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

If velocity (v), acceleration (a) and force (F) are taken as fundament

1.	If velocity (v), acceleration (a) and force (F) are taken as fundamental quantities, the dimensions of Young's modulus (Y) would be
Answer» `[Fa^(2)v^(-2)]` `[Fa^(2)v^(-3)]` `[Fa^(2)v^(-4)]` `[Fa^(2)v^(-5)]` Solution :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)]`