## Grade 11 Physics Note

# Unit and measurement

**Unit:**

A unit is defined as a convention to define an amount of physical property in a specific system of units. There are several systems of units with different conventions to express these units. There are units defined for every physical quantity and that physical quantity is expressed in terms of that specific unit. For example, in International System of units we have kilogram for mass, meter for length, seconds/minutes/hours for time, etc. Without a unit a physical property cannot be distinguished or described.

For writing, units are expressed in terms of alphabets, referred to as unit symbols. Like meter is written ‘m’, second as ‘s’, grams as ’g’, etc.

To expresses larger quantities of units in terms of power of tens, certain prefixes are used.

Examples: kilogram for 1000 grams, centimeters for 1/100 of a meter meters, milliseconds for 1/1000 of a second, etc.

**International system of units:**

Initially there were several systems of units in practice such as the MKS system which used meter for length, kilogram for mass and second for time, CGS system which used centimeter for length, gram for mass and second for time, British Engineering system which used foot for length, slugs for mass and second for time. The diversity in the use of system of units by different fields caused problems in coordination between the units and hence there was a need of a system of units to be used internationally. Therefore, the MKS system was promoted to be called as The International System of Units abbreviated as the SI system. The Si system has seven fundamental physical quantities given below:

Physical Quantity |
Unit |
Symbol |

mass |
kilogram |
kg |

length |
Meter |
m |

time |
second |
s |

temperature |
Kelvin |
K |

electric current |
ampere |
A |

amount of matter |
Mole |
mol |

luminous intensity |
candela |
Cd |

**Derived units:**

Derived units are composed of the fundamental units and are originated from the products and ratios of the fundamental units. Derived units belong to physical quantities that are derived from basic physical properties. Examples are:

Velocity – a physical quantity derived from fundamental quantities of length and time

${\rm{velocity}} = \frac{{{\rm{length}}}}{{{\rm{time}}}} = \frac{{\rm{m}}}{{\rm{s}}}$

So the units of velocity is ms^{-1}

Acceleration – a physical quantity derived from velocity and time

${\rm{acceleration}} = \frac{{{\rm{velocity}}}}{{{\rm{time}}}} = \frac{{\rm{m}}}{{{{\rm{s}}^2}}}$

The units of acceleration is ms^{-2}

Force – a physical quantity derived from mass and acceleration

${\rm{force}} = {\rm{mass}}* {\rm{acceleration}} = {\rm{kg\

: m\: }}{{\rm{s}}^{ - 1}}{\rm{\: or\: N}}$

The units of force are kg.m.s^{-1} or N (Newton)

Work – a physical quantity derived from force and length

${\rm{work}} = {\rm{force}}* {\rm{length}} = {\rm{N\: m\: or\: J}}$

The units of work are Nm or J (Joules)

**Principle of homogeneity of dimensions:**

The principle of homogeneity of dimensions is used to verify the physical laws involving physical quantities. For a relation to be valid and correct the dimensional form of the physical law must be balanced, i.e. the dimensions at both side of the equation must be same. Every physical law that is valid obeys the principle of homogeneity of dimensions.

For example consider following equations:

Equations of motion:

${{\rm{v}}_{\rm{f}}} = {{\rm{v}}_{\rm{i}}} + {\rm{a\: t}}$

${\rm{L}}{{\rm{T}}^{ - 1}} = {\rm{L}}{{\rm{T}}^{ - 1}} + {\rm{L}}{{\rm{T}}^{ - 2}}{\rm{\: T}}$

${\rm{L}}{{\rm{T}}^{ - 1}} = {\rm{L}}{{\rm{T}}^{ - 1}} + {\rm{L}}{{\rm{T}}^{ - 1}}$

${\rm{s}} = {{\rm{v}}_{\rm{i}}}{\rm{\: t}} + \frac{1}{2}{\rm{a\: }}{{\rm{t}}^2}$

${\rm{L}} = {\rm{L}}{{\rm{T}}^{ - 1}}{\rm{\: T}} + \frac{1}{2}{\rm{L}}{{\rm{T}}^{ - 2}}{{\rm{T}}^2}$

${\rm{L}} = {\rm{L}} + \frac{1}{2}{\rm{L}}$

$2{\rm{\: a\: s}} = {\rm{v}}_{\rm{f}}^2 - {\rm{v}}_{\rm{i}}^2$

$2{\rm{\: L}}{{\rm{T}}^{ - 2}}{\rm{\: L}} = {\left( {{\rm{L}}{{\rm{T}}^{ - 1}}} \right)^2} - {\left( {{\rm{L}}{{\rm{T}}^{ - 1}}} \right)^2}$

$2{{\rm{L}}^2}{{\rm{T}}^{ - 2}} = {{\rm{L}}^2}{{\rm{T}}^{ - 2}} - {{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

Mass – Energy equivalence:

${\rm{E}} = {\rm{m}}.{{\rm{c}}^2}$

${\rm{J}} = {\rm{kg}}.{\left( {{\rm{m}}{{\rm{s}}^{ - 1}}} \right)^2}$

${\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}} = {\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

All the equations that are given above are dimensionally correct i.e. the dimensions are same at both sides of the equation. It is to be noted that all the constants are ignored in dimensional form and also dimensions are never to be canceled by subtraction, just like we have two similar terms with opposite signs in last equation of motion.

**Dimensional analysis:**

Dimensional analysis involves checking of a given physical law to check for its validity. Any physical law that is valid follows principle of dimensional homogeneity, and hence this analysis is used to check and verify physical laws by expressing them in the form of dimensional equations.It is a useful technique to check the validity of derived relations between physical quantities.

Dimensional analysis is also used to find out relationship between physical quantities. For example, let’s derive an expression for kinetic energy just with the knowledge that it depends upon the mass and velocity of a body,

${\rm{K}}.{\rm{E}} \propto {{\rm{m}}^{\rm{x}}}{{\rm{v}}^{\rm{y}}}$

${\rm{K}}.{\rm{E}} = {\rm{k\: }}{{\rm{m}}^{\rm{x}}}{{\rm{v}}^{\rm{y}}}$

Converting in dimensional form

${\rm{J}} = {\rm{k\: k}}{{\rm{g}}^{\rm{x}}}{\left( {{\rm{m}}{{\rm{s}}^{ - 1}}} \right)^{\rm{y}}}$

${\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}} = {\rm{k\: }}{{\rm{M}}^{\rm{x}}}{{\rm{L}}^{\rm{y}}}{{\rm{T}}^{ - {\rm{y}}}}$

Comparing powers of dimensions we get

Comparing M |
$1 = {\rm{x}}$ |
Equation – 1 |

Comparing L |
$2 = {\rm{y}}$ |
Equation – 2 |

Comparing T |
$ - 2 = - {\rm{y}}$ |
Equation – 3 |

Using values obtained through the above equations, the kinetic energy can be written as,

${\rm{K}}.{\rm{E}} = {\rm{k\: }}{{\rm{m}}^1}{{\rm{v}}^2}$

And kinetic energy is given as,

${\rm{K}}.{\rm{E}} = \frac{1}{2}{\rm{\: m\: }}{{\rm{v}}^2}$

Hence the dimensional analysis is used to find relations between the physical quantities.

**Coherent system of unit:**

A coherent system of units is the one which have its derived units made up of from base unit without any other numerical factor other than 1 or some power of ten. The Si system is the coherent system of units. For example consider the unit of power in SI system i.e. watt. A watt is expressed as,

$1{\rm{\: watt}} = \frac{{1{\rm{\: Joule}}}}{{1{\rm{\: second}}}}$

Hence the SI unit of power ‘watt’ is coherent.

The unit of volume ‘liter’ is expressed as,

$1{\rm{\: liter\: }}\left( {{\rm{d}}{{\rm{m}}^3}} \right) = 1000{\rm{\: c}}{{\rm{m}}^3}$

Hence ‘liter’ is also a coherent unit.

Example of non-coherent unit is horsepower which is expressed as,

$1{\rm{\: horsepower}} = \frac{{550{\rm{\: foot}} - {\rm{pound}}}}{{1{\rm{\: second}}}}{\rm{or\: \: }}746{\rm{\: watts}}$

As the numerical factor is not 1 neither some whole power of ten, it is considered as non-coherent unit.

**Dimensional equations:**

Dimensional equations are equations of physical laws expressed in terms of fundamental quantities. When correctly written, these equations are also dimensionally correct, i.e. the physical quantities are balanced at both sides of the equation. This method is known as dimensional analysis and is used to verify analytical relationship obtained from experimental data.

For example:

${{\rm{v}}_{\rm{f}}} = {{\rm{v}}_{\rm{i}}} + {\rm{at}}$

expressing each quantity in dimensional form

${\rm{L}}{{\rm{T}}^{ - 1}} = {\rm{L}}{{\rm{T}}^{ - 1}} + {\rm{L}}{{\rm{T}}^{ - 2}}{\rm{*T}}$

${\rm{L}}{{\rm{T}}^{ - 1}} = {\rm{L}}{{\rm{T}}^{ - 1}} + {\rm{L}}{{\rm{T}}^{ - 1}}$

Hence the quantities on both sides of the equation have same dimensions of LT^{-1}, therefore the equation is dimensionally correct. One might think that right side may be written as

${\rm{L}}{{\rm{T}}^{ - 1}} = 2{\rm{L}}{{\rm{T}}^{ - 1}}$

In this case the numbers will not prove the equations as invalid. In dimensional analysis the numbers are not significant and only dimensions are considered.

For another example, time period is given as,

${\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} $

Expressing in dimensional form

${\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{{\rm{L}}}{{{\rm{L}}{{\rm{T}}^{ - 2}}}}} $

Where g is acceleration due to gravity and has same dimensions as that of acceleration

${\rm{T}} = 2{\rm{\pi \: T}}$

Which is again dimensionally correct and constant term ‘2π’ is ignored.

**Pressure and stress have the same dimensions;**

Pressure is defined as the applied external force normal to the unit area:

${\rm{P}} = \frac{{\rm{F}}}{{\rm{A}}}$

Dimensionally, force can be expressed as ${\rm{ML}}{{\rm{T}}^{ - 2}}$

And area can be expressed as ${{\rm{L}}^2}$ so,

${\rm{P}} = \frac{{{\rm{ML}}{{\rm{T}}^{ - 2}}}}{{{{\rm{L}}^2}}}$

${\rm{P}} = {\rm{M}}{{\rm{L}}^{ - 1}}{{\rm{T}}^{ - 2}}$

Now, stress is the internal force exerted by the body on a unit area:

$\sigma = \frac{{\rm{F}}}{{\rm{A}}}$

Dimensionally, force can be expressed as ${\rm{ML}}{{\rm{T}}^{ - 2}}$

And area can be expressed as ${{\rm{L}}^2}$ so,

$\sigma = \frac{{{\rm{ML}}{{\rm{T}}^{ - 2}}}}{{{{\rm{L}}^2}}}$

$\sigma = {\rm{M}}{{\rm{L}}^{ - 1}}{{\rm{T}}^{ - 2}}$

**Momentum and Impulse:**

Momentum is defined as the product of mass and velocity. Momentum is a vector quantity and has a direction along with magnitude. The SI unit for momentum is kgms^{-1} or N-s Momentum can be written as,

${\rm{Momentum\: }}\left( {\rm{p}} \right) = {\rm{m\: v}}$

Converting in dimensional form we get,

${\rm{p}} = {\rm{ML}}{{\rm{T}}^{ - 1}}$

Impulse is defined as the integral of force acting on a body in a time interval during which the force is acting on the body. The unit of impulse is N-s. As it involves force which is a vector, impulse is also a vector quantity. Impulse is expressed as,

${\rm{Impulse\: }}\left( {\rm{J}} \right) = {\rm{F\: t}}$

Converting in dimensional form we get,

${\rm{J}} = {\rm{ML}}{{\rm{T}}^{ - 2}}{\rm{\: T}} = {\rm{ML}}{{\rm{T}}^{ - 1}}$

Since it is proved that the momentum (p) and impulse (J) have same dimensions.

**Torque and Energy**

Torque can be defined as the force with a tendency of rotating a body about an axis. Torque causes the body to rotate. Torque is a vector quantity and defined as the cross product of force and the distance of force from the axis of rotation, also called as turning radius. The units of torque is N-m and expressed as,

${\rm{Torque\: }}\left( \tau \right) = {\rm{F}}* {\rm{r}}$

Converting in dimensional form we get,

$\tau = {\rm{ML}}{{\rm{T}}^{ - 2}}{\rm{\: L}} = {\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

Energy can be defined as the ability of a system to perform work. As it is something that makes a system able to do work, the energy is expressed in units of work i.e. Joules. Writing energy in dimensional form we get,

${\rm{Energy\: }}\left( {\rm{E}} \right) = {\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

Hence it is proved that the torque and energy have same dimensional forms.

**Work and Energy**

Work is said to be perform when a force acts on a body and displaces it through some distance. This physical quantity has the units of N-m or Joules. Work is expressed as,

${\rm{Work\: }}\left( {\rm{W}} \right) = {\rm{F\: d}}$

Converting into dimensional form we get,

${\rm{W}} = {\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

The term energy has been already defined and can be expressed in the dimensional form as

${\rm{Energy\: }}\left( {\rm{E}} \right) = {\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}$

Hence energy and work has same physical dimensional form.