A second pendulum is taken to a planet where the acceleration due to gravity is 9 times that on earth. Calculate the time period of the pendulum on that planet.

Dear student,

$T=2\pi \sqrt{\frac{l}{g}}$

 ​If the value of g is 1/9 the time period will be increase  to three times of the original, as time period is inversely proportional to g.
As the time period for seconds pendulum is 2 sec, so now it will become 6 sec

Regards