If `I` is moment of inertia, `F` is force, `v` is velocity, `E` is energy and `L` is length then, dimension of `(IFv^2/(EL^4))` will be:

Options

A. `["MLT"^(-2)]`
B. `["MT"^(-2)]`
C. `["ML"^(2)"T"^(-3)]`
D. `["LT"^(-1)]`

Answer

B

Explanation

We knows,
Dimensions of moment of inertia = `["ML"^(2)]`
Dimensions of moment of force `F=["MLT"^(-2)]`
Dimensions of moment of velocity `(v)=["LT"^(-2)]`
Dimensions of moment of work `(W)=["ML"^(2)"T"^(-2)]`
Dimensions of moment of length `(l)=["L"]`
Here, Dimensions of `[X]=("Dimensions of "[IFv^(2)])/("Dimensions of "[WL^(3)])`
`" "=(["ML"^(2)]["MLT"^(-2)]["LT"^(-2)]^(2))/(["ML"^(2)"T"^(-2)]["L"^(3)])`
`(["M"^(2)"L"^(5)"T"^(-4)])/(["ML"^(-5)"T"^(-2)])=["MT"^(-2)]`