- A$T$
- ✓$\frac{T}{2}$
- C$2 T$
- D$\frac{T}{4}$
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$\vec{F}_{\text {rot }}=\vec{F}_{\text {in }}+2 m\left(\vec{v}_{\text {rot }} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega},$
where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\bar{r}$ is the position vector of the particle with respect to the centre of the disc.
Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the $x$-axis along the slot, the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$. A sm a $1$ block of mass $m$ is gently placed in the slot at $\vec{r}=(R / 2) \hat{i}$ at $t=0$ and is constrained to move only along the slot.
(Image)
($1$) The distance $r$ of the block at time $t$ is
($A$) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$ ($B$) $\frac{R}{2} \cos \omega t$ ($C$) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$
($D$) $\frac{F}{2} \cos 2 \omega t$
($2$) The net reaction of the disc on the block is
($A$) $\frac{1}{2} m \omega^2 R\left(e^{2 \omega t}-e^{-2 \omega t}\right) \hat{j}+m g \hat{k}$
($B$) $\frac{1}{2} m \omega^2 R\left(e^{\omega t}-e^{-a t t}\right) j+m g k$
($C$) $-m \omega^2 R \cos \omega t \hat{j}-m g \hat{k}$
($D$) $m \omega^2 R \sin \omega t \hat{j}-m g \hat{k}$
Give the answer quetioin ($1$) ($2$)
Statement $I$ : The law of gravitation holds good for any pair of bodies in the universe.
Statement $II$ : The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below.