The acceleration due to gravity at a place is $\pi^2 \, m/s^2$. Then the time period of a simple pendulum of length $1 \, m$ is:

Frequency $f$ of a simple pendulum depends on its length $\ell$ and acceleration $g$ due to gravity according to the following equation $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$. $A$ graph between which of the following quantities is a straight line?

$A$ person of mass $M$ is sitting on a swing of length $L$ and swinging with an angular amplitude $\theta_0$. If the person stands up when the swing passes through its lowest point,the work done by him,assuming that his centre of mass moves by a distance $l$ $(l << L)$,is close to

What is the velocity of the bob of a simple pendulum at its mean position,if it is able to rise to a vertical height of $10 \, cm$ ......... $m/s$? (Take $g = 9.8 \, m/s^2$)

If the length of a simple pendulum is increased by $300\%$,then the time period will be increased by ..... $\%$

If a second's pendulum is moved to a planet where the acceleration due to gravity is $4$ times that on Earth,the length of the second's pendulum on the planet should be made .......... times to keep the time period constant.