$A$ 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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Inner and outer radii of a spool are $r$ and $R$ respectively. $A$ thread is wound over its inner surface and placed over a rough horizontal surface. The thread is pulled by a force $F$ as shown in the figure. Then,in the case of pure rolling:

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$A$ particle of mass $M=0.2 \ kg$ is initially at rest in the $xy$-plane at a point $(x=-l, y=-h)$,where $l=10 \ m$ and $h=1 \ m$. The particle is accelerated at time $t=0$ with a constant acceleration $a=10 \ m/s^2$ along the positive $x$-direction. Its angular momentum and torque with respect to the origin,in $SI$ units,are represented by $\vec{L}$ and $\vec{\tau}$,respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x, y$ and $z$-directions,respectively. If $\hat{k}=\hat{i} \times \hat{j}$,then which of the following statement$(s)$ is(are) correct?
$(A)$ The particle arrives at the point $(x=l, y=-h)$ at time $t=2 \ s$.
$(B)$ $\vec{\tau}=2 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(C)$ $\vec{L}=4 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(D)$ $\vec{\tau}=\hat{k}$ when the particle passes through the point $(x=0, y=-h)$.

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then,the rise in temperature of the sphere is

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