IIT-JEE 2012 · Paper 2, Physics question 6

Ground frame. Once airborne the pebbles move with their launch velocity (relative to the disc) plus the disc's velocity at their launch points. The disc rotates counter-clockwise, so the velocity of the surface along the diameter PQR points to the right ($+x$) on the lower side of O and to the left above O; P (at the rim, below O) and Q (just above O) therefore get these sideways velocities $\omega r$.

Where the boundary goes. In the flight time the dividing diameter turns counter-clockwise by $\varphi=\omega t$: points of it below O move right by about $|y|\varphi$, points above O move left.

• P starts at $y=-r$ with sideways velocity $r\omega$ to the right and lands at height $y=-r+d$ ($d<r/2$). It moves right by $r\varphi$, while the boundary at its landing height has moved right by only $(r-d)\varphi$. P lands to the right of the boundary: unshaded.
• Q starts just above O, and the boundary above O moves left, while Q's own small sideways velocity points left by less than the boundary's at its landing height (which is further from O). Q also lands to the right of the boundary: unshaded.

Both pebbles land in the unshaded region. The official key accepts (C) or (D).

(C) (key: C or D)