Physical Quantities Class 11 Physics | Notes

Physical quantities:

All measurable quantities are called physical quantities. For example; mass, length, time, velocity, temperature etc.

Types of physical quantities:

There are two types of physical quantities.

1. Fundamental physical quantity:

The physical quantity which is independent of other physical quantities is called fundamental physical quantity. For example; mass, length, time, temperature etc.

2. Derived physical quantity:

The physical quantity which is derived from other physical quantities is called derived physical quantity. For example; velocity, acceleration, force etc.

Unit:

The standard physical quantity taken as reference to express the physical quantity is called unit. For example; kg, meter, ms^-1, Newton (N) etc.

Types of unit:

1. Fundamental unit:

The unit which is independent of other units is known as fundamental unit. For example; kg, m, sec. etc.

2. Derived unit:

The unit which is derived from other units is known as derived unit. For example; kgm^-3, ms^-1etc.

System of unit	Length	Mass	Time
F.P.S.	foot	pound	sec
C.G.S	cm	gm	sec
M.K.S.	m	kg	sec

SI system of unit:

In 1960, the general conference of weight and measurement held in Paris gave a worldwide accepted system of unit is called SI system of unit.

Physical quantities and their units given by S.I. System of unit

Physical quantity	Unit	Symbol
1. Mass	Kilogram	kg
2. Length	meter	m
3. Time	second	S
4. Temperature	Kelvin	K
5. Current	Ampere	A
6. Luminous intensity	Candela	cd
7. Amount of substance	mole	mol

Supplementary SI Units:

These are dimensionless units.

Physical quantity	Unit	Symbol
1. Angle	radian	rad
2. Solid angle	steradian	Sr

Dimension:

The dimension of a physical quantity is defined as the power raised to the fundamental quantities to express the physical quantity.

Dimensional formula:

It is defined as the expression of physical quantity in terms of fundamental physical quantity with proper dimension. The dimensional formula of mass, length, time, temperature and current are [M], [L], [T], [K] & [A] respectively.

Find the dimensional formula of the following:

1. Velocity, v = $\frac{d}{t}$

$\therefore $ Dimensional formula of v = $\frac{[L]}{[T]}$= [LT^-1] = [M⁰LT^-1]

Note: The dimensional formula of Velocity is [M⁰LT^-1] and dimensions of velocity are 0 in mass, 1 in length and –1 in time.

2. Momentum, P = mv

$\therefore $ Dimensional formula of P = [M] [LT^-1] = [MLT^-1]

3. Acceleration, a = $\frac{dv}{dt}$

$\therefore $ Dimensional formula of a = $\frac{[L{{T}^{-1}}]}{[T]}$= [LT^-2]= [M⁰LT^-2]

4. Force, F= ma

$\therefore $ Dimensional formula of F = [M] [LT^-2] = [MLT^-2]

5. Kinetic Energy, K.E.= $\frac{1}{2}$mv²

$\therefore $ Dimensional formula of K.E. = [M] [LT^-1]²= [ML²T^-2]

6. Potential Energy, P.E.= mgh

$\therefore $ Dimensional formula of P.E. = [M] [LT^-2][L] = [ML²T^-2]

7. Work done, W = F. d

$\therefore $ Dimension of W = [MLT^-2] [M] = [ML²T^-2]

8. Specific heat capacity (S)

Heat loss/gain, dQ = mS$\Delta \theta $

or, S = $\frac{dQ}{m\Delta \theta }$

$\therefore $ Dimensional formula of S = $\frac{[M{{L}^{2}}{{T}^{-2}}]}{[M][K]}$ = [M⁰L²T^-2K^–1]

9. Charge (Q)

Current, I = $\frac{Q}{t}$, Q = It

$\therefore $ Dimensional formula of Q = [A] [T] = [M⁰L⁰TA]

Q. Find the dimensional formula of the following physical quantities:

10. Power, P {Hint: P = $\frac{Workdone}{time}$ , Ans: [ML²T^-3]}

11. Pressure, P {Hint: P = $\frac{Force}{area}$ , Ans: [ML^–1T^-2]}

12. Surface tension, T {Hint: T = $\frac{Force}{length}$, Ans: [ML⁰T^-2]}

13. Coefficient of viscosity ($\eta $) {Hint: F = $6\pi \eta r{{v}_{t}}$, Ans: [M¹L^–1T^-1]}

14. Specific latent heat capacity (L), {Hint: dQ = mL, Ans: [M⁰L²T^-2]}

Types of physical quantities:

Physical quantities may be variables or constants. They may or may not have dimension.

(i) Dimensional variable:

These are the quantities which are variable and have dimensions as well. For example; velocity, momentum, force etc.

(ii) Dimensional constants:

These are the quantities which have constant values and yet have dimensions. For example; gravitational constant, coefficient of viscosity, Planck’s constant, velocity of light etc. Their numerical value does not change.

(iii) Non-dimensional Variables:

These are variable quantities and have no dimensions. For example; specific gravity, strain, angle etc. They are typically ratio of two similar physical quantities.

(iv) Dimensionless constants:

These are numbers like 2, 3, 4, $\pi $. The numerical value of dimensionless physical quantities remains unchanged in any system of units.

Dimensional equation:

The equation of physical quantities expressed in terms of their dimensional formula is called dimensional equation.

Principle of homogeneity:

It states that for any correct physical relation (formula), the dimensions of physical quantity on L.H.S. is equals to that on R.H.S.

1. For any correct physical relation, the dimensions of each term on L.H.S. are equal to the dimensions of each term on R.H.S.

2. Two physical quantities can be added or subtracted only when their dimensions are same.

Application (uses) of dimensional equation:

1. To check the correctness of physical relation (formula)

2. To derive the physical relation between various physical quantities.

3. To convert a physical quantity from one system of unit to another system of unit.

4. To determine the dimensional formula of physical constant from a given relation.

Application No: 1

To check the correctness of physical relation (formula):

Example 1. Check the correctness of v² = u² + 2as.

Ans: The dimension of L.H.S., v²= [M⁰LT^-1]² = [M⁰L²T^-2]

The dimension of R.H.S., u² + 2as = [M⁰LT^-1]² + 2 [M⁰LT^-2] [L] = [M⁰L²T^-2] + 2[M⁰L²T^-2]

= 3[M⁰L²T^-2]

= [M⁰L²T^-2] ∵ 3 has no dimension.

Here, dimension of L.H.S. = Dimension of R.H.S.

$\therefore $The given relation v² = u² + 2as is dimensionally correct.

Frequently Asked Questions in Exam:

Example 2. Check the correctness of physical relation v² = u² + 5as. [Ans: Correct]

Example 3. Check the correctness of physical relation v = u + at². [Ans: Incorrect]

Example 5. Check the correctness of formula T = 2$\pi $$\sqrt{\frac{{l}}{{g}}}$. [Ans: Correct]

Example 6. A student writes $\sqrt{\frac{{R}}{{2GM}}}$for escape velocity. Check the correctness of the formula by using dimensional method. [Ans: Incorrect]

Example 7. Check the correctness of formula t = 2$\pi $$\sqrt{\frac{{m}}{{k}}}$, where ‘t’ be the time period, ‘m’ is the mass and ‘k’ is the force per unit displacement. [Ans: Correct]

Example 8. A student writes an expression of force causing a body of mass ‘m’ to move in a circular motion with radius ‘r’ and velocity ‘v’ as F = $\frac{{m}{{{v}}^{2}}}{{r}}$. Use dimensional method to check its correctness. [Ans: Correct]

Example 9. Check the correctness of the relation h = $\frac{{2Tcos}\theta }{r\rho g}$, where ‘h’ is height, ‘T’ is surface tension, ‘r’ is the radius and ‘$\rho $’ is density of liquid. [Ans: Correct]

Example 10. The density $\rho $ of the earth with radius R is given by $\rho $ = $\frac{{3g}}{{4}\pi RG}$ where g is acceleration due to gravity and G is gravitational constant. Check the dimensional consistency of this relation. [Ans: Correct]

Q. Is dimensionally correct relation be always a correct physical relation? Justify your answer. What about dimensionally wrong equation?

Application No: 2

To derive the physical relation between various physical quantities:

Example 1. The time period of simple pendulum depends upon its mass, length and acceleration due to gravity. Derive formula for time period using dimensional method.

Ans: Let us consider a simple pendulum of mass ‘m’ and effective length ‘l ‘ and the value of acceleration due to gravity is ‘g’.

According to question, the time period of simple pendulum depends upon

1. its mass i.e. T $\alpha $ m^a………………….(1)

2. its length i.e. T $\alpha $ l^b………………….(2)

3. and ‘g’ i.e. T $\alpha $ g^c………………….(3)

Where a, b and c are constants which are to the determined.

Combining (1), (2) and (3) we get,

T $\alpha $ m^a l^bg^c

T = k m^a l^bg^c………………..(4)

where ‘k’ is proportionality constant which has no dimension.

The dimensional formula of T = [M⁰L⁰T]

The dimensional formula of m = [ML⁰T⁰]

The dimensional formula of l = [M⁰LT⁰]

The dimensional formula of g = [M⁰LT^-2]

Rewriting equation (4) in dimensional form,

[M⁰L⁰T] = [M]^a [L]^b [LT^-2]^c

Or, [M⁰L⁰T] = [M^a L^b+c T^-2c]

Equating the powers of corresponding quantities,

(i) a = 0,

(ii) -2c = 1 $\therefore $ c = – $\frac{1}{2}$ and

(iii) b + c = 0

b – $\frac{1}{2}$= 0 $\therefore $b = $\frac{1}{2}$

Equation (4) becomes,

T = k m⁰ l^1/2g^-1/2

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

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

Experimentally the value of ‘k’ is is found to be 2$\pi $

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

This is the required expression for the time period of simple pendulum.

Frequently Asked Questions in Exam:

Example 2. Using dimensional method, derive an expression for the centripetal force ‘F’ acting on a particle of mass ‘m’ moving with velocity ‘v’ in a circle of radius ‘r’. Ans: F = $\frac{m{{v}^{2}}}{r}$

Example 3. Using dimensional method, derive Stoke’s law F_v = $6\pi \eta r{{v}_{t}}$ where the symbols have their usual meaning.

Example 4. The force acting on an object depends upon area (A), velocity (v) and density ($\rho $). Derive formula for force dimensionally. Ans: F = Av²$\rho $

Example 5. Using dimensional method, derive V_e= $\sqrt{\frac{2GM}{R}}$ where the symbols have their usual meaning.

Example 6. Using dimensional consideration, deduce Poiseulle’s formula, V = $\frac{\pi }{8}$$\left( \frac{P}{l} \right)$$\frac{{{r}^{4}}}{\eta }$ for the rate of flow of liquid through a capillary tube.

Application No: 3

To convert a given value of physical quantity from one system of unit to another system of unit.

Derivation of conversion formula from one system of unit to another system of unit

A physical quantity ‘Q’ can be expressed as

Q = nu where ‘n’ is numerical value and ‘u’ is the unit of given physical quantity.

[eg. mass, m= 5kg, where 5 is numerical value and kg is unit]

Let n₁, u₁and n₂, u₂ be the numerical value and unit of given physical quantity in first and second unit system respectively.

Then Q₁= n₁u₁ and Q₂= n₂u₂

For a given value of physical quantity, the total quantity remains same in both the unit systems.

$\therefore $ Q₁= Q₂

i.e. n₁u₁= n₂u₂ …………….(i)

Let the dimensional formula of u₁= [M₁^aL₁^bT₁^c] and the dimensional formula of u₂ =[M₂^aL₂^bT₂^c]

Eqⁿ (i) becomes, n₁ [M₁^aL₁^bT₁^c] = n₂ [M₂^aL₂^bT₂^c]

$\therefore $n₂= n_{1 ${{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}$${{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}$${{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}$}

This is the required relation to convert a given value of physical quantity from one system of unit to another system of unit.

Example 1. Convert 10 Newton into dyne.

Ans: Here,

‘N’ and ‘dyne’ are the units of force in SI and CGS system respectively.

The dimensional formula of force = [MLT^-2]

Given system (SI) new system

n₁= 10 n₂= ?

M₁= 1kg M₂= 1g

L₁= 1m L₂= 1cm

T₁= 1sec T₂= 1sec

We have, n₂= n_{1 ${{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}$${{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}$${{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}$}

where, a =1, b = 1 & c = –2 ∵ dimensional formula of F = [MLT^-2]

$\therefore $ n₂= 10 _{${{\left[ \frac{1kg}{1g} \right]}^{1}}$${{\left[ \frac{1m}{1cm} \right]}^{1}}$${{\left[ \frac{1sec}{1sec} \right]}^{2}}$}

Or, n₂= 10 ${{\left[ \frac{1000g}{1g} \right]}^{1}}$${{\left[ \frac{100cm}{1cm} \right]}^{1}}$[1]^–2

Or, n₂= 10×1000×100

Or, n₂= 1×10⁶

Since, n₁u₁= n₂u₂

$\therefore $ 10N = 10⁶dyne.

Frequently Asked Questions in Exam:

Example 2. Convert 10 ergs into joules. Ans: 10^–6J

Example 3. The density of gold is 19.3 gcm^-3. Express this value in SI unit. Ans: 19300 kgm^-3

Example 4: Convert 36 kmhr^-1 to ms^-1. Ans: 10 ms^-1

Example 5: Convert 3×10^–7J into ergs. Ans: 3 ergs

Example 5: Convert density of water 2 g/cm³into kg/m³. Ans: 2000 kg/m³

Application No: 4

To determine the dimensional formula of physical constant from a given relation.

Example 1: Find the dimensional formula of gas constant ‘R’ from PV = nRT or PV = RT

Solⁿ:

We have, PV= nRT

R = $\frac{PV}{nT}$

= $\frac{F}{A}$$\frac{V}{nT}$

= $\frac{[ML{{T}^{-2}}]{{[L]}^{3}}}{{{[L]}^{2}}[K]}$ [ ∵ n is a number and has no dimension]

$\therefore $The dimensional formula of R = [M L²T^-2K^-1]

Frequently Asked Questions in Exam:

Example 2: Find the dimensions of Planck’s constant ‘h’ at from the given equation $\lambda $ = $\frac{h}{p}$ where ‘$\lambda $’ is wavelength and ‘p’ is momentum of photon. Ans: [ML²T^-1]

Example 3: Find the dimensional formula of Gravitational constant (G).

Hint: { F = G $\frac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$ , Ans: [M^–1L³T^-2]}

Limitations/ Draw backs / Shortcomings of dimensional analysis:

Following are the limitations of dimensional analysis:

1. It does not give information about a physical quantity whether it is vector or scalar quantity.

2. It gives no information about dimensionless constant.

3. It is not applicable to derive a formula having more than one term either in L.H.S. or in R.H.S.

4. It cannot differentiate between two different physical quantities having same dimensions.

5. It is not applicable to trigonometric function, logarithmic function, and exponential function etc.

Some important questions:

Q. Taking force length and time as fundamental quantity find the dimensional formula of density and gravitational constant. Ans: [FL^-4T²] & [F^-1L⁴T^-4]

Q. In Van der Waals equation, (P + $\frac{a}{{{V}^{2}}}$) ( v–b) = RT, find the dimensional formula for ‘a’ and ‘b’.

Ans: [M⁰L³T⁰] & [ML⁵T^-2]

Q. The force ‘F’ is given in terms of time ‘t’ and displacement ‘x’ by equation

F = AsinBx + CsinDt. What is the dimension of $\frac{D}{B}$? Ans: [LT^-1]

Q. The displacement ‘y’ is expressed as y = a +bt + ct²+dt³, where‘t’ is time. Find the dimensions of a, b, c and d. Ans: [L], [LT^-1], [LT^-2], [LT^-3]

Significant figure:

The number of meaningful digits in a number is called number of significant figure (digit).

Rules:

1. All non-zero digits are significant figure. For example; 5563$\to $ 4 significant figure.

2. The zeros between two non-zero digits are significant figure.

For example; 2018$\to $4 significant figure.

3. The zeros left to non-zero digit are not significant figure.

For example; 005$\to $ 1 significant figure,

0.2cm$\to $ 1 significant figure

05563$\to $ 4 significant figure,

0.001cm$\to $ 1 significant figure.

4. The zeros right to the decimal and right to non-zero digit are significant figure.

For example; 0.001500$\to $ 4 significant figure.

5. The zeros right to the non-zero digit are not significant figure.

For example; 71000g$\to $2 significant figure.

6. The powers of 10 are not significant figures.

For example; 71×10^{3$\to $}2 significant figure.

Precise and accurate measurement:

Precise measurement:

It refers to the closeness of measured values to each other. In other words the limit or resolution to which a physical quantity is measured by the instrument is called precision. It is associated with small random errors.

$\to $ Less the least count of an instrument, more is the preciseness in the measurement.

$\to $ A precise measurement is not necessarily an accurate measurement.

Accurate measurement:

It refers to the closeness of measured values to the true value. It is associated with small symmetric uncertainties.

$\to $ Good accuracy means the reading or mean of set of readings is very close to the true value.

$\to $ Greater the number of significant figures, greater is the accuracy.

Example: Let us consider a man is 70 kg (true value).

1. The set of measurements 90 kg, 80 kg, 60 kg and 50 kg is accurate but not precise.

2. The set of measurements 85.23kg, 85.25kg, 85.26kg is precise but not accurate.

3. The set of measurements 69.81kg, 69.94kg, 70.00kg, 75.11kg is both accurate and precise.