The time period of a simple pendulum is $4 \ s$ at a place on the Earth where the acceleration due to gravity is $\pi^2 \ ms^{-2}$. The length of the pendulum in meters is:

Two simple pendulums are given: the first has a bob of mass $M_1$ and length $L_1$,and the second has a bob of mass $M_2$ and length $L_2$. Given $M_1 = M_2$ and $L_1 = 2L_2$. If the vibrational energy of both is the same,which of the following statements is correct?

$A$ 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 $T'$.

$A$ lift is descending with an acceleration of $g/3$. What will be the time period of a simple pendulum suspended from its ceiling if its time period in a stationary lift is $T$?

The length of the seconds pendulum is $1 \,m$ on Earth. If the mass and diameter of a planet are $1.5$ times that of the Earth, the length of the seconds pendulum on the planet will be nearly (in $\,m$)

$A$ string of a pendulum of length $l$ is displaced through $90^{\circ}$ from its vertical and released. Then,the minimum strength of the string needed to withstand the tension as the pendulum passes through its mean position is