$A$ bucket full of water is revolved in a vertical circle of radius $2\,m$. What should be the maximum time-period of revolution so that the water does not fall off the bucket? (Take $g = 10\,m/s^2$) ......... $\sec$.

$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)

$A$ stone of mass $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are:

Lowest Point	Highest Point
$(a) \ mg - T_1$	$mg + T_2$
$(b) \ mg + T_1$	$mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$	$mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$	$mg + T_2 + \frac{mv_2^2}{R}$

$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

$A$ particle is whirled in a vertical circle of radius $1.0 \ m$ using a string with one end fixed. If the ratio of maximum and minimum tension in the string is $\frac{5}{3}$,the minimum velocity of the particle during circular motion is ......... $ms^{-1}$.

In the given figure,for an initial velocity $u = u_0/3$,find the height from the ground at which the block leaves the hemisphere,where $u_0 = \sqrt{gr}$.

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. Choose the correct alternative$(s)$ if $H = 2R$.