LECTURE: 2 : CLASS XI: PHYSICS : UNITS & MEASUREMENTS

CLASS XI   |    PHYSICS    |    CHAPTER 2

      notes prepared by Subhankar Karmakar

Conversion table from degree to radian:

a.   1° = 1.745 x 10⁻² rad

b.   1' = 2.91 x 10⁻⁴ rad

c.   1" = 4.85 x 10⁻⁶ rad

 

Q1. The moon is observed from two diametrically opposite points A and B on the earth. The angle θ subtended at the moon by the two directions of observation is 1°54'. Given the diameter of the earth to be 1.276 x 10⁷ m, compute the distance of the Moon from the Earth. 

 

Soln. Here the parallactic angle 

θ = 1°54' = 1.745 x 10⁻² + 54 x 2.91 x 10⁻⁴ rad

                = 3.32 x 10⁻² rad.

Here, b = AB = 1.276 x 10⁷ m

The distance of the Moon from the Earth,

S = b/θ = 1.276 x 10⁷/3.32 x 10⁻²

             = 3.84 x 10⁸ m

 

Q2. The angular diameter of the sun is 1920". If the distance of the sun from the earth is 1.5 x 10¹¹ m, what is the linear diameter of the sun?

 

Soln. Distance of the sun from the earth

          S = 1.5 x 10¹¹ m 

          Angular diameter of the sun

          θ = 1920" = 1920 x 4.85 x 10⁻⁶ rad

                           = 9312 x 10⁻⁶ rad

Linear diameter of the sun

          D = Sθ = 1.5 x 10¹¹ x 9312 x 10⁻⁶ m

                           = 13968 x 10⁵ m

                           = 1.4 x 10⁶ km

 

 

• DIMENSION OF A PHYSICAL QUANTITY:

 

All the derived physical quantities can be expressed in terms of some combination of the seven fundamental or base quantities. We call these fundamental quantities as the seven dimensions of the world, which are denoted with square brackets [ ]. 

 

• Dimension of length = [L]

• Dimension of mass = [M]

• Dimension of time = [T]

• Dimension of electric current = [A]

• Dimension of thermodynamic temperature = [K]

• Dimension of luminous intensity = [cd]

• Dimension of amount of substance = [mol]

 

The dimensions of a physical quantity are the powers to which the fundamental quantities must be raised to represent that quantity completely. 

For example, 

Density = Mass/Volume = Mass/ (Length x breadth x height) 

Dimensions of density = [M]/([L] x [L] x [L])

= [M¹L⁻³T⁰]

 

·        Area = [M⁰L²T⁰] = m²

·        Volume = [M⁰L³T⁰] = m³

·        Density = [M¹L⁻³T⁰] = kg m⁻³

·        Speed or Velocity = [M⁰L¹T⁻¹] = m/s

·        Acceleration = [M⁰L¹T⁻²] = m/s²

 

DIFFERENT TYPES OF VARIABLES AND CONSTANTS: 

 

There are two types of variables

1. Dimensional variables: 

 

The physical quantities which possess dimensions and have variable values are cal dimensional variables. For example, area, volume, velocity, force, power, energy etc.

 

2. Dimensionless variables: 

 

The physical quantities which have no dimensions but have variable values are called dimensionless variables. For example, angle, specific gravity, strain etc.

 

There are two types of constants:

 

1. Dimensional constants: 

 

The physical quantities which possess dimensions and have constant values are called dimensional constants. For examples, gravitational constant, Planck's constant, electrostatic constant etc.

 

2. Dimensionless constants: 

 

The constant quantities having no dimensions are called dimensionless constants. For example, π, e etc. 

 

Application of dimensional analysis: 

 

The method of studying a physical phenomenon on the basis of dimensions is called dimensional analysis. 

 

Following are the three main uses of dimensional analysis: 

 

1. To convert a physical quantity from one system of units to other. 

2. To check the correctness of a given physical relation.

3. To derive a relationship between different physical quantities.

 

1. Conversion of one system of units to other:

 

As the magnitude of physical quantities remain same and does not depend upon our choices of units, therefore, 

                   Q = n₁u₁ = n₂u₂

where Q is the magnitude of the physical quantity, u₁ and u₂ are the units of measurement of that quantity and n₁ and n₂ are the corresponding numerical values. 

u₁ = M₁^aL₁^bT₁^c

u₂ = M₂^aL₂^b T₂^c

∴ n₁[M₁^aL₁^bT₁^c] = n₂[M₂^aL₂^b T₂^c]

⟹  n₂ = n₁ [M₁/M₂]^a [L₁/L₂]^b [T₁/T₂]^c   

 

Q1. Convert 1 Newton into dyne.

 

Soln. Newton is the SI unit of force and dyne the CGS unit of force. Dimensional formula of force is M¹L¹T⁻²

∴ a = 1, b = 1, c = -2

In SI system; 

M₁ = 1 kg = 1000 g

L₁ = 1 m = 100 cm

T₁ = 1 s and n₁ = 1 (Newton)

In CGS system;

M₂ = 1 g ; L₂ = 1 cm ; T₂ = 1 s

∴ n₂ = n₁ [M₁/M₂]^a [L₁/L₂]^b [T₁/T₂]^c 

         = 1 x [1000/1]¹ x [100/1]¹ x [1/1]⁻²

        = 1 x 10³ x 10²

        = 10⁵ 

∴ 1 N = 10⁵ dyne

 

Q2. Convert 1 erg into Joule.

 

Soln. Erg is CGS unit of energy whereas joule is SI unit of energy. Dimensional formula of energy is M¹L²T⁻².

∴ a = 1, b = 2, c = -2

In CGS system;

M₁ = 1 g ; L₁ = 1 cm ; T₁ = 1 s ; n₁ = 1

In SI system; 

M₂ = 1 kg = 1000 g

L₂ = 1 m = 100 cm

T₂ = 1 s and n₂ = ?

∴ n₂ = n₁ [M₁/M₂]^a [L₁/L₂]^b [T₁/T₂]^c 

         = 1 x [1/1000]¹ x [1/100]² x [1/1]⁻²

        = 1 x 10⁻³ x 10⁻⁴

        = 10⁻⁷

∴ 1 erg =  10⁻⁷ N

 

Q3. The density of Mercury is 13.6 g/cm³ in CGS system. Find its value in SI system.

 

Soln. The dimensional formula of density is

M¹L⁻³T⁰

∴ a = 1, b = - 3, c = 0

In CGS system;

M₁ = 1 g ; L₁ = 1 cm ; T₁ = 1 s ; n₁ = 13.6

In SI system; 

M₂ = 1 kg = 1000 g

L₂ = 1 m = 100 cm

T₂ = 1 s and n₂ = ?

∴ n₂ = n₁ [M₁/M₂]^a [L₁/L₂]^b [T₁/T₂]^c 

        = 13.6 x [1/1000]¹ x [1/100]⁻³ x [1/1]⁰

        = 13.6 x 10⁻³ ⁺ ⁽⁻²⁾⁽⁻³⁾ 

        = 13.6 x 10³

∴ The density of Mercury in SI unit is 13.6 x 10³ kg/m³

 

Q4. If the value of atmospheric pressure is 10⁶ dyne / cm², find its value in SI units.

 

Q5. If the value of universal gravitational constant in SI unit is 6.6 x 10⁻¹¹ N m² kg⁻², then find its value in CGS unit.

 

 

 

2. CHECKING THE DIMENSIONAL CONSISTENCY OF EQUATIONS:

 

• Principle of homogeneity of dimensions:

 

According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both side of the equation are the same. 

 

Q6. Check the dimensional accuracy of the equation of motion s = ut + ½at².

 

Soln. Dimensions of different terms are

[s] = [L],

[ut] = [LT⁻¹] x [T] = [L],

[½at²] =  [LT⁻²] x [T²] = [L]

 

As all the terms on both sides of the equation have the same dimensions, show the given equation is dimensionally correct. 

 

Q7. Check the correctness of the equation

       FS = ½mv² - ½mu²

       Where F is a force acting on a body of mass m and S is the distance moved by the body when its velocity changes from u to v.

 

Soln. 

    [FS] = [M¹L¹T⁻²][L] = [M¹L²T⁻²]

    [½mv²] = [M][LT⁻¹]² = [M¹L²T⁻²]

    [½mu²] = [M][LT⁻¹]² = [M¹L²T⁻²]

Since the dimensions if all the terms in the given equation are same, hence the given equation is dimensionally correct. 

 

Q8. The Vander Waal's equation for a gas is

       ( P + a/V²)(V - b) = RT

Determine the dimensions of a and b. Hence write the SI units of a and b.

 

Soln. Since the dimensionally similar quantities can be added or subtracted, therefore, 

[P] = [a/V²] 

⟹ [a] = [ PV²] = [ M¹L⁻¹T⁻²] [L³]² = [M¹L⁵T⁻²]

Also, [b] = [V] = [L³]

∴ The SI unit of a is kg m⁵/s² and that of b is m³

 

3. DEDUCING RELATION AMONG THE PHYSICAL QUANTITIES:

 

By making use of the homogeneity off dimensions, we can derive an expression for a physical quantity if we know the various factors on which it depends

 

Q9. 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.

 

Soln. Centripetal force F depends upon mass M, velocity V and radius r.

Let F ∝ mᵃ vᵇ rᶜ 

∴ F = K mᵃ vᵇ rᶜ --------(1)

where K is a dimensionless constant. 

Dimensions of the various quantities are

 [m] = [M],  [v] = [LT⁻¹],  [r] = [L]

Writing the dimensions of various quantities in equation 1, we get

 [M¹L¹T⁻²] = 1 [M]ᵃ [LT⁻¹]ᵇ [L]ᶜ

⟹  [M¹L¹T⁻²] = [M]ᵃ [L]ᵇ ⁺ ᶜ [T]⁻ᵇ

Comparing the dimensions of similar quantities on both sides, we get

      a = 1

      b + c = 1 and 

      - 2 = - b ⟹ b = 2

∴ c = 1 - b = 1 - 2 = - 1

∴ a = 1, b = 2 and c = - 1

∴ F = K m v² r⁻¹ = K mv²/r

This is the required expression for the centripetal force.

 

Q10. The velocity  v of water waves depends on the wavelength λ, density of water ρ, and the acceleration due to gravity g. Did use by the method of dimensions the relationship between these quantities. 

 

Soln. Let  v = K λᵃ ρᵇ gᶜ -------(1)

where  K = a dimensionless is constant

Dimensions of the various quantities are

[v] = [LT⁻¹],  [λ] = [L],  [ρ] =  [M¹L⁻³], [g] = [LT⁻²]

Substituting these dimensions in equation (1), we get

[LT⁻¹] = [L]ᵃ  [M¹L⁻³]ᵇ [LT⁻²]ᶜ

∴ [M⁰ L¹T⁻¹] = [Mᵇ Lᵃ⁻³ᵇ⁺ᶜ T⁻²ᶜ]

Equating the powers of M, L and T on both sides, 

b= 0 ; a - 3b + c =1 ; - 2c = - 1

On solving,  a= ½ ; b = 0, c = ½

∴ v = K √(λg)

 

Q11. The frequency "ν" off vibration of a a stretched string depends up on:

a. Its length l

b. Its mass per unit length m and

c. The tension T in the string.

Obtain dimensionally an expression for frequency ν.

 

Soln. Let the frequency of vibration of the string be given by

       ν = K lᵃ Tᵇ mᶜ ----------(1)

where K is a dimensionless constant.

Dimension of the various quantities are

[ν] = [T⁻¹] ; [l] = [L]; [T] =  [M¹L¹T⁻²] ; [m] = [M¹L⁻¹]

Substituting this dimensions in equation 1,  we get

 [T⁻¹] = [L]ᵃ  [M¹L¹T⁻²]ᵇ  [M¹L⁻¹]ᶜ

⟹ M⁰ L⁰ T⁻¹ = Mᵇ ⁺ ᶜ Lᵃ ⁺ ᵇ ⁻ ᶜ T⁻²ᵇ

Equating the dimensions of M, L and T , we get

b + c = 0;  a + b - c = 0; - 2b = - 1

On solving, a = - 1, b = ½, c = - ½

∴ ν = K l⁻¹√(T/m) = (K/l)√(T/m)

 

Q12. The period of vibration of A tuning fork depends on the length l of its prong, density d and Young's modulus Y of its material. Deduce an expression for the period of vibration on the basis of dimensions.