What is the velocity of the bob of a simple pendulum at its mean position,if it is able to rise to a vertical height of $10 \, cm$ ......... $m/s$? (Take $g = 9.8 \, m/s^2$)

$Assertion:$ The time period of a simple pendulum on a satellite orbiting the Earth is infinity.
$Reason:$ The time period of a simple pendulum is inversely proportional to $\sqrt{g}$.

$A$ simple pendulum has a time period $T$. The bob is given a negative charge and the surface below it is given a positive charge. The new time period will be

$A$ simple pendulum,suspended from the ceiling of a lift,has a period of oscillation $T$,when the lift is at rest. If the lift starts moving upwards with an acceleration $a = 3g$,then the new period 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 $= 10^3\, kg/m^3$).

The bob of a simple pendulum of length $L$ has a mass $m$ and charge $q$. The pendulum is suspended between the plates of a charged parallel plate capacitor. The direction of the electric field is shown in the figure. The period of oscillations of the simple pendulum is (acceleration due to gravity $g > qE/m$):