$A$ roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20\, m$. The speed of the car at the top of the hill is between

$A$ string of length '$L$' fixed at one end carries a mass '$m$' at the other end. The string makes $\frac{3}{\pi}$ r.p.s. around the vertical axis through the fixed end. The tension in the string is (in $mL$)

$A$ particle moves in a horizontal circle on the smooth inner surface of a hemispherical bowl of radius $R$. The plane of motion is at a depth $d$ below the centre of the hemisphere. The speed of the particle 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:

$A$ particle moves in a circular path inside a cone with speed $v$. Then:

One end of a string of length $l$ is connected to a particle of mass $m$ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$,the net force on the particle (directed towards the centre) is :
$(i) \; T$
$(ii) \; T - \frac{m v^{2}}{l}$
$(iii) \; T + \frac{m v^{2}}{l}$
$(iv) \; 0$
$T$ is the tension in the string. [Choose the correct alternative].