$A$ bob of a simple pendulum of mass $m$ is displaced through $90^{\circ}$ from the mean position and released. When the bob is at the lowest position,the tension in the string 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 $s$)

Two pendulums have time periods $T$ and $\frac{5T}{4}$. They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has completed one oscillation (in $^o$)?

From a fixed support,two small identical spheres are suspended by means of strings of length $1 \,m$ each. They are pulled aside as shown and then released. $B$ is the mean position. Then the two spheres collide

$A$ seconds pendulum is placed in a space laboratory orbiting round the Earth at a height $3R$ from the Earth's surface. The time period of the pendulum will be ($R =$ radius of Earth).

At which position in the string of a simple pendulum is the tension maximum?