Obtain an expression for the time period T of a simple pendulum. [The time period T depend upon (i) mass l of the bob (ii) length m of the pendulum and (iii) acceleration due to gravity g at the place where pendulum is suspended.
Assume the constant `k=2pi`]

Explanation

`Tprop m^(a)l^(b)g^(c),T=k.m^(a)l^(b)g^(c)`
Here k is the dimensiionless constant.
Rewriting the above equation with dimensions.
`[T^(1)]=[M^(a)][L^(b)][LT^(-2)]^(c)`
`[M^(0)L^(0)T^(1)]=[M^(a)L^(b+c)T^(-2c)]`
Comparing the powers of M,L and T on both sides `a=0,b+c=0,-2c=1`
Solving for a,b and c we get `a=0,b=1//2` and `c=-1//2`
From the above equation `T=k.m^(2)l^(1//2)g^(-1//2)`
`T=k(l/g)^(1//2)=ksqrt(l//g)`
Experimentally `k=2pi ` hence `T=2pi sqrt(l//g)`