If the length of a simple pendulum is increased by $44\%$,then what is the percentage change in the time period of the pendulum (in $\%$)?

$A$ light balloon filled with helium of density $\rho_{He}$ is tied to a light string of length $L.$ The string is tied to the ground forming an "inverted" simple pendulum (figure). If the balloon is displaced slightly from equilibrium as in figure and released, the period of the motion is. Take the density of air to be $\rho_{air}$. Assume the air applies a buoyant force on the balloon but does not otherwise affect its motion.

$A$ simple pendulum oscillates with an angular amplitude $\theta$. If the maximum tension in the string is $4$ times the minimum tension,then the value of $\theta$ is

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

$A$ 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:

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: