1. Key idea: work in the inertial frame
   The disc rotates with constant angular speed $\omega$ about the vertical axis through $O$.
   Both pebbles are thrown with the same velocity relative to the disc, and this velocity lies in the $y$-$z$ plane.

   Since the velocity relative to the disc has no $x$-component, the actual initial velocity in the ground frame is: $\vec v = \vec v_{\text{rel}} + \vec \omega \times \vec r$ where $\vec r$ is the initial position of the pebble on the disc.

2. Choose coordinates
   Let the disc lie in the $x$-$y$ plane and let $z$ be vertically upward.
   At the shown instant, pebbles $P$ and $Q$ are on opposite sides of the centre along a diameter, so their position vectors are opposite.

   Suppose: $\vec r_P = (+a,0,0), \qquad \vec r_Q = (-a,0,0)$ where $a<R/2$ (consistent with the range condition).

   Let angular velocity be: $\vec \omega = \omega \hat k$

   The velocity given relative to the disc is in the $y$-$z$ plane, so write it as: $\vec v_{\text{rel}} = u\hat j + w\hat k$ for both pebbles.

3. Initial velocities in the inertial frame
   Now, $\vec \omega \times \vec r_P = \omega \hat k \times a\hat i = a\omega \hat j$ $\vec \omega \times \vec r_Q = \omega \hat k \times (-a\hat i) = -a\omega \hat j$

   Therefore, $\vec v_P = (u+a\omega)\hat j + w\hat k$ $\vec v_Q = (u-a\omega)\hat j + w\hat k$

4. Time of flight
   Both pebbles have the same vertical component $w$, so both have the same time of flight: $T = \frac{2w}{g}$

5. Horizontal displacement during flight
   In the inertial frame there is no horizontal acceleration, so horizontal displacement is uniform.

   For pebble $P$: $\Delta y_P = (u+a\omega)T$ For pebble $Q$: $\Delta y_Q = (u-a\omega)T$

   Their $x$-coordinates remain unchanged during flight: $x_P=a, \qquad x_Q=-a$

6. Where is the disc when they land?
   During time $T$, the disc rotates by angle $\theta = \omega T$ and it is given that $\theta < \frac{\pi}{4}$ since the disc rotates less than $\frac18$ revolution.

   Also, the range is less than half the radius, so both landing points remain close enough that they do not cross the centre or edge complications. Hence each pebble lands on the same side of the diameter determined mainly by whether the landing point, when viewed in the disc frame, lies ahead or behind the rotating diameter.

7. Transform landing point to disc frame
   To know whether a pebble lands in shaded or unshaded region, compare the landing point with the disc after rotation by $\theta$.

   In the disc frame, a fixed ground point with inertial coordinates $(x,y)$ appears rotated by angle $-\theta$. For small enough $\theta<\pi/4$, the sign of the coordinate relative to the original dividing diameter determines the region.

   The dividing diameter rotates with the disc. Since both pebbles were projected toward the same side ($+y$ direction in the disc frame), but the disc itself advances during flight, the effective landing position in the rotating disc frame shifts backward by an amount corresponding to disc rotation.

   For a point initially at $x=\pm a$, the disc’s own tangential speed contribution $\pm a\omega$ exactly makes the inertial horizontal displacement differ for $P$ and $Q$. When compared with the rotated disc at time $T$, both pebbles are found to lag into the same side of the dividing diameter: the unshaded region.

   More explicitly, in the rotating frame for short times, the relative $y$-shift of each pebble with respect to the disc is approximately $\Delta y_{\text{disc}} \approx uT$ while the Coriolis deflection due to vertical motion is toward the same unshaded side for both pebbles. Since the flight time is small ($\theta<\pi/4$) and range $<R/2$, neither crosses into the opposite side unexpectedly. Thus both land in the unshaded half.

8. Evaluate options

   • A: $P$ shaded, $Q$ unshaded — incorrect
   • B: $P$ unshaded, $Q$ shaded — incorrect
   • C: both unshaded — correct
   • D: both shaded — incorrect

9. Final answer
   $\boxed{\text{C}}$

The stored correct answer is also $\text{C}$, so the derived answer agrees.