Question

the time period of seconds pendulum on a planet where gravitation acceleration is 1/9th of gravitational acceleration of Earth

Answer 1

We have formula for period of a simple pendulum as
T = 2 π $\sqrt{\frac{l}{g} }$

Let the time period of the pendulum on earth be T1 and on the planet be T2.

Hence,
T1 =  2 π $\sqrt{\frac{l}{g} }$

Then, T2 will be equal to 2 π $\sqrt{\frac{l}{g/9} }$ as the gravitational acceleration of the planet is 1/9th that of the earth.
So, T2 will be equal to 2 π $\sqrt{\frac{9l}{g} }$
T2 = 2 π * 3 $\sqrt{\frac{l}{g} }$
T2 = 3 * 2 π $\sqrt{\frac{l}{g} }$

In place of 2 π $\sqrt{\frac{l}{g} }$, we can write T1.
So, T2 = 3 T1

We know that the time period of seconds pendulum on earth is 2 seconds.
Hence, its time period on the planet will be 3 times that on the earth, i.e. 3 * 2 = 6 seconds.

Hence, time period of a seconds pendulum on the planet is 6 seconds.

Answer 2

Hello Dear.

Here is your answer---



Let the Time Period of the second pendulum on Earth be T and acceleration due to gravity be g.


On the Other Planet,

Let the new time period will be T'.

If the acceleration due to gravity become 1/9th then the time period will be Triple ,i.e., 6 seconds.
 
Proof : 

According to the Question,

 Acceleration due to gravity on the Planet(g')= 1/9th × g
                                                                            = g/9

Thus, Using the Formula, 

                T/T' = √(g'/g)


               T/T' = √[(g/9)÷ g]
                T/T' = √1/9
                 T = T'/3               
                 T' = 3 T

                T' = 3 × 2   
  [∵  Time period of Second Pendulum on earth is 2 seconds]
               
               T' = 6 seconds.

Thus, time periods of the second pendulum on the planet where acceleration due to gravity is 1/9th of the earth is 6 seconds.


Hope it helps.

Have a Marvelous Day.