$A$ uniform rod $AB$ of mass $2 \ kg$ and length $30 \ cm$ is at rest on a smooth horizontal surface. An impulse of force $0.2 \ Ns$ is applied to end $B$. The time taken by the rod to turn through a right angle will be $\frac{\pi}{X} \ s$,where $X = \text{ . . . . . . }$.

Find the increase in the moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

$A$ ring of radius $R$ rolls without sliding with a constant velocity $u$. The radius of curvature of the path followed by any particle of the ring at the highest point of its path will be

Two discs of moment of inertia $I_1 = 4 \ kg \ m^2$ and $I_2 = 2 \ kg \ m^2$ about their central axes and normal to their planes,rotating with angular speeds $10 \ rad/s$ and $4 \ rad/s$ respectively,are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is . . . . . . $J$.

One end of a straight uniform $1\; m$ long bar is pivoted on a horizontal table. It is released from rest when it makes an angle $30^{\circ}$ with the horizontal (see figure). Its angular speed when it hits the table is given as $\sqrt{n}\; s^{-1},$ where $n$ is an integer. The value of $n$ is