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?

The average speed of the bob of a simple pendulum oscillating with a small amplitude $A$ and time period $T$ is

$A$ simple pendulum is suspended in a lift which is going up with an acceleration $5 \ m/s^2$. An electric field of magnitude $5 \ N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1 \ mC$ and mass is $1 \ mg$. Taking $g = \pi^2 \ m/s^2$ and length of the simple pendulum $1 \ m$,the time period of the simple pendulum is ...... $s$.

At a place,the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping the amplitude the same. Then,the total energy will be:

$A$ small sphere of radius $r$ is placed on a concave surface of radius of curvature $R$ a little away from the center. When the sphere is released, it oscillates. Assuming the oscillation to be simple harmonic motion and $r << R$, then the time period is

Given below are two statements:
Statement $I :$ $A$ second's pendulum has a time period of $1$ second.
Statement $II :$ It takes precisely one second to move between the two extreme positions.
In the light of the above statements,choose the correct answer from the options given below: