### Physical Quantity:

A physical quantity is a quantity that can be measured i.e. a physical quantity is properly defined, has proper units, and its value can be measured by an instrument.

Physical quantities are classified as fundamental and derived quantities.

### Fundamental Quantities :

Fundamental quantities are those that are defined directly by the process of measurement only. They are not defined in terms of other quantities; their units are not defined in terms of other units. In mechanics we treat length, mass and time as basic or fundamental quantities.

**Derived quantities** : The quantities, which can be expressed in terms of the fundamental quantities are called derived quantities. For example volume, velocity, Pressure etc.

**Derived Units :**

The units of all other physical quantities, which can be obtained from fundamental units, are called derived unit.

**Different **System of Units:

**(a) FPS system:** The fps. system is the system of unit. In this system foot is the unit of length, pound is unit of mass and second is the unit of time.

**(b) C.G.S. system:** It is the Gaussian system in which centimeter, gram and second are taken as three basic unit for length, mass and time respectively.

**(c) M.K.S. system:** In it metre, kilogram and second are taken as the fundamental units of length, mass and time respectively.

**The drawback of f.p.s system**

Inconvenient multiples and submultiples involved in it for conversion. ex-(1 feet = 12 inch)

**The drawback of c.g.s system**

Many of the derived units on this system are inconveniently small.

**Advantages of m.k.s. system**

some of the derived units are of convenient size.

In this system ,electrical units are of practical use.

**Note :**

When m.k.s. system is extended to electricity, then

(i) If current is taken as fundamental quantity and ampere (A) as its unit, it is called MKSA system.

(ii) If charge (Q) is taken as fundamental quantity and coulomb as its unit, it is called MKSQ system.

**International system of Units (SI)**

(i) Also known as rationalized m.k.s. system.

(ii) SI system can be used to any branch of physics while m.k.s is used only in mechanics.

(iii) It has seven fundamental and two basic supplementary

Illustration : Find the unit of speed.

Solution : $ \displaystyle Speed = \frac{distance}{time} = m s^{-1}$

__Definitions of Base Units:__

Meter:

Since 1983, the standard metre is defined as the length of the path travelled by light in vacuum in (1/299,792,458) th part of a second.

Kilogram:

Nowadays the standard kilogram is the mass of a cylinder made of platinum-iridium alloy and stored in a special vaule in the International Bureau of Weights and Measures at Sevres in France.

Second:

At present second is defined on the basis of an atomic clock, which uses the energy difference between the two lowest energy states of the cesium atom. When bombarded by microwaves of precisely the proper frequency, cesium atoms undergo a transition from one of these states to other. One second is defined as the time required for 9,192,631,770 cycles of this radiation

In physics SI system is based on seven fundamental and two derived units.

Basic Physical Quantities: | Fundamental Unit: |

Mass | kilogram |

Length | meter |

Time | second |

Temperature | kelvin |

Electric current | ampere |

Luminous intensity | candela |

Quantity of matter | mole |

Supplementary Physical Quantities | Supplementary unit |

Plane angle | radian |

Solid angle | steradian |

### Some Practical units:

For the measurement of very large distance, the following three units are used.

(i) Astronomical Unit (AU) : It is the average distance of the centre of the sun from the centre of the earth.

1 AU = 1.5 × 10^{11 }metre

(ii) Light year : It is defined as the distance traveled by light in vaccum in one year

1 light year = 3 × 10^{8} × (365 × 24 × 60 × 60) metre

1 ly = 9.4 × 10^{15 }metre

(iii) Par sec : It is defined as the distance at which an arc 1 AU long subtends an angle of 1’’ .

1 par sec = 3.1 × 10^{16 }metre

Solved Example : Fill in the blank by suitable conversion of units

1 kg m^{2}s^{-2} =….. g cm^{2}s^{-2}

Solution : 1 kg m^{2}s^{-2} = 1×10^{3} g (10^{2}cm)^{2 }s^{-2}

= 10^{7}g cm^{2 }s^{-2}

Solved Example : Calculate the angle of (a) 1° (degree) (b) 1′ (minute of arc or arc min) and (c) 1” (second of arc or arc sec) in radian. Use 360° = 2 π , 1° = 60′ and 1′ = 60”.

Sol: (a)$\large 1^o = \frac{2\pi}{360} rad = \frac{\pi}{180} rad$

$\large = \frac{22}{7 \times 180} rad = 1.746 \times 10^{-2} rad$

(b) 1 arc min = 1′ $\large = \frac{1}{60} deg = \frac{1}{60} \times \frac{\pi}{180} rad$

= 2.91 × 10^{-4} rad

(c) 1 arc sec = (1/60)’ $\large = \frac{1}{60 \times 60} deg = \frac{1}{60 \times 60} \times \frac{\pi}{180} rad $

= 4.85 × 10^{-6} rad

Solved Example : Fill in the blanks:

(a)The volume of a cube of side 1 cm is equal to …….. m^{3}

(b)The surface area of a solid cylinder of radius 2 cm and height 10 cm is equal to …….. (mm)^{2}

(c)A vehicle moving with a speed of 18 km h^{-1} cover …… m in 1 s

(d)The relative density of load is 11.3 Its density is ……. g cm^{-3} or ……. kg m^{-3}

Solution: (a)Length L = 1 cm = 10^{-2} m,

Volume, L^{3} = (10^{-2})^{3} = 10^{-6} m^{3}

(b)Given, r = 2 cm = 20 mm, h = 10 cm = 100 mm

Surface area of solid cylinder = (2πr)h

= 2 ×3.14 × 20 × 100 = 1.26 × 10^{4} mm^{2}

(c) Speed, v = 18 kmh^{-1 }

$ = \frac{18 \times 1000}{60 \times 60} $

v = 5 ms^{-1}

Distance covered in 1 sec = 5 m

(d) Relative density = 11.3

density = 11.3 g/cc = = 11.2 × 10^{3} kgm^{-3}

Solved Example : Fill in the blank by suitable conversion of units:

(a)1 kgm^{-2} s^{-2} = —–g cm2 s^{-2}

(b) 1 m = —– light year

(c) 3 ms^{-2} = —- kmh^{-2}

(d) G = 6.67 × 10^{-11} Nm^{2} kg^{-2} = —-cm^{3} s^{-2} g^{-1}

Solution: (a)1 kgm^{2} s^{-2} = 1 × 10^{3} g (10^{2 }cm)^{2} s^{-2} = 10^{7} g cm^{-2} s^{-2}

(b)1 light year = 9.46 × 10^{15} m

1 m = 1/ 9.46 × 10^{15} = 1.053 × 10^{-16} light year

(c) 3 ms^{-2} = 3 × 10^{-3} km s^{-2} = 3 × 10^{-3} × 3600 × 3600 kmh^{-2} = 3.888 × 10^{4} kmh^{-2}.

(d) G = 6.67 × 10^{-11} Nm^{2} kg^{-2} = 6.67 × 10^{-11} (kg m s^{-1}) m^{2} kg^{-2}

= 6.67 × 10^{-11} m^{3} s^{-1} kg^{-1} = 6.67 × 10^{-8} cm^{3} s^{-1} g^{-1}

Solved Example : A calorie is unit of heat energy and it equal about 4.2 j, where 1 J = kgm^{2} s^{-2}. Suppose, the employ a system of units in which the unit of mass equals α kg, the unit of length equal β m and the unit of time is γ s . show that a calorie has a magnitude of 4.2 α^{-1} β^{-2} γ^{2} in terms of new units.

Solution: 1 calorie = 4.2 J = 4.2 kg m^{2} s^{-2}

If α kg = new units of mass

Then, 1 kg =1/α new unit of mass = α^{-1} new unit of mass

Similarly, 1 m = β^{-1} new unit of length

1 s = γ^{-1 }new unit of time

Now, 1 calorie = 4.2 (α^{-1} new unit of mass) (β^{-1} new unit of length)^{2} (γ^{-1} new unit of time)^{-2 }

= 4.2 α^{-1} β^{-2} γ^{2} unit of energy. (Proved)