I have tried 9.81 and 9.82 but says its wrong. **Pendulum Period and Gravity**A pendulum with a period of 2.00000 seconds in one location (where the acceleration due to gravity, \( g = 9.80 \, \text{m/s}^2 \)) is moved to a new location where the period is now 1.99801 seconds. What is the change in acceleration (in \(\text{m/s}^2\)) due to gravity at its new location?The calculated gravity at the new location is shown as \(9.82 \, \text{m/s}^2\), which is marked as incorrect.**Note:**- Variables: - Initial period: 2.00000 s - Initial gravity: 9.80 m/s² - New period: 1.99801 sThe aim is to determine the new gravitational acceleration that results in the reduced period of 1.99801 seconds.

Given: Time period of pendulum in one location is 2.0000 s. Acceleration due to gravity is 9.80 m/s2. Time period of pendulum in other location is 1.99801 s. Formula Used: The formula to calculate time period of pendulum is: T=2πlg Here, T is the time period of pendulum, l is the length of the string of the pendulum and g is the acceleration due to gravity at different locations. Since length of the string remains same at both of the location, therefore time period will only depend on acceleration due to gravity. Take ratio of the time period at two location and solve to calculate acceleration at second location. T1T2=2πlg12πlg2T1T2=g2g1T1T22=g2g1g2=T1T22g1Substitute the values in the equation. g2=2.0000 s1.99801 s29.80 m/s2=9.8195 m/s2 Therefore acceleration due to gravity at other location is 9.8195 m/s2. Change in acceleration due to gravity is: ∆g=g2-g1=9.8195 m/s2-9.80 m/s2=0.0195 m/s2 Therefore, change in acceleration due to gravity is 0.0195 m/s2.