Question

Three quantities are given: A = 5.0 m3/kg·s 2 , B = 2.0 m/s, and C = 8.0 kg·m2/s. Using dimensional analysis determine the simplest algebraic combination of these three quantities that has the dimensions of length and compute its value in meters (Answer: 2.2 m

Answer 1

here Length(L) A^P   ( L is directly proportional to A^P)

L B^q

L   C^r

By combining all, L A^P B^q C^r

L = K A^P B^q C^r ------------------1 ( K = constant of proportionality)

Now write dimension formula on both sides of eq.1 ( for A = [ L³M^-1T^-2 ] , B = [ L T^-1 ] , C = [ M L² T^-1 ] . here K being constant has no dimension)

[ M⁰ L¹T⁰ ] = [ L³ M^-1 T^-2 ] ^P  [ L T^-1 ] ^q   [ M L²T^-1 ]^r

on rearranging this:

[ M⁰ L¹ T⁰ ] = [ M^-P+r   L^3P+q+2r   T^-2P-q-r  ]

Now comparing the powers of M, L and T on both sides:

- P +r = 0 -----------------2 , P = r

3P +q +2r = 1 --------------------------3

-2P - q -r = 0 ----------------------4

on solving eq. 2,3,& 4 we will get:

P = 1/2, q = -3/2, r = 1/2

Now put this values in eq.1:

L = K A^1/2 B^-3/2 C ^1/2

L = ( AC/ B³ ) ^1/2   ( Let K =1 being constant)

This is the algebraic equation that has the dimension of length.

Now put the value of A,B,C to find Length

L = ( 5*8 / 2³ ) ^1/2

L = 2.23 m