The time period T of simple pendulum depends upon length l of the pendulum and gravitational acceleration. The formula for time period of simple pendulum is given by

Let $T\propto l^{a}andT\propto g^b$

where a and b are dimensionless constants

$T=k{l}^a{g}^b$

where, k is dimensionless constant.

$\left[LHS\right]\left[T\right]=\left[M^0{L}^0{T}^{-1}\right]$

and $\left[RHS\right]\left(l^a{g}^b\right)={\left[L\right]}^a{\left[L{T}^{-2}\right]}^b$

$=\left[L^{a+b}{T}^{-2b}\right]=\left[M^0{L}^{a+b}{T}^{-2b}\right]$

According to homogeneity principle,

[LHS] = [RHS]

or $\left[M^0{L}^{0}T\right]=\left[M^0{L}^{a+b}{T}^{-2b}\right]$

For dimensional balance, dimensions on both sides should be same.

$\therefore a+b=0$

and -2b =1

$\therefore b=-\frac{1}{2}$

and $a=\frac{1}{2}$

$\therefore T=2\pi \sqrt{\frac{l}{g}}$

[since, numerical value of k in case of simple pendulum is $2\pi$]