A particle is moving on a circular path with constant speed, then its acceleration will be

An object moves at a constant speed along a circular path in horizontal $XY$ plane with centre at origin. When the object is at  $x = -2\,m$ , its velocity is $-(4\,m/ s)\hat j$ . What is object's acceleration when it is at $y = 2\,m$ ?

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first $2$ sec, it rotates through an angle ${\theta _1}$. In the next $2$ sec, it rotates through an additional angle ${\theta _2}$. The ratio of ${\theta _2}\over{\theta _1}$ is

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

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

The ratio of period of oscillation of the conical pendulum to that of the simple pendulum is : (Assume the strings are of the same length in the two cases and $\theta$ is the angle made by the string with the verticla in case of conical pendulum)