$A$ particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance $r$. The speed of the particle is

$A$ motorcycle driver doubles their velocity when taking a turn. The force exerted outwardly will be

$A$ train is moving with a speed of $12 \,m/s$ on rails which are $1.5 \,m$ apart. To negotiate a curve of radius $400 \,m$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $g = 10 \,m/s^2$): (in $\,cm$)

$A$ particle is moving with a uniform speed in a circular orbit of radius $R$ in a central force inversely proportional to the $n^{th}$ power of $R$. If the period of rotation of the particle is $T$,then

$A$ 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:

$A$ particle is released on a vertical smooth semicircular track from point $X$ so that $OX$ makes an angle $\theta$ from the vertical (see figure). The normal reaction of the track on the particle vanishes at point $Y$ where $OY$ makes an angle $\phi$ with the horizontal. Then