If density$\left(\rho \right)$, acceleration (a) and force (F) are taken as basic quantities, then time period has

            dimensions

$$\begin{gathered} \text{T }\propto {\rho }^{\text{a}}{\text{a}}^{\text{b}}{\text{F}}^{\text{c}} \\ \text{ Dimensionally are equal} \\ \left[\text{T}\right]={\left[{\text{ML}}^{-\text{3}}\right]}^{\text{a}}{\left[{\text{LT}}^{-\text{2}}\right]}^{\text{b}}{\left[{\text{MLT}}^{-\text{2}}\right]}^{\text{c}} \\ \left[{\text{T}}^{\text{1}}\right]=\left[{\text{M}}^{\text{a}+\text{c}}\right]\left[{\text{L}}^{-\text{3a}+\text{b}+\text{c}}\right]\left[{\text{T}}^{-\text{2b}-\text{2c}}\right] \\ \text{a}+\text{c}=\text{0}\to \left(1\right) \\ -\text{3a}+\text{b}+\text{c}=\text{0}\to \left(2\right) \\ \text{4c}+\text{b}=\text{0 4c }=-\text{b} \\ -\text{2b}-\text{2c}=\text{1} \\ \text{2}\left(-\text{b}\right)-\text{2c}=\text{1 a}=-\frac{\text{1}}{\text{6}} \\ \text{2}\left(\text{4c}\right)-\text{2c}=\text{1 b}=-\text{4}\times \frac{\text{1}}{\text{6}} \\ \text{6c}=\text{1 b}=-\frac{\text{2}}{\text{3}} \\ \text{c}=\frac{\text{1}}{\text{6}} \end{gathered}$$