Answer: µ = ρ¹ * A¹
Where x=1 and y=1
Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below
µ = ρ * A
The dimension for each of these quantities is given below
Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1
ρ is density with unit kg/m³ and the dimension is ML^3
A is area with unit m², thus the dimension is M^2
Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.
Their dimensions below
Mass = M
Length = L
Time = T
Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below
µ = ρ^x * A^y.
By getting the power of x and y we will be able to get the formula that relates the quantities.
This is done by slotting in the dimensions of the respective quantities.
ML^-1 = (ML^-3)^x * (L²) ^y
By using law of indices on the right hand side of the equation, we have that
ML^-1 = (M^x * L^-3x) * (L^2y)
Also applying law of indices on the right hand side, we have that
ML^-1 = (M^x) * (L^-3x +2y)
The next step is to relate equal variables on both sides
For the M variable
M¹ = M^x which results to
x = 1
For the L variable
L^-1 = L^-3x+2y which results to
-1 = - 3x +2y
But x = 1
We have that
-1 = - 3(1) +2y
-1 = - 3 + 2y
-1 +3= 2y
2 = 2y
y = 1
Thus x=1 and y=1 and the formulae that relates the quantities is
µ = ρ¹ * A¹
