Question

If velocity, acceleration and force are chosen as fundamental quantities, then find the dimensions of linear momentum, angular momentum and Young's Modulus of Elasticity.

Answer 1

Dear Student ,

Young’s Modulus is given by,

Y = Stress/Strain

Stress = Force/Area

Strain = change in length/length = dimensionless

Therefore, dimension of Y is [ML^-1T^-2]

Let,

[F^xA^yV^z] = [ML^-1T^-2]

=> [MLT^-2]^x [LT^-2]^y [LT^-1]^z = [ML^-1T^-2]

=> [M^x L^x+y+z T^-2x-2y-z] = [ML^-1T^-2]

Thus,

x = 1

x+y+z = -1

-2x-2y-z = -2

Solving we get,

x = 1, y = 2, z = -4

So, the required dimension is [FA²V^-4]

We know,
(L is angular momentum)
$$\begin{gathered} u\to \left[M^0{L}^1{T}^{-1}\right] \\ A\to \left[M^0{L}^1{T}^{-2}\right] \\ F\to \left[M^1{L}^1{T}^{-2}\right] \\ L\to \left[M^1{L}^3{T}^{-1}\right] \\ Let, \\ L=u^x{A}^y{F}^z \end{gathered}$$
$\left[M^1{L}^2{T}^{-1}\right]\to [M$${}^0{L}^1{T}^{-1}$${]}^x{\left[M0L1T-2\right]}^y{\left[M1LT-2\right]}^z$
$\left[M^1{L}^2{T}^{-1}\right]\to \left[M^z{L}^{x+y+z}{T}^{-x-2y-2z}\right]$
Here we get three equations on comparison ,
z=1
x+y+z=2
-x-2y-2z=-1

Solving we get,

x = 3, y = -2, z = 1

Thus the required dimensional formula is, u³A^-2F¹.

​​​​​Regards