If 'L' is length of simple pendulum and 'g' is acceleration due to gravity then the dimensional formula for (l/g)^(1/2) is same as that for

If 'L' is length of sample pendulum and 'g' is acceleration due to gravity then the dimensional formula for ((l)/(g))^(1/2) is same that for

How time period of a simple pendulum depends on the acceleration due to gravity ?

A simple pendulum can be used to determine acceleration due to gravity at a given place.

How is the length of seconds pendulum related with acceleration due to gravity of any planet?

The time period of a simple pendulum is given by T = 2pi sqrt((l)/(g)) where 'l' is the length of the pendulum and 'g' is the acceleration due to gravity at that place. (a) Express 'g' in terms of T (b) What is the length of the seconds pendulum ? (Take g = 10 m s^(-2) )

If g is the acceleration due to gravity and R is the radius of earth, then the dimensional formula for g R is

Given that T stands for time and l stands for the length of simple pendulum . If g is the acceleration due to gravity , then which of the following statements about the relation T^(2) = ( l// g) is correct?

Consider a simple pendulum. The period of oscillation of the simple pendulum depends on its length and acceleration due to gravity. Then the expression for its time period is

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 ?