Solution:
The physical relationship of density (D) is given as:
\(D=\frac{Mass}{Volume}\)
\(=\frac{Force/Acceleration}{Volume}\)
\(=\frac F{a\times V}\)
Dimension of F = [F]
Dimension of acceleration (a) = [LT-1]
Dimension of volume,
\(V=\left[L^3\right]\)
\(=\frac{\left[F\right]}{\left[LT^{-2}\right]\left[L^3\right]}\)
\(=\left[FL^{-4}T^2\right]\)
Hence, the dimensions of D are 1 in F, -4 in L and 2 in T ie, [F L-4 T2]