Question #7f6ae

1 Answer
Feb 24, 2018

Dimension of time is #[M^0 L^0 T]#

Let, #[M^0 L^0 T] = c^a G^ b h^ c'#

so, #[M^0 L^0 T] = [L T^(-1)]^a [M^(-1) L^3 T^(-2)]^b [ML^2T^(-1)]^(c') = [L^a T^-a] [M^(-b) L^(3b) T^(-2b)] [M^c' L^(2c') T^(-c')]#

So,comparing both sides,

#0=-b+c'# , #0=a+3b+2c'# and, #1=-a-2b-c'#

Solving these set of equations we get,

#a=-5/2 , b=1/2 & c'=1/2#

So,we get, #[T] = c^(-5/2) b^(1/2) c^(1/2)=(sqrt(bc))/sqrt(c^5)#