Physical Quantities

Measurement:
Need of measurement:
The prime goal of physics is to study the nature by developing suitable theories based on experiment which is possible through qualitative and quantitative measurement. We need measurement to develop suitable theories and study the nature. In the measurements certain amount of matter is taken as standard and we find the number that express how many times that standard unit is in that physical quantity. In general, physical quantity is written as Q = nu,
where, n = number, u = unit of the measurement
Certain units (quality of matter) which is accepted and used throughout the world is called as standard unit.

Characteristics:
i. They are independent and well defined.
ii. They are same throughout the world.
iii. They shouldn’t affected by external factor.
iv. They must be reproducible.

Supplementary Units:

Dimension:
Dimensions of a physical quantity are the powers to be raised on the fundamental units in order to obtain the desired unit of that physical quantity.
For dimension of a physical quantity mass, length, time, temperature, amount of substance, luminous intensity and electric current are represented by M, L, T, K, Mol, Cd, and A respectively.
Examples:
1) Area (A)
A = l x b
Dimensional formula of area (A) = [L] x [L]

[latex][M^oL^2T^o][/latex]
Dimensions of area are 0, 2 and 0 in mass, length and time respectively.
2) Volume (V)
V = l x b x h
= [latex][M^oL^3T^o][/latex]
Dimension of volume are 0, 3 and 0 in mass, length and time respectively.
3) Density (D) = [latex]\frac{M}{V}[/latex]
D = [latex][ML^{-3}T^o][/latex]
Dimensional formula of a physical quantity is an expression that tells us how and which fundamental units are needed in order to obtain the desired units of that physical quantity.
4) Force (F)
F = m x a
= [latex][MLT^{-2}][/latex]
The dimensional formula of force is [latex][MLT^{-2}][/latex].
Two things are clear from the expression that unit of force depends on mass, length and time.
Unit of force varies as directly as mass and length and inversely as square of time.
Dimensional formula of physical quantity:

Demerit:

1. It is not clear about the constant of the physical quantity.
2. It is not clear about scalar or vector product.

Dimensional equations:

If the given relation is expressed in terms of dimension, then, it is called dimensional equation.

For e.g.  s = [latex]ut+\frac{1}{2}at^2[/latex]

L = [LT^-1] [T] + [LT^-2][T²]

[M⁰LT⁰] = [M⁰LT⁰] + [M⁰LT⁰]

Dimensionally correct relation may not be physically correct relation.

Principle of homogeneity:

It states that a given relation is dimensionally correct if and only if each and every terms on both side has the same dimensions.

i. S = ut + at²

Dimension of LHS = [M⁰LT⁰]

Dimension of RHS = [LT^-1] [T] + [LT^-2] [T²]

                        = [M⁰LT⁰] + [M⁰LT⁰]

Since, the dimension of each and every terms on both side are same, the given relation is dimensionally correct.

Q. If dimensionally correct relation is physically correct, what about dimensionally incorrect/wrong relation?

No, dimensionally correct relation does not necessarily be correct relation because in the above example, the terms of LHS and RHS are correct in dimension but it is not correct relation rather the correct relation is, [latex]S = ut + \frac{1}{2}[/latex] at²

ii. v³ = u² + 2at

Dimension of LHS = [M^oLT^-1]³ = [M^oL³T^-3]

Dimension of RHS = [LT^-1]² + [LT^-2] [T]

                        = [M^oL²T^-2] + [M^oLT^-1]

Since, the dimensions of both sides are not the same. It is incorrect relation.

Uses of dimensional analysis:

A. To check the correctness of the given relation.

By using homogenetic principle, we can check the correctness of the given physical relation. If the dimension of each and every term on both side are the same, then, the given relation is dimensionally correct.

For e.g. s = ut + [latex]\frac{1}{2}at^2[/latex]

L = [LT^-1] [T] + [LT^-2][T²]

[M⁰LT⁰] = [M⁰LT⁰] + [M⁰LT⁰]

B. To obtain dimension of constant.

By using homogenetic principle, we can find dimensional constant from the given relation.

a. Dimension of G

We have,

F = [latex]\frac{GM_1M_2}{R^2}[/latex]

G = [latex]\frac{FR^2}{M_1M_2}[/latex]

Now, G = [latex]\frac{[MLT^{-2}][L^2]}{[M][M]}[/latex] = [M^-1L³T^-2]

b. Dimension of coefficient of viscosity ([latex]\eta[/latex]):

We have, F = 6π[latex]\eta[/latex]rv where,

F = viscous force

[latex]\eta[/latex] = coefficient of viscosity

r = radius

v = velocity

Or, [latex]\eta = \frac{F}{6\pi rv}[/latex]

            = [latex]\frac{[MLT^{-2}]}{[L][LT^{-1}]}[/latex]

            = [ML^-1T^-1]

C. To obtain relation between different physical quantities.

By using homogeneity principle, we can obtain the relation between different physical quantities from the given statement.

Example: The time taken by a freely falling body from a certain height depends on its mass (m), height (h) and acceleration due to gravity (g). Mathematically,

t ∝ m^ah^bg^c

Or, t = K m^ah^bg^c ………….. (1)

Where, K is dimensionally dimensionless proportionality constant. a, b and c are unknowns to be determined. Dimensionally eqⁿ(1) must be correct, i.e.

Dimension of LHS = Dimension of RHS

[T] = [M]^a[L]^b[M^oLT^-2]^c

Or, [M^oL^oT] = [M^aL^b+cT^-2c]

Equating value of M, L and T

a = 0, b + c = 0

Or, b = – c

And, c = -1/2

So, a = 0, b = ½ and c = -1/2

Substituting the values of a, b and c in equation (1), we get,

t = Km^oh^1/2g^-1/2

            = k [latex]\sqrt{\frac{h}{g}}[/latex]

Experimentally, value of k is [latex]\sqrt{2}[/latex]. Therefore, t = [latex]\sqrt{\frac{2h}{g}}[/latex]

D. To convert one system of measurement into another.

Generally, a physical quantity is written as Q = n.u ……………… (1)

Where, n is numeric value

U is unit of measurement

Let, n₁ and n₂ be the numerical value of physical quantity in two system of measurements.

Where, u₁ and u₂ are units respectively.

Q = n₁u₁ = n₂u₂………….. (2)

Let, a, b and c be the dimensions of physical quantity in terms of MLT. Then,

[u₁] = [M₁^aL₁^bT₁^c]

[u₂] = [M₂^aL₂^bT₂^c]

Then, from eq^n. (2), we get,

n₁ [M₁^aL₁^bT₁^c] = n₂ [M₂^aL₂^bT₂^c]

E.g.: Convert 1 Newton into dynes:

In SI system

n₁= 1

M₁ = 1 Kg = 1000 g

L₁ = 1m = 100cm

T₁ = 1s

In CGS system

n₂ = ?

M₂ = 1g

L₂ = 1cm

T₂ = 1s

Dimensional formula of force is

[F] = [MLT^-2]

a = 1, b = 1 and c = -2

We have,

n₂= n₁ [latex][\frac{M_1}{M_2}]^a[\frac{L_1}{L_2}]^b[\frac{T_1}{T_2}]^c[/latex]

            = 1 [latex][\frac{1000g}{1g}]^1[\frac{100cm}{1cm}]^1[\frac{1s}{1s}]^{-2}[/latex]

            = 10⁵

∴ 1N = 10⁵ dynes.

Limitations of dimensional analysis:

i. It doesn’t tell us the value of constants.

ii. It does not tell us whether the physical quantity is scalar or vector.

iii. We cannot obtain the relation between physical quantities by dimensional analysis if the proportionality constant has dimensions.

iv. By dimensional analysis, we cannot obtain the relation between different physical quantities, if the relation contains trigonometric ratios.

v. By dimensional analysis, we cannot obtain the relation between different physical quantities if the relation contains more than four different physical quantities.

Classification of physical quantities:

On the basis of dimensional analysis, all physical quantities are classified into four group. i.e.

1. Dimensional variables: Those physical quantities which have fixed dimension but not fixed value. For e.g. velocity, force, acceleration, energy, mass, etc.

2. Dimensionless variable: Those physical quantities which do not have dimension and no fixed value. E.g. ratio, angle, strain.

3. Dimensionless constant: Those physical quantities which have no dimension and fixed value is called dimensionless constant. E.g., relative permittivity of water.

1. Differentiate between accuracy and precision of measurement. (2075 ‘A’, 2074 ‘B’, 2067)

Ans: Accurate measurement is the degree of agreement between experimental value and the true value. Good accuracy means the reading or the mean of a set of readings is very close to the true value and it associated with small systematic uncertainties.

The limit or resolution to which a physical quantity is measured by the instrument is known as its precision.

e.g., If one student finds acceleration due to gravity as 7.91 m/s², 7.92 m/s², 7.91 m/s², 7.71 m/s² in several experiments, then the measurement is precise, since all the values are very close to their mean.

However, if one finds 9.81 m/s², 9.80 m/s², 9.79 m/s², 9.80 m/s² then measurement is accurate and precise, which are very close to standard value 9.8 m/s².

2. In one of the printed documents, the unit of universal gravitational constant is given as NmKg^-2. Check its correctness from dimensional analysis. (2075 ‘B’)

Ans: Gravitational constant (G) = [latex]\frac{Fd^2}{M_1M_2}[/latex]

Or, G = [latex][\frac{MLT^2L^2}{MM}][/latex]

              = [latex][M^{-1}L^3T^{-2}][/latex]

              = [latex][Nm^2Kg^{-2}][/latex]

Hence, the given unit is dimensionally incorrect.

3. Is dimensionally correct equation necessarily be a correct physical relation? What about dimensionally wrong equation? (2074 ‘S’, 2063)

Ans: No, dimensionally correct equation may not be a correct physical relation. Consider the following relation,

v = u + 10at ……………… (i)

Dimensional formula of v is LT^-1 and that of (u + 10at) is also LT^-1. Hence, the relation (i) is dimensionally correct. But, it is established fact that the relation (i) isn’t correct physical relation.

Dimensionally wrong equation can’t be correct physical relation. For example, v = u + at² is dimensionally wrong equation and it is also incorrect physical relation.

4. A student writes an expression for the momentum (p) of a body of mass (m) with total energy (E) and considering the duration of time (t) as p = [latex]\sqrt{\frac{2mE}{t}}[/latex]. Check its correctness on dimensional analysis. (2073 ‘S’)

Ans: p = [latex]\sqrt{\frac{2mE}{t}}[/latex]

Dimension of p = [MLT^-1]

Dimension of m = [M]

Dimension of E = [mL²T^-2]

Dimension of t = [T]

Dimension of right hand side = [latex]\sqrt{\frac{MML^2T^{-2}}{T}}[/latex]    [∵ 2 is dimensionless]

                             = [latex]\frac{\sqrt{M^2L^2T^{-2}}}{\sqrt{T}}[/latex]

= [latex]\frac{MLT^{-1}}{T^{1/2}} = MLT^{-1-1/2}[/latex]

                             = MLT^-3/2

Since, dimension of RHS [latex]\neq[/latex] Dimension of LHS, the given expression is dimensionally incorrect.

5. Check dimensionally correctness of the Stoke’s formula, F = [latex]6\pi \eta rv[/latex]  where symbols have their usual meanings. (2073 ‘C’)

Ans: F = [latex]6\pi \eta rv[/latex]

Dimension of F = [MLT^-2]

Dimension of  = [ML^-1T^-1]

Dimension of r = [L]

Dimension of v = [LT^-1]

∴ [MLT^-2] = [latex]6\pi [ML^{-1}T^{-1}][L][LT^{-1}] = 6\pi [ML^{-1+1+1}T^{-1-1}][/latex]

[latex][MLT^{-2}] = 6\pi [MLT^{-2}][/latex]

Here, [latex]6\pi[/latex] is constant and unitless. Since dimension of LHS is equal to dimension of RHS. Hence, the relation is correct.

6. Check the correctness of the formula, PV = RT. (2072 ‘S’)

Ans: Here, PV = RT

Dimension of P = [ML^-1T^-2]

Dimension of V = [L³]

Dimension of R = [ML²T^-2mol^-1K^-1]

Dimension of T = [K]

∴ [latex][ML^{-1}T^{-2}L^3] = [ML^2T^{-2}mol^{-1}K^{-1}K][/latex]

Or, [ML²T^-2] = [ML²T^-2mol^-1]

But, for 1 mole of gas,

[ML²T^-2] = [ML²T^-2]

Hence, for 1 mole of gas, the equation is correct.

7. The diameter of a steel rod is given as 56.47 [latex]\pm[/latex] 0.02 mm. What does it mean? (2072 ‘C’)

Ans: The precision of a number is generally indicated by following the number with the symbol [latex]\pm[/latex]; and the second number indicating the maximum likely error.

If the diameter of a steel rod is given as 56.47 [latex]\pm[/latex] 0.02 mm then the true diameter lies between 56.45 mm and 56.49 mm. The precession can also be expressed in terms of maximum likely percentage error.

8. The length of rod is exactly 1 cm. An observer records the readings as 1.0 cm, 1.00 cm and 1.000 cm, which is the most accurate measurement? (2072 ‘D’)

Ans: The digits whose values are accurately known in a particular measurement are called its significant figures. In 1.0 cm, there are 2 significant figures, in 1.00 cm, there are 3 and in 1.000 cm there are 4 significant figures. Greater the no. of significant figures obtained when making a measurement, more accurate the measurement is. Hence, in the given three observations, 1.000 cm is more accurate.

9. What do you mean by significant figure? (2071 ‘S’)

Ans: The number of meaningful digits in a number is called the number of significant figures. The significant figures in a measured physical quantity indicate the number of digits for which we have more confidence.

For example, if an internal radius of a hollow cylinder is 2.71 cm by using a vernier callipers having the least count 0.01 cm, the significant figures in the above example are 3. In the significant figures, the right most digit must be rounded to the figures of the least count of the instrument.

10. A student writes an expression of the force causing a body of mass (m) to move in a circular motion with a velocity (v) as F = mv². Use the dimensional method to check its correctness. (2071 ‘D’)

Ans: Given, F = mv²

Dimension of F = [MLT^-2]

Dimension of m = [M]

Dimension of v = [LT^-1]

 Dimension of RHS = [MLT^-1]

Since, dimension of LHS is not equal to dimension of RHS, the given equation is dimensionally incorrect.

11. Name any two physical quantities which have the same dimensions. Can a quantity have unit but no dimension? Explain. (2070 ‘C’)

Ans: 1^st Part: The two physical quantities which have same dimension are work and torque.

Work = Force x displacement = [MLT^-2] x [L] = [ML²T^-2]

Torque = Force x perpendicular distance = [MLT^-2] x [L] = [ML²T^-2]

2^nd Part: Yes, a quantity may have unit but no dimension. Eg. Angle. The angle has no dimension but its S.I. unit is radian.

Angle = [latex]\frac{arc length}{radius} = \frac{m}{m}[/latex] = dimensionless

12. If y = a + bt + ct², where y is the distance and t is time. What is the dimension and unit of c? (2070 ‘D’)

Ans: Here, y = a + bt + ct²

From the principle of homogeneity, the dimension of physical quantity of left side must be equal to dimension of physical quantity on right side.

 dimension of y = dimension of ct²

Or, [L] = c[T²]

∴ C = [latex][LT^{-2}][/latex]

Unit: ms^-2

13. Check the correctness of the formula t = [latex]2\pi \sqrt{\frac{l}{g}}[/latex] using dimensional analysis, t is the time period of simple pendulum, [latex]l[/latex] is the length of simple pendulum and g is the acceleration due to gravity. (2069 ‘B’)

Ans: Given, formula is

              t = [latex]2\pi \sqrt{\frac{l}{g}}[/latex]

Dimensional formula of L.H.S. = [T]

Dimensional formula of R.H.S. = [latex]\frac{L}{LT^{-2}} = \sqrt{T^{-2}} = [T][/latex]

[∵ constant [latex]2\pi[/latex] can be neglected in dimensional analysis]

Since, L.H.S. = R.H.S., the given formula is dimensionally correct.

14. Write dimensional formula of gravitational constant and latent heat. (2067 ‘S’, 2056)

Ans: (i) We have, F = G [latex]\frac{m_1m_2}{d^2}[/latex]

Or, G = [latex]\frac{Fd^2}{m_1m_2}[/latex] ………………. (1)

Where, F = force, d = distance, m₁ and m₂ = masses of two bodies

 ∴ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]} = [M^{-1}L^3T^{-2}][/latex]

(ii) Dimensional formula of latent heat,

Latent heat of a substance is given by

L = [latex]\frac{Q}{m}[/latex]

Now, dimension of Q = [ML²T^-2]

Dimension of m = [M]

 ∴ Dimension of L = [latex]\frac{ML^2T^{-2}}{M} = [M^oL^2T^{-2}][/latex]

15. Check the correctness of formula t = [latex]2\pi \sqrt{\frac{m}{k}}[/latex] where t be the time period, m is the mass and k is the force per unit displacement. (2064)

Ans: Given, t = [latex]2\pi \sqrt{\frac{m}{k}}[/latex] ………………….. (i)

Where, t = time period, m = mass, k = force/displacement

In dimensional form, the equation (i) becomes

Or, T = [latex]\sqrt{\frac{M}{MLT^{-2}}\times L}[/latex]           [∵ Constant, [latex]2\pi[/latex] can be neglected in dimensional analysis]

Or, T = [latex]\sqrt{T^2}[/latex]

Or, T = T which is true.

Hence, the equation (i) is dimensionally correct. Notice that the equation (i) remains dimensionally correct if 2[latex]\pi[/latex] is replaced by other non-dimensional constants like 3[latex]\pi , 4\pi , 100, etc.[/latex]

16. Derive the dimensional equation of power. (2063 ‘S’)

Ans: Power = [latex]\frac{Work done}{Time} = \frac{Force\times Displacement}{Time}[/latex]

              = [latex]\frac{(Mass\times Acceleration)\times Displacement}{Time}[/latex]

∴ [latex][P] = \frac{[M]\times [LT^{-2}]\times [L]}{[T]} = [ML^2T^{-3}][/latex]

Hence, the dimensional equation of power is [P] = [ML²T^-3]

Note: Power has 1 dimension in mass, 2 dimensions in length and -3 dimensions in time. [ML²T^-3] is the dimensional formula of power and [P] = [ML²T^-3] is the dimensional equation of power. Carefully notice the difference among dimension, dimensional formula and dimensional equation.

17. A student writes [latex]\sqrt{\frac{R}{2GM}}[/latex] for escape velocity. Check the correctness of the formula by using dimensional analysis. (2062)

Ans: Given, v = [latex]\sqrt{\frac{R}{2GM}}[/latex] ………………… (i)

Where, v is the escape velocity, R is the radius, G is universal gravitational constant and M is the mass.

In dimensional form, the equation (i) becomes,

[v] = [latex][\sqrt{\frac{R}{GM}}][/latex] ………………….. (ii) [∵ numerical constant 2 can be neglected]

Here,

Dimensional formula of [v] = LT^-1

Dimensional formula of [R] = L

Dimensional formula of [G] = M^-1L³T^-2              [∵ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]} = [M^{-1}L^3T^{-2}[/latex]]]

Dimensional formula of [M] = M

Now, the equation (ii) becomes,                        

LT^-1 = [latex]\sqrt{\frac{L}{M^{-1}L^3T^{-2}\times M}}[/latex]

Or, LT^-1 = [latex]\sqrt{L^{-2}T^2}[/latex]

Or, LT^-1 = L^-1T which is inconsistent.

Hence, the equation (i) is incorrect dimensionally.

18. Convert 10 ergs in joules. (2061)

Ans: We can convert 10 ergs in joules by the method of dimensions. Since ergs and joules are both units of energy and dimension formula for energy is MLT^-2. Let, 10 ergs = n₂ joules

∴ [latex]n_1[M_1L_1^2T_1^{-2}] = n_2[M_2L_2^2T_2^{-2}][/latex] ……………………… (i)

Here,

n₁ = 10                              n₂ =?

M₁ = 1 g                           M₂ = 1 kg

L₁ = 1 cm                          L₂ = 1 m

T₁ = 1 s                             T₂ = 1 s

From (i), we have,

10 x 1 g x (1 cm)² x (1s)^-2 = n₂ x 1 kg x (1 m)² x (1s)^-2

Or, 10 x 1 g x (1 cm²) = n₂ x (100g) x (100 cm)²

Or, 10 = n₂ x 1000 x 10000

Or, n₂ = 10^-6 joules

Hence, 10 ergs equal 10^-6 joules.

19. Check the correctness of the formula v² = u² + 2as using dimensional analysis.

Ans: Given formula is

v² = u² + 2as …………… (i)

Dimensional formula of L.H.S. = [LT^-1]² = [L²T^-2]

Dimensional formula of R.H.S. = [LT^-1]² + [LT^-2L]= [L²T^-2]

(Notice that numerical constant can be neglected in dimensional analysis)

Since, L.H.S. = R.H.S., therefore, the given formula (i) is dimensionally correct.

20. Taking force, length and time to be fundamental quantities, find the dimensional formula for the density.

Ans: Force, length and time are taken as fundamental quantities. Let, F, L and T be the dimensional formula of force, length and time respectively.

Since, Density = [latex]\frac{Mass}{Volume}[/latex]

              = [latex]\frac{Force}{Acceleration\times Volume}[/latex]     (∵ [latex]Mass = \frac{Force}{Acceleration}[/latex])        

              = [latex]\frac{Force}{\frac{Velocity}{time}\times Volume}[/latex]                     (∵ [latex]Acceleration = \frac{Velocity}{Time}[/latex])

              = [latex]\frac{Force}{\frac{Displacement}{(time)^2}\times (length)^3}[/latex]               (∵ [latex]Velocity = \frac{Displacement}{Time}[/latex])

∴ [Density] = [latex]\frac{F}{LT^{-2}\times L^3} = [FL^{-4}T^2][/latex]

Hence, dimensional formula of density is [FL^-4T²].

21. The escape velocity of a body is v = [latex]\sqrt{\frac{2GM}{R}}[/latex]. Check the correctness of the formula using dimension.

Ans: v = [latex]\sqrt{\frac{2GM}{R}}[/latex] ……………… (i) where, v is the escape velocity, G is the universal gravitational constant, M is for mass of the earth and R is the radius of the earth.

L.H.S. = [V] = [LT^-1]

R.H.S. = [latex]\sqrt{\frac{2[G][M]}{[R]}}[/latex]

Since, 2 is numerical constant, which can be neglected in dimensional form.

              = [latex]\sqrt{\frac{M^{-1}L^3T^{-2}M}{L}} = \sqrt{L^2T^{-2}}=[LT^{-1}][/latex]        [∵ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]}] = [M^{-1}L^3T^{-2}[/latex]]

L.H.S. = R.H.S., so the equation (i) is dimensionally correct.

22. The density of gold is 19.3 g/cc. Express its value in SI unit.

Ans: 19.3 g/cc = [latex]\frac{19.3\times 10^{-3}kg}{10^{-6}m^3}[/latex]                     {∵ 1000 g = 1 kg & 1 cc = [latex]1 cm^3 = 10^{-6} m^3[/latex]}

              = 19.3 x 10³ kgm^-3 = 19300 kgm^-3

23. Obtain dimensions of specific heat capacity and gravitational constant.

Ans: Dimensional formula of specific heat capacity:

Specific heat capacity of a substance is defined by the relation:

Q = [latex]ms\Delta \theta[/latex]

Or, s = [latex]\frac{Q}{\Delta \theta m}[/latex]

Where, m = mass, [latex]\Delta \theta = [/latex] change in temperature,

Q = Heat energy supplied

 ∴ Dimension of Q = [latex][ML^2T^{-2}][/latex]

Dimension of m = [M]

Dimension of [latex]\Delta \theta = [/latex] [K]

Dimension of s = [latex]\frac{[ML^2T^{-2}]}{[MK]} = [M^oL^2T^{-2}K^{-1}][/latex]

Dimensional formula of gravitational constant:

We have, F = G [latex]\frac{m_1m_2}{d^2}[/latex]

Or, G = [latex]\frac{Fd^2}{m_1m_2}[/latex] ………………. (1)

Where, F = force, d = distance, m₁ and m₂ = masses of two bodies  

∴ Dimension of G = [latex]\frac{[MLT^{-2}\times L^2]}{[M\times M]} = [M^{-1}L^3T^{-2}][/latex]