{1} = 1s | T_{2} = 1s |

According to conversion formula,

${\mathrm{N}}_{2}={\mathrm{N}}_{1}[\frac{{\mathrm{M}}_{1}}{{\mathrm{M}}_{2}}{]}^{\mathrm{a}}[\frac{{L}_{1}}{{L}_{2}}{]}^{\mathrm{b}}[\frac{{T}_{1}}{{T}_{2}}{]}^{c}$

Putting the above values, we get

${\mathrm{N}}_{2}=10[\frac{1\mathrm{g}}{1\mathrm{kg}}{]}^{1}[\frac{1\mathrm{cm}}{1\mathrm{m}}{]}^{1}[\frac{1\mathrm{s}}{1s}{]}^{\u20132}$

=${\mathrm{N}}_{2}=10[\frac{1\mathrm{g}}{1000\mathrm{g}}{]}^{1}[\frac{1\mathrm{cm}}{100\mathrm{cm}}{]}^{1}[\frac{1\mathrm{s}}{1s}{]}^{\u20132}$

$=10\times {10}^{\u20133}\times {10}^{\u20132}\times 1$$={10}^{\u20134}N$

$Hence,10dyne={10}^{\u20134}Nproved$

**# Determine the dimension of universal gravitational constant (G)**

Solution,

We have

$F=\frac{Gm1m2}{{r}^{2}}$

$G=\frac{F{r}^{2}}{m1m2}$

$G=\frac{\left[ML{T}^{\u20132}\right]\left[{L}^{2}\right]}{\left[M\right]\left[M\right]}$

$G=\left[{M}^{\u20131}{L}^{3}{T}^{\u20132}\right]$

Hence the dimension of G are -1 in a mass, 3 in length and -2 in time.

__Limitation of Dimensional analysis:__

- It cannot give information about dimensionless constant involved in a physical relation.
- It is very difficult to derive the dimensional relation between more than three physical quantities.
- It cannot be use to derive the relation involving trigonometric and logarithmic functions.
- It can’t tell us whether a quantity is vector or scalar.
- The dimensionally correct relation may not be physically correct.

__Error:__

The difference between standard value and observed value of physical quantity is called error.

i.e Error = Standard value – observed value

__Note:__

The measurement of a physical quantity is expressed as x y where x is observed value and y is error which means the standard value of this measurement lies between (x + y) to (x – y).

**Q. The measurement of resistance of a conductor is expressed as (10 **** 5%) ****Ω****. What does it means?**

**ans: **The measurement of resistance of a conductor is expressed as (10 5%) Ω which means 10 Ω is observed value and 5% is error, so the standard value of this resistance lies between (10 – 0.05) Ω to (10 + 0.05) Ω.

__Types of error :__

__Systematic error__

The error introduced due to the Limitation of the formula used and fault in the instrument is called systematic error.

__Random error__

The error introduced due to the carelessness of the experiment and unfavorable condition of the environment such as temperature, pressure, and humidity etc is called Random error.

__Accurate Measurement:__

The measurement in which observed value of any physical quantity is closer to the standard value of that physical quantity is called accurate measurement.

Example: Standard value of g in lab is 9.8 m/s^{2}. The measured value of g is obtained as 9.67 m/s^{2 }and 9.79 m/s^{2}. The value so, it is more accurate than the value 9.67 m/s^{2}.

__Precise Measurement:__

The measurement in which the observed value of a physical quantity can be reproduced again and again by repeated experiment and produce is called precise measurement.

If the reading are very close to each other, they are called precise reading. The thickness of a glass plate measured by spherometer are 2.43mm, 2.44mm, 2.42mm and 2.44mm. These reading are very closed to each other. So they are précised measurement.

__Significant Figure :__

The meaningful digits of a number are called significant figure. Significant figure depends on the least count of a measuring device.

Example:

Let us take the length of rod by a ruler for three times. Reading are obtained as 10.2cm, 10.3cm and 10.3cm. so mean length

$=\frac{10.2+10.3+10.2}{3}$

= 10.2666666 cm

Hence, all digits are not significant only first three digits are significant. The Length of rod is taken 10.2cm or 10.3 cm

Q. The length of rod is exactly 1 cm. An observe records the reading as 1.0cm, 1.00cm and 1.000cm which is most accurate measurement?

Solution

$\mathrm{Length}=\frac{1.0+1.00+1.000}{3}$

$=\frac{3.000}{3}$

=1.000

Hence, the length of rod = 1.0 or 1.00