$A$ uniform rod of length $2L$ has one end on a horizontal floor. It is inclined at an angle $\alpha$ to the horizontal floor. It falls without slipping,rotating about the point of contact. Its angular velocity when it hits the horizontal floor will be:

  • A
    $\omega = \sqrt{\frac{3g\sin\alpha}{2L}}$
  • B
    $\omega = \sqrt{\frac{2L}{3g\sin\alpha}}$
  • C
    $\omega = \sqrt{\frac{6g\sin\alpha}{L}}$
  • D
    $\omega = \sqrt{\frac{L}{g\sin\alpha}}$

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Two equal masses each of mass $M$ are joined by a massless rod of length $L$. Now an impulse $J = MV$ is given to one of the masses,making an angle of $30^{\circ}$ with the length of the rod. The angular velocity of the rod just after imparting the impulse is:

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$A$ sphere of mass $M$ and radius $R$ is attached by a light rod of length $l$ to a point $P$. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about $P$ when the rod becomes vertical is:

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Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2I$ respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2\omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.
$1.$ The ratio of $x_1/x_2$ is
$(A)$ $2$ $(B)$ $1/2$ $(C)$ $\sqrt{2}$ $(D)$ $1/\sqrt{2}$
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$(A)$ $\frac{2I\omega}{3t}$ $(B)$ $\frac{9I\omega}{2t}$ $(C)$ $\frac{9I\omega}{4t}$ $(D)$ $\frac{3I\omega}{2t}$
$3.$ The loss of kinetic energy during the above process is
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