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
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.
