$A$ body of mass $1\,kg$ is rotating in a vertical circle of radius $1\,m$. What will be the difference in its kinetic energy at the top and bottom of the circle? $(g = 10\,m/s^2)$

$A$ body of mass $m$ connected to a massless and unstretchable string moves in a vertical circle of radius $R$ under gravity $g$. The other end of the string is fixed at the center of the circle. If the velocity at the top of the circular path is $n\sqrt{gR}$,where $n \geq 1$,then the ratio of the kinetic energy of the body at the bottom to that at the top of the circle is:

$A$ hollow sphere has a radius of $6.4 \, m$. The minimum velocity required by a motorcyclist at the bottom to complete the vertical circle is ........... $m/s$.

$A$ mass attached to one end of a string crosses the topmost point on a vertical circle with critical speed. Its centripetal acceleration when the string becomes horizontal will be ($g =$ gravitational acceleration).

$A$ particle of mass $m$,attached to a string of length $l$,is moving in a vertical circle. If the particle is just looping the loop without the string slackening,and $v_A, v_B, v_D$ are the speeds at positions $A, B, D$ shown in the figure,then:

$A$ stone of mass $40$ g is tied to a light inextensible string of length $50$ cm and whirled in a vertical circle at the rate of $30$ revolutions per minute. The tension in the string when it is at the lowest point of the circle is (Take gravitational acceleration, $g = 10 \text{ m/s}^2$ and $\pi^2 = 10$) (in $N$)