CBSE- Class XI - Physics - CH2 - Dimension Analysis #class11Physics #eduvictors

NCERT Chapter Solutions and Other Numerical Problems

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Joseph Fourier -
How is he related to Dimensions?

Q1: Define the term 'Dimension' 

Answer: The term 'dimension' is used to refer to the physical nature of a quantity and the type of unit used to specify it. Mathematically dimensions of a physical quantity are the powers to which the fundamental quantities must be raised.

e.g. Dimension of velocity = Displacement / time = [L]/[T] = [M⁰][L¹][T^-1]





Q2: What are dimensional constants?

Answer:  Constants which possess dimensions are called dimensional constants. E.g. Planck's Constant.

Q3: What are dimensional variables?

Answer: Those physical quantities which possess dimensions but do not have a fixed value are called dimensional variables. E.g. Displacement, Force, velocity etc.

Q4: What are dimensionless quantities?

Answer: Physical quantities which do not possess dimensions are called dimensionless quantities. E.g. Angle, specific gravity, strain. In general, a physical quantity which is a ratio of two quantities of the same dimension will be dimensionless.

Q5: Define the principle of homogeneity of dimensions. On What principle is it based?

Answer: The principle of homogeneity of dimensions states that an equation is dimensionally correct if the dimensions of the various terms on either side of the equation are the same.

This principle is based on the fact that two quantities of the same dimension only can be added up, and the resulting quantity also possesses the same dimension.

i.e. In equation X + Y = Z is valid if the dimensions of X, Y and Z are the same.

Q6: Who introduced Dimension Analysis

Answer: Fourier (Joseph Fourier - French Mathematician)

Q7: List the basic dimensions.
Answer:

• Length - L
• Time - T
• Mass - M
• Temperature - K or  θ
• Current - A

Q8:  What are the uses (applications) of dimensional analysis?

Answer: The applications of dimensional analysis are:

1. To convert a physical quantity from one system of units to another.
2. To check the dimensional correctness of a given equation.
3. establish a relationship between different physical quantities in an equation.

Q9 (NCERT): A book with many printing errors contains four different formulas for the displacement
y of a particle undergoing a certain periodic motion:

(a) y = a sin 2π t/T
(b) y = a sin vt
(c) y = (a/T) sin t/a
(d) y = (a 2) (sin 2πt / T + cos 2πt / T )

(a = maximum displacement of the particle, v = speed of the particle. T = time-period
of motion). Rule out the wrong formulas on dimensional grounds.

Answer:
Given,
      Dimension  of a = displacement = [M⁰L¹T⁰]
      Dimension of v (speed) =  distance/time = [M⁰L¹T^-1]
      Dimension of  t or T (time period) = [M⁰L⁰T¹]

Trigonometric function sine is a ratio, hence it must be dimensionless.

(a)  y = a sin 2π t/T (correct ✓ )
       Dimensions of RHS =  [L¹] sin([T].[T^-1] ) =  [M⁰L¹T⁰] = LHS  (eqation is correct).

(b) y = a sin vt  (wrong ✗)
      RHS =  [L¹] sin([LT^-1] [T¹]) =  [L¹] sin([L]) = wrong, since trigonometric function must be dimension less.

(c) y = (a/T) sin t/a (wrong ✗)
     RHS = [L¹] sin([T].[L^-1] ) = [L¹] sin([TL^-1] ) = wrong, sine function must be dimensionless.



(d)  y = (a 2) (sin 2πt / T + cos 2πt / T )  (correct ✓ )
      RHS =  [L¹] ( sin([T].[T^-1] + cos([T].[T^-1] ) = [L¹] ( sin(M⁰L¹T⁰) + cos(M⁰L¹T⁰) )
               = [L¹]  = RHS  = equation is dimensionally correct.

Q10(NCERT): A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes

Answer:
Dimension of m (mass) = [M¹L⁰T⁰]
Dimension of m₀ (mass) = [M¹L⁰T⁰]

Dimension of v (velocity) = [M⁰L¹T^-1]

∴ Dimension of v² = [M⁰L²T^-2]

Dimension of c (velocity) = [M⁰L¹T^-1]

Applying principle of homogeneity of dimensions,  [LHS] = [RHS] = [M¹L⁰T⁰]
⇒ The equation (1- v²)^½  must be dimension less, which is possible if we have the expressions as:
(1 -  v²/c²) The equation after placing 'c' will be:

Q11: Check the following equation for calculating whether the displacement is dimensionally correct or not
(a) x = x0 + ut + (1/2) at²
where x is the displacement at given time t
            xo is the displacement at t = 0
            u is the velocity at t = 0
            a represents the acceleration.

(b) P = (ρgh)^½
      where P is the pressure,
                 ρ is the density
                 g is the gravitational acceleration
                 h is the height. 

Answer:
(a)  x = x0 + ut + (1/2) at²
Applying the principle of homogeneity, all the sub-expressions  of the equation must have the same dimension and be equal to [LHS]
Dimension of  x = [M⁰L¹T⁰]

Dimensions of  sub-expressions of [RHS] must be [M⁰L¹T⁰]
⇒ Dimension of  x0 (displacement) = [M⁰L¹T⁰] = [LHS]

Dimension of ut  = velocity x time = [M⁰L¹T^-1][M⁰L⁰T¹] = [M⁰L¹T⁰] = [LHS]

Dimension of  at² = acceleration x (time)²  = [M⁰L¹T^-2][M⁰L⁰T^-2] = [M⁰L¹T⁰] = [LHS]

∴ The equation is dimensionally correct.

(b) P = (ρgh)^½

Dimensions of LHS i.e. Pressure [P] = [M¹L^-1T^-2]

Dimensions of ρ = mass/volume =  [M¹L^-3T⁰]
Dimensions of g (acceleration) = [M⁰L¹T^-2]
Dimensions of h (height) = [M⁰L¹T⁰]


Dimensions of RHS = [(ρgh)^½] = ([M¹L^-3T⁰]. [M⁰L¹T^-2].[M⁰L¹T⁰])^½ = ([M¹L^-1T^-2])^½
 = [M^½L^-½T^-1]  ≠ [LHS]
The given equation is not dimensionally correct.

Q12: Prove that trigonometric functions like sin θ is dimensionless.

Answer:  As shown in the figure, sin θ = y/r. = [L]/[L] = 1
⇒ sin θ is dimensionless





Q13 (NCERT): A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v : tan θ = v and checks that the relation has a correct limit: as v → 0, θ →0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.

Answer: Given, v = tanθ
Dimensions of  LHS = [v] = [M⁰L¹T^-1]
Dimension of  RHS = [tanθ] = [M⁰L⁰T⁰]      (trigonometric ratios are dimensionless)

Since [LHS] ≠ [RHS]. The equation is dimensionally incorrect.

To make the equation dimensionally correct, LHS should also be dimensionless. It may be possible if consider the speed of rainfall (V_r) and the equation will become:
    tan θ = v/V_r

Q14: Hooke’s law states that the force, F, in a spring extended by a length x is given by F = −kx.
According to Newton’s second law F = ma, where m is the mass and a is the acceleration.
Calculate the dimension of the spring constant k.

Answer: Given, F = -kx
⇒ k = - F/x

F = ma, the dimensions of force is:
[F] = ma = [M¹L⁰T⁰].[M⁰L¹T^-2] = [M¹L¹T^-2]
Therefore, the dimension of spring constant (k) is:
[k] = [F]/[x] = [M¹L¹T^-2].[M⁰L^-1T⁰] = [M¹L⁰T^-2] or [MT^-2] .....     (answer)


Q15: Compute the dimensional formula of electrical resistance (R).

Answer: According to Ohm's law
V = IR or R = V/I
Since Work done (W) = QV  where Q is the charge
⇒ R = W/QI = W/I²t                           (I = Q/t)

Dimensions of Work [W] = [M¹L²T^-2]
∴Dimension of R =  [R] = [M¹L²T^-2][A^-2T^-1] = [M¹L²T^-3A^-2]   ... (answer)

Note: Formula used to convert a physical quantity from one system of units to another.
If n₁ and n₂ are numerical units of physical quantities and U₁ and U₂ represent different system units, then
 

Q16: Convert 76 cm of mercury pressure into N m-2 using the dimensional formula. (Density of Hg = 13.6 gm/mL)

Answer:
Pressure P = hdg            (h = height, d = density, g = acceleration due to gravity).
In cgs system, 76 cm of Hg will exert pressure (P1), i.e.
P1 = 76 x 13.6 x 980 = 1012928 dynes cm^-2.

Dimension of Pressure in [M¹L^-1T^-2]
Applying to the formula (shown above in note), a = 1, b = -1 and c = -2
⇒

        









Q17: A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m² s^–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α^–1 β^–2 γ² in terms of the new units.

Answer:  Considering the unit conversion formula,
                   n₁U₁ = n₁U₂ 
               n₁[M₁^aL₁^bT₁^c] = n₂[M₂^aL₂^bT₂^c]
Given here,  1 Cal = 4.2 J = 4.2  kg m² s^–2.

n₁ = 4.2,  M₁ = 1kg,  L₁ = 1m,  T₁ = 1 sec
and 
n₂ = ?,  M₂ = α kg,  L₂ = βm,  T₂ = γ sec

The dimensional formula of energy is =  [M¹L²T^-2]
 ⇒ a = 1, b =1 and c = -2  Putting these values in above equation,

n₂= n₁[M₁/M₂]^a[L₁/L₂]^b[T₁/T₂]^c

    = n₁[M₁/M₂]¹[L₁/L₂]²[T₁/T₂]^-2

    = 4.2[1Kg/α kg]¹[1m/βm]²[1sec/γ sec]^-2 = 4.2 α^–1 β^–2 γ²              ... (answer)


Q18: The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ω. Considering the relation to be K = kI^aω^b where k is a dimensionless constant. 
Find a and b. The moment of Inertia of a sphere about its diameter is (2/5)Mr²

Answer:
Dimensions of Kinetic energy K = [M¹L²T^-2]
Dimensions of Moment of Inertia (I) = [ (2/5)Mr²] = [ML²T⁰]
Dimensions of angular speed ω = [θ/t] = [M⁰L⁰T^-1]
Applying the principle of homogeneity in dimensions in the equation K = kI^aω^b

[M¹L²T^-2] = k ( [ML²T⁰])^a([M⁰L⁰T^-1])^b
[M¹L²T^-2] = k [M^aL^2aT^-b]
⇒ a = 1 and b = 2
⇒ K = kIω²                                                          ... (answer)


Q19: What are the limitations of Dimensional Analysis?

Answer: Limitations of Dimensional Analysis are:

1. It cannot determine the value of dimensionless constants.
2. We cannot use this method to equations involving exponential and trigonometric functions. 
3. It cannot be applied to an equation involving more than three physical quantities.
4. It is a too not a solution i.e.  It can check only if the equation is dimensionally correct or not. But cannot say the equation is absolutely correct.

Q20: Convert 1 Newton into dyne using the method of dimensions.

Answer: Dimensions of Force = [M¹L¹T^-2]
Considering dimensional unit conversion formula i.e.  n₁[M₁^aL₁^bT₁^c] = n₂[M₂^aL₂^bT₂^c]
 ⇒ a = 1, b = 1 and c = -2
In SI system,  M₁ = 1kg,        L₁ = 1m   and T₁ = 1s
In cgs system, M₂ = 1g,    L₂ = 1cm  and T₂ = 1s

Putting the values in the conversion formula,
n₂ = n₁(1Kg/1g)¹.(1m/1cm)¹(1s/1s)^-2 = 1.(10³/1g)(10²cm) = 10⁵dyne   ...(answer)

Q21: The centripetal force  (F) acting on a particle (moving uniformly in a circle) depends on the mass (m) of the particle, its velocity (v) and the radius (r) of the circle. Derive dimensionally formula for force (F).

Answer:  Given,   F ∝ m^a.v^b.r^c
 ∴ F = km^a.v^b.r^c      (where k is constant)

Putting dimensions of each quantity in the equation,

     [M¹L¹T^-2] = [M¹L⁰T⁰]^a. [M⁰L¹T^-1]^b. [M⁰L¹T⁰]^c  = [M^aL^b+cT^-b]

⇒  a =1, b +c = 1, -b = -2
⇒  a= 1, b = 2, c = -1
∴ F = km¹.v².r^-1     =  kmv²/r

Q22: If the velocity of light c, gravitational constant G and planks constant h be chosen as fundamental units, find the value of a gram, a cm and a sec in terms of a new unit of mass, length and time respectively.
(Take c = 3 x 10¹⁰ cm/sec, G = 6.67 x 10⁸ dyn cm²/gram² and  h = 6.6 x 10^-27 erg sec)

Answer: Given,
           c = 3 x 10¹⁰ cm/sec
           G = 6.67 x 10⁸ dyn cm²/gm²  
           h = 6.6 x 10^-27 erg sec
Putting respective dimensions,
           Dimension formula for c = [M⁰L¹T^-1] = 3 x 10¹⁰ cm/sec                             .... (I)
           Dimensions of G             =  [M^-1L³T^-2] = 6.67 x 10⁸ dyn cm²/gm²                 ...(II)
           Dimensions of h              = [M¹L²T^-1] =  6.6 x 10^-27 erg sec                       ...(III)

(Note: Applying newton's law of gravitation, you can find dimensions of G i.e. G = Fr²/(mM)
 Similarly, Planck's Constant (h) = Energy / frequency)
To get M, multiply eqn-I and III and divide by eqn.-II,
⇒ [M⁰L¹T^-1].[M¹L²T^-1].[M¹L^-3T²]
            = ( 3 x 10¹⁰ cm/sec).( 6.6 x 10^-27 erg sec)/ 6.67 x 10⁸ dyn cm²/gm2
 ⇒[M²] = 2.968 x 10^-9
 ⇒[M] =  0.5448 x 10^-4 gm
or 1gm = [M]/0.5448 x 10^-4 = 1.835 x 10^-4 unit of mass

To obtain length [L],  eqn.-II x eqn-III / cube of eqn.-I i.e.
[M^-1L³T^-2].[M¹L²T^-1].[M⁰L^-3T³]
             = (6.67 x 10⁸ dyn cm²/gm² ).( 6.6 x 10^-27 erg sec)/(3 x 10¹⁰ cm/sec)³

⇒ [L²] =  1.6304 x 10^-65 cm²
⇒ [L]  =  0.4038 x 10^-32 cm
or  1cm = [L]/ 0.4038 x 10^-32 = 2.47 x 10^-32unit of length

In eqn-I,  [M⁰L¹T^-1] = 3 x 10¹⁰ cm/sec
⇒  [T] = [L] ÷ 3 x 10¹⁰cm/s
⇒  [T] =  0.4038 x 10^-32 cm  ÷ 3 x 10¹⁰cm/s = 0.1345 x 10^-42 s
or 1s = [T]/0.1345 x 10^-42 s = 7.42 x 10⁴² unit of time

Q 23: A student while doing an experiment finds that the velocity of an object varies with time and it can be expressed as an equation:
                                    v = Xt² + Yt +Z . 
If units of v and t are expressed in terms of SI units, determine the units of constants X, Y and Z in the given equation.

Answer: Given,  v = Xt² + Yt +Z
Dimensions of velocity v =  [M⁰L¹T^-1]
Applying the principle of homogeneity in dimensions, terms must have the same dimension.
                        [v] = [Xt²] + [Yt] + [Z]
 ∴ [v] = [Xt²]
⇒ [X] = [v] /[t²]  =  [M⁰L¹T^-1] / [M⁰L⁰T²] = [M⁰L¹T^-3]              ....(i)


Similarly, [v] = [Yt]
⇒  [Y] = [v] / [t] = [M⁰L¹T^-1]/ [M⁰L⁰T^-1] = [M⁰L¹T^-2]             ...(ii)

Similarly, [v]= [Z]
[Z] =  [M⁰L¹T^-1]      ...(iii)

⇒ Unit of X = m-s^-3
⇒ Unit of Y = m-s^-2
⇒ Unit of Z = m-s^-1

Q24: Express Capacitance in terms of dimensions of fundamental quantities i.e. Mass (M), Length(L), Time(T) and Ampere(A)

Answer: Capacitance(C) is defined as the ability of an electric body to store electric charge.
∴ Capacitance (C) = Total Charge(q) / potential difference between two plates (V)
       = Coulomb/ Volt
  ∵  Volt = Work done (W)/ Charge(q) = Joule/Coulomb
⇒  Capacitance (C) = Charge(q)²/ Work(W)
  ∵ Charge (q) = Current (I) × Time(t)
Dimension of [q] = [AT]                                                 ----------- (I)
Dimension of Work = Force × distance = [MLT^-2][L] = [ML²T^-2]            --------- (II)

Putting values of I and II,
[C] = ([AT])²/ [ML²T^-2] = [M^-1L^-2T²⁺²A²] = [M^-1L^-2T⁴A²]

Physical Quantities have the same dimensional formula:
a. impulse and momentum.
b. force, thrust.
c. work, energy, torque, the moment of force, energy
d. angular momentum, Planck’s constant, rotational impulse
e. force constant, surface tension, surface energy.
f. stress, pressure, modulus of elasticity.
g. angular velocity, frequency, velocity gradient
h. latent heat, gravitational potential.
i. thermal capacity, entropy, universal gas constant and Boltzmann’s constant.
j. power, luminous flux.

Q25: If Force (F), velocity (V) and acceleration (A) are taken as the fundamental units instead of mass, length and time, express pressure and impulse in terms of F, V and A.

Answer: We know that Force = mass ✕ acceleration
⇒ mass = FA^-1
and  length = velocity ✕ time = velocity ✕  velocity ÷  acceleration = V²A^-1
and time = VA^-1

∵ Pressure = Force  ÷ Area = F ÷ (V²A^-1)² = FV^-4A²
 Impulse = Force ✕ time = FVA^-1