$A$ 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 $X = \text{ . . . . . . }$.

  • A
    $4$
  • B
    $5$
  • C
    $6$
  • D
    $7$

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$A$ rod of mass $M$ and length $l$ is at rest on a smooth horizontal surface. $A$ particle of the same mass $M$ strikes one end of the rod with velocity $u$ perpendicular to the rod,elastically. Just after the collision,what is the kinetic energy of the upper half part of the rod?

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Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ is

$ABC$ is an equilateral triangle with $O$ as its centre. $\vec F_1, \vec F_2$ and $\vec F_3$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about $O$ is zero,then the magnitude of $\vec F_3$ is

Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.
$(a)$ Does the law of conservation of angular momentum apply to the situation? Why?
$(b)$ Find the angular speed of the two-disc system.
$(c)$ Calculate the loss in kinetic energy of the system in the process.
$(d)$ Account for this loss.

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