The graph between the time period $(T)$ and the length $(l)$ of a simple pendulum is

The time period of a simple pendulum is $T$. When the length is increased by $10 \ cm$,its period is $T_1$. When the length is decreased by $10 \ cm$,its period is $T_2$. Then,the relation between $T, T_1$,and $T_2$ is

$A$ simple pendulum with a bob of mass $m$ and length $x$ is held in position at an angle $\theta_1$ and then at an angle $\theta_2$ with the vertical. When released from these positions,the speeds with which it passes the lowest position are $v_1$ and $v_2$ respectively. Then,the ratio $\frac{v_1}{v_2}$ is .............

$A$ simple pendulum with a bob of mass $40 \ g$ and charge $+2 \ \mu C$ makes $20$ oscillations in $44 \ s$. $A$ vertical electric field of magnitude $4.2 \times 10^4 \ NC^{-1}$ pointing downward is applied. The time taken by the pendulum to make $15$ oscillations in the electric field is (acceleration due to gravity $= 10 \ ms^{-2}$) (in $s$)

The length of a simple pendulum is $l$ and its maximum angular displacement is $\theta$. What is its maximum kinetic energy $(K.E.)$?

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