$A$ string of length $2 \ m$ can withstand a maximum tension of $16000 \ N$. $A$ particle of mass $2 \ kg$ is tied to one end of the string and rotated in a horizontal circle. What is the maximum number of rotations per second the particle can make so that the string does not break? (Take $\pi^2 = g = 10 \ m/s^2$)

Two masses $M$ and $m$ are attached to a vertical axis by weightless threads of combined length $l$. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity $\omega$. If the tensions in the threads are the same during motion,the distance of $M$ from the axis is

$A$ string of length ' $\ell$ ' fixed at one end carries a mass 'm' at the other end. The string makes $\frac{3}{\pi}$ revolutions per second around the vertical axis through the fixed end as shown in the figure. The tension '$T$' in the string is:

Two particles of the same mass are moving in circular orbits because of a force given by $F(r) = -\frac{16}{r} - r^3$. The first particle is at a distance $r = 1$,and the second is at $r = 4$. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to:

$A$ mass of $100\, g$ is tied to one end of a string $2\, m$ long. The body is revolving in a horizontal circle making a maximum of $200$ revolutions per minute. The other end of the string is fixed at the center of the circle of revolution. The maximum tension that the string can bear is .......... $N$ (approximately).

$A$ string of length $L$ fixed at one end carries a body of mass $m$ at the other end. The mass is revolved in a circle in the horizontal plane about a vertical axis passing through the fixed end of the string. The string makes an angle $\theta$ with the vertical. The angular frequency of the body is $\omega$. The tension in the string is