If velocity (v), acceleration (a) and force (F) are taken as fundamental quantities, the dimensions of Young's modulus (Y) would be

Options

A. `[Fa^(2)v^(-2)]`
B. `[Fa^(2)v^(-3)]`
C. `[Fa^(2)v^(-4)]`
D. `[Fa^(2)v^(-5)]`

Answer

C

Explanation

Let `Y= kv^(x)a^(y)F^(z)`
where k is a dimensionless constant
`:. [ML^(-1)T^(-2)]= [LT^(-1)]^(x)[LT^(-2)]^(y)[MLT^(-2)]^(z)= [M^(z)L^(x+y+z)T^(-x-2y-2z)]`
Equating the powers of M, L and T we get
z= 1, x+y+z=-1, -x-2y-2z= -2
Solving we get x= -4, y=2, z=1
`:. Y= v^(-4)a^(2)F^(1)` or `[Y]= [Fa^(2)v^(-4)]`