For a simple pendulum, the graph of time period $(T)$ versus length $(l)$ is depicted by:

Savitha, a $XI$ standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$, notes down the data of time taken to complete $30$ oscillations as $60 \text{ s}$ and hence calculates the length of the simple pendulum as: (Take $\pi^2 = 9.8$, and $g = 9.8 \text{ m/s}^2$) (in $\text{ m}$)

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:

The length of a seconds pendulum on the surface of the Earth is $1\, m$. The length of a seconds pendulum on the surface of the Moon,where $g$ is $1/6$th the value of $g$ on the surface of the Earth,is:

The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44 \%$,then its new time period of oscillation will be ......... $s$.

$A$ simple pendulum oscillating in air has period $T.$ The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$ of the material of the bob. If the bob is inside the liquid all the time,its period of oscillation in this liquid is: