The acceleration due to gravity at a place changes from `9.8 m//s^(2) " to " 9.5 m//s^(2)`. Then the length of the second's pendulum changes by

The acceleration due to gravity changes from 9.8 m//s^(2) to 9.5 m//s^(2) . To keep the period of pendulum constant, its length must changes by

The acceleration due to gravity is 9.8m//s^(2)

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) )

The acceleration due to gravity at a place is pi^(2)m//s^(2) . Then, the time period of a simple pendulum of length 1 m is

The acceleration due to gravity at a place is pi^(2)m//s^(2) . Then, the time period of a simple pendulum of length 1 m is

The acceleration due to gravity is 9.8 m s^(-2) . Give its value in ft s^(-2)

The acceleration due to gravity is 9.8 m s^(-2) . Give its value in ft s^(-2)

The acceleration due to gravity is 9.8 m s^(-2) . Give its value in ft s^(-2)