The mass of the bob of a simple pendulum of length $L$ is $m$. If the bob is released from its horizontal position,then the speed of the bob and the tension in the thread at the lowest position of the bob will be respectively:

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{x(u^{2}-gL)}$. The value of $x$ is .............

$A$ simple pendulum with a bob of mass $m = 1 \ kg$, charge $q = 5 \ \mu C$ and string length $l = 1 \ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 \times 10^6 \ V/m$ at its bottom-most point $A$, as shown in the figure. It is given a speed $u$ such that the particle leaves the circular path at its topmost point $C$. Find the speed $u$. (Take $g = 10 \ m/s^2$)

$A$ simple pendulum has a length of $75 \, cm$. It describes a vertical semi-circle. What will be the velocity of the bob when it passes through the mean position?

$A$ stone is tied to a string of length $L$ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is