$A$ $2 \, kg$ stone at the end of a string $1 \, m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \, ms^{-1}$. The tension in the string will be $52 \, N$ when the stone is

$A$ particle of mass $m$ is tied to one end of a string of length $l$. The particle is held horizontal with the string taut. It is then projected upward with a velocity $u$. The tension in the string is $\frac{mg}{2}$ when it is inclined at an angle $30^o$ to the horizontal. The value of $u$ is

$A$ bullet of mass $m$ and velocity $v$ strikes the bob of a pendulum of mass $M$ and emerges out with velocity $v/2$. What is the minimum value of $v$ such that the bob of the pendulum will swing through a complete circle? (Take the length of the pendulum $= \ell$)

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

As shown in the figure, a mass $m$ and another block of mass $10m$ are connected with a string of length $L$. Friction is sufficient to prevent the slipping of the $10m$ block. Mass $m$ is given a velocity $u$ in the vertical direction. For the complete circular motion of mass $m$:

$A$ particle is kept at rest at the top of a sphere of diameter $42\, m$. When disturbed slightly,it slides down. At what height $H$ from the bottom will the particle leave the sphere? $(...\, m)$