Time period of a simple pendulum in a stationary lift is ' $T$ '. If the lift accelerates with $\frac{ g }{6}$ vertically upwards then the time period will be .....

(where $g =$ acceleration due to gravity)

The acceleration due to gravity on the surface of moon is $1.7 \;ms ^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period (in $sec$) on the surface of earth is $3.5\; s ?( g$ on the surface of earth is $9.8\; ms ^{-2} )$

Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g / 2,$ the time period of pendulum will be

In an experiment to determine the period of a simple pendulum of length $1\, m$, it is attached to different spherical bobs of radii $r_1$ and $r_2$ . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $5\times10^{-4}\, s$, the difference in radii, $\left| {{r_1} - {r_2}} \right|$ is best given by .... $cm$

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4,$ then the new time period of the pendulum is

Time period of pendulum, on a satellite orbiting the earth, is