$A$ 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 $s$)

If a simple pendulum can reach a height of $10 \ cm$,what is its velocity when it is at its mean position (in $m/s$)? $(g = 9.8 \ m/s^2)$

The time period of a simple pendulum inside a stationary lift is $T$. When the lift starts accelerating upwards with an acceleration of $\frac{g}{3}$,the time period of the pendulum will be

The acceleration due to gravity on the surface of the moon is $1.7 \; m s^{-2}$. What is the time period of a simple pendulum on the surface of the moon if its time period (in $s$) on the surface of the earth is $3.5 \; s$? ($g$ on the surface of the earth is $9.8 \; m s^{-2}$)

$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.

In order to find time,the astronaut orbiting in an earth satellite should use