A seconds pendulum is one that takes one second to swing to an extreme position and return to the equilibrium position. This gives the time period of a seconds pendulum as 2 seconds. The time period of any pendulum of length L is equal to 2*pi*sqrt(L/g) where g is the free fall acceleration.
As the length of the seconds pendulum in Tokyo, Japan is 0.9914 m, it gives:
2 = 2*pi*sqrt(L/g)
=> sqrt(L/g) = 1/pi
=> L/g = 1/pi^2
=> g = L*pi^2
=> g = 0.9914*pi^2
=> g = 9.78473 m/s^2
The correct answer is option 3, 9.78473 m/s^2
the answer is 9.78473 m/s2.
the formula for the time period of a simple pendulum is T=2 pi square root of (l/g).
using the given data,
solving the above equation for'g',we get the value of 'g' as 9.78473 m/s2 which is option (3).