Choose the correct length $(L)$ versus square of time period $(T^2)$ graph for a simple pendulum executing simple harmonic motion.

$A$ simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$,the percentage increase in the time period of the pendulum of increased length is ..... $\%$

Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$,then the ratio of lengths of the two pendulums will be

$A$ cylindrical plastic bottle of negligible mass is filled with $310\, ml$ of water and left floating in a pond with still water. If pressed downward slightly and released,it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is $2.5\, cm$,then $\omega$ is close to ..... $rad\, s^{-1}$ (density of water $= 10^3\, kg/m^3$).

$A$ simple pendulum of length $l$ has a brass bob attached at its lower end. Its period is $T$. $A$ steel bob of the same size,having density $x$ times that of brass,replaces the brass bob. Its length is then changed such that the period becomes $2T$. What is the new length?

In a stationary lift,the time period of a simple pendulum is $T$. If the lift starts accelerating downwards with an acceleration of $\frac{g}{4}$,then the new time period of the pendulum will be: