$A$ 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

$A$ stone of mass $1 \ kg$ tied to a light inextensible string of length $L = \frac{5}{3} \ m$ is rotating in a circular path of radius $L$ in a vertical plane. If the ratio of maximum tension in the string to the minimum tension in the string is $3$,the speed of the stone at the highest point of the circle is ($g =$ acceleration due to gravity).

$A$ particle is moving along a vertical circle of radius $R$. At point $P$,what will be the velocity of the particle? (Assume the critical condition at the highest point $C$).

$A$ 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 $g = 10 \ m/s^2$)

$A$ 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 $N$)

$A$ ball of mass $0.1 \ kg$ is suspended by a string. It is displaced through an angle of $60^\circ$ and released. When the ball passes through the mean position,the tension in the string is ........ $N$.