Show that the expression of the time period T of a simple pendulum of length l given by `T = 2pi sqrt((l)/(g))` is dimensionally correct

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally currect

The graph of time period (T) of a simple pendulum versus its length (I) is

Show that the time period of a simplw pendulam of infinite length is given by t=2pi sqrt((R)/(g)) where R is the radius of the earth

The periodic time (T) of a simple pendulum of length (L) is given by T=2pisqrt((L)/(g)) . What is the dimensional formula of Tsqrt((g)/(L)) ?

How does time period (T) of a seconds pendulum vary with length(l)?

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

Time period T of a simple pendulum of length l is given by T = 2pisqrt((l)/(g)) . If the length is increased by 2% then an approximate change in the time period is

The time period 'T' of a simple pendulum of length 'l' is given by T = 2pisqrt(l/g) .Find the percentage error in the value of 'T' if the percentage error in the value of 'l' is 2%.