The time period of a simple pendulum is $T$. If its length is increased by $21\%$,what is the percentage increase in its time period?

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 $5 \times 10^{-4}$,the difference in radii,$|r_1 - r_2|$,is best given by .... $cm$.

$A$ piece of wood has length,breadth and height,'$a$','$b$' and '$c$' respectively. Its relative density is '$d$'. It is floating in water such that the side '$a$' is vertical. It is pushed down a little and released. The time period of $S.H.M.$ executed by it is ($g=$ acceleration due to gravity).

$A$ brass cube of side $a$ and density $\rho$ is floating in mercury of density $\sigma$. If the cube is displaced a bit vertically,it executes $S.H.M.$ Its time period will be

$A$ pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same angular amplitude as in the first case,its maximum kinetic energy is $K_2$,then:

$A$ simple pendulum of length $L$ is suspended from the roof of a trolley. The trolley moves in a horizontal direction with an acceleration $a$. What would be the period of oscillation of the simple pendulum? [$g$ is the acceleration due to gravity]