If `T = 2pi sqrt(l/g)` is the time period of a simple pendulu, then the unit of `4pi^(2) l/T^(2)` in the SI system is ________ .

T=2pisqrt(l/g) is the time period of a simple pendulum, then the unit of 4pi^(2)l/T^(2) in SI system is ______ .

Establish the relation T = 2pi sqrt1//g for the time period of a simple pendulum with the help of dimensional analysis.

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

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

For simple pendulum, T = 2pi sqrt((l)/(g)) where T is the periodic time and l is the length of pendulum. If there is 4% error in the measure of periodic time then find the error percentage in the length of pendulum.

The time period of a simple pendulum is given by the formula, T = 2pi sqrt(l//g) , where T = time period, l = length of pendulum and g = acceleration due to gravity. If the length of the pendulum is decreased to 1/4 of its initial value, then what happens to its frequency of oscillations ?

If the time period of a simple pendulum of effective length L is T, then the effective length of a simple pendulum having time period 2T will be