If density (ρ), acceleration (a) and force (F) are taken as basic quantities, then time period has dimensions.

Steps to Solve

1. Identify Units of The Basic Quantities

   The dimensions of the basic quantities are as follows:

   • Density ($\rho$): $[M L^{-3}]$ (mass per unit volume)
   • Acceleration ($a$): $[L T^{-2}]$ (change in velocity over time)
   • Force ($F$): $[M L T^{-2}]$ (mass times acceleration)

2. Determine the Dimensional Formula for Time Period

   The time period ($T$) has the following dimensional formula: $$ [T] $$

3. Express Time (T) in Terms of The Basic Quantities

   We can relate these basic quantities to the time period. We want to find the relationship such that: $$ [T] = [F]^{m}[\rho]^{n}[a]^{p} $$ (where $m$, $n$, and $p$ are constants to be determined).

4. Set Up The Equation for Dimensions

   Substituting in the dimensions of $F$, $\rho$, and $a$ gives us: $$ [T] = [M L T^{-2}]^{m} [M L^{-3}]^{n} [L T^{-2}]^{p} $$

   Breaking it down, we get: $$ = [M^{m+n} L^{m - 3n + p} T^{-2m - 2p}] $$

5. Equate Dimensions

   Set the equation for each dimension (mass, length, time) equal to their corresponding dimensions in $[T]$:

   • For mass: $m + n = 0$
   • For length: $m - 3n + p = 0$
   • For time: $-2m - 2p = 1$

6. Solve The System of Equations

   Use the three equations to solve for $m$, $n$, and $p$.

   From $m + n = 0$, we have $n = -m$.

   Substituting $n$ into the other two equations:

   • For length: $m - 3(-m) + p = 0 \Rightarrow 4m + p = 0 \Rightarrow p = -4m$.
   • For time: $-2m - 2(-4m) = 1 \Rightarrow -2m + 8m = 1 \Rightarrow 6m = 1 \Rightarrow m = \frac{1}{6}$.

   Thus, $n = -\frac{1}{6}$ and $p = -\frac{2}{3}$.