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

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 h is height and g is acceleration due to gravity, then the dimensional formula of sqrt((2h)/(g)) is the same as that of

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

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?

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?

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 ?