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

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 correct

Find the expression of time period of a simple pendulum of length l.

(i) Show dimensionally that the equation of the time period of a simple pendulum of length l. Given relation T=2pil//g is incorrect. (ii) Find its correct form.

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)?

The time period T of oscillation of a simple pendulum of length l is given by T=2pi. sqrt((l)/(g)) . Find the percentage error in T corresponding to 2% in the value of l.

The time period T of oscillation of a simple pendulum of length l is given by T=2pi. sqrt((l)/(g)) . Find the percentage error in T corresponding to (i) on increase of 2% in the value of l. (ii) decrease of 2% in the value of l.