$A$ 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:

The bob of a simple pendulum is a spherical hollow ball filled with water. $A$ plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation,till water is coming out,the time period of oscillation would:

Two pendulums of length $121 \ cm$ and $100 \ cm$ start vibrating in phase. At some instant,the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

$A$ simple pendulum of length $l$ and having a bob of mass $M$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$. If the pendulum makes small oscillations in a radial direction about its equilibrium position,what will be its time period?

$A$ simple pendulum of length $1\,m$ is allowed to oscillate with an amplitude of $2^o$. It collides elastically with a wall inclined at $1^o$ to the vertical. Its time period will be: (use $g = \pi^2$)

Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $\theta_1$ and $\theta_2$ from their mean positions,respectively. If the angular accelerations of both pendulums are same,then which expression is correct?