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 = M₁aL₁bT₁c
u₂ = M₂aL₂b T₂c
∴
n₁[M₁aL₁bT₁c] =
n₂[M₂aL₂b T₂c]
⟹ 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.