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:

Length of a simple pendulum is $1 \,m$. When its bob is at its lowest point its velocity is $7 \,ms^{-1}$. If the bob leaves its circular path at a height $h$ above the centre of the circle, then the value of $h$ is $(g=10 \,ms^{-2})$. (in $\,m$)

$A$ bullet of mass $m$ moving with velocity $v$ passes through a pendulum bob of mass $M$ and emerges with a velocity $v/2$. What is the minimum value of $v$ so that the pendulum bob completes a full vertical circle of length $\ell$?

$A$ steel wire can withstand a load up to $2940 \ N$. $A$ load of $150 \ kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position,so that the wire does not break when the load passes through the position of equilibrium,is (in $^{\circ}$)

$A$ wire,which passes through the hole of a small bead,is bent in the form of a quarter of a circle. The wire is fixed vertically on the ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from $A$ to $B$,the force it applies on the wire is

$A$ 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 .............