If a simple pendulum is taken to a place where $g$ decreases by $2\%$,then the time period:

$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.

Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed a number of oscillations equal to

$A$ hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flows out of the hole at the bottom, the period of oscillation will

$A$ positively charged pendulum is oscillating in a uniform electric field pointing upwards. How does its time period compare to the time period when it oscillates without an electric field?

$A$ bob of a simple pendulum has mass $m$ and is oscillating with an amplitude $a$. If the length of the pendulum is $L$,then the maximum tension in the string is (given $\cos 0^{\circ}=1, g=$ acceleration due to gravity):