DIMENSIONAL ANALYSIS PROBLEMS AND SOLUTIONS

 Problem #1

Find thedimensional formulae of follwoingquantities :
(a) The surface tension S,
(b) The thermal conductivityk and
(c) The coefficient of vescosity h.
Some equation involvingthese quntities are
(i) S = ρgrh/2
(ii) Q = kA(θ₂ – θ₁)t/d
(iii) F = –ηA(v₂ – v₁)/(x₂ – x₁)

where the symbols have their usual meanings. (r - density, g - acceleration due to gravity, r - radius, h - height,A- area, q₁ & q₂ - temperatures, t - time, d - thickness, v₁ & v₂ - velocities, x₁ & x₂ - positions.)

Answer:
(a) S = ρgrh/2
or
[S] = [ρ][g][r][h]
     = [ρ][g][r][h]

     = ML^-3.LT^-2.L²
[S] = MT^-2

(b)  Q = kA(θ₂ – θ₁)t/d or k = Qd/[A(θ₂ – θ₁)t]
Here, Q is the heat energyhaving dimension ML²T ^–2 , q₁ – q₁ is temperature, A is area, d is thickness and t is time. Thus,
[k] = MLT^-2/L²KT = MLT^–3K^–1.

(c)  F = –ηA(v₂ – v₁)/(x₂ – x₁)
Or MLT^-2 = [η]L².LT^-1/L
[η] = ML^-1T^-1

Problem #2
Consider the following three equations:
(i) x = vt² + 2at
(ii) max + ½ mv² = Fx
(iii) E = p²/2m + ½ kx²
where position x, velocity v, acceleration a, time t, mass m, momentum p, spring constant k and
mechanical energy E. Determine the correct equations in dimensions.

Answer:
Review equation (i):
x = vt² + 2at
The dimensions of the left section are L
The dimensions of the right segment are LT^-1 . T² + LT^-2T = LT + LT^-1
The dimensions of each tribe are different, then equation (i) is wrong.

Review equation (ii):
max + ½ mv² = Fx
The dimensions of the left section are MLT^-2.L + M(LT^-1)² = ML²T^-2 + ML²T^-2
The dimensions of the right segment are MLT^-2 .L = ML²T^-2
The dimensions of each term are the same, then equation (ii) is true.

Review equation (ii):
E = p²/2m + ½ kx²
The dimensions of the left section are ML²T^-2
The dimensions of the right segment are (MLT^-1)² .M^-1 + MT^-2.L² = ML²T^-2 + ML²T^-2
The dimensions of each term are the same, then equation (iii) is true.

Problem #3
The viscosity force F acting on the surface of the ball that moves through the fluid depends on the velocity (v), the radius of the ball (r) and the coefficient of fluid viscosity 'η'. Using the dimension method specify the formula 'F'.

Answer:
Let F α v^a, F α r^b and F α η^c ...... (1)
So, F = Kv^ar^bη^c
Where 'k' is a dimensional constant.
Dimension formula F = [MLT^-2]
Dimension formula v = [LT^-1]
Dimension formula r = [L]
Dimension formula from η = [ML^-1T^-1]
Substitute for the dimensional formula in equation (1),
[MLT^-2] = [LT^-1]^a [L]^b [ML^-1T^-1]^c
[MLT^-2] = [M^cL^{a + b + c} T^-a-c] ... ... (2)
In accordance with the principle of homogeneity, the dimensions of the two sides of the relation (2) must be the same.
So,
c = 1 ... ... (3)
a + b - c = 1 ... ... (4)
-a - c = -2 ... ... (5)
Placing c = 1 in (5), we get a = 1 Placing a = 1 and c = 1 in (4), we get b = 1 By substituting a, b and c at (1),]
we get relationships
F = kηrv

Problem #4
Let us find an expression for the time period T of a simple pendulum. The time period t may depend upon (i) mass m of the bob of the pendulum, (ii) length l of pendulum, (iii) acceleration due to gravity g at the place where the pendulum is suspended.

Answer:
Let (i) T µ m^a (ii)  T µ L^b (iii) T µ g^C
Combining all the three factors, we get
T µ m^aL^bg^c or  T = km^aL^bg^c
where k is a dimensionless constant of proportionality. Writing down the dimensions on either side of equation (i), we get
[T] = [M]^a [L]^b[LT^–2]^c = [M^aL^b+cT^–2c]
Comparing dimensions,
a = 0, b + c = 0 , – 2c = 1
\ a = 0, c = – 1/2, b = ½
From equation (i)
T = k m⁰L^1/2g ^-1/2
or
T = k√[L/g]   

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