**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} T^{2}]**