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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2012 · Paper 2 · Physics · Rotational Motion

Consider a disc rotating in the horizontal plane with a constant...

Consider a disc rotating in the horizontal plane with a constant angular speed ω\omega about its centre O\mathrm{O}. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles PP and QQ are simultaneously projected at an angle towards RR. The velocity of projection is in the yy - zz plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/81 / 8 rotation, (ii) their range is less than half the disc radius, and (iii) ω\omega remains constant throughout. Then Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O . The disc has a

Options

  1. APP lands in the shaded region and QQ in the unshaded region.
  2. BPP lands in the unshaded region and QQ in the shaded region.
  3. CBoth PP and QQ land in the unshaded region.
  4. DBoth PP and QQ land in the shaded region.

Correct answer

C. Both PP and QQ land in the unshaded region.

Step-by-step solution

For 18\frac{1}{8} of rotation of disc, t=18×2πω=π4ωx-coordinate of P=ωRT=πR4>Rsin⁡45∘\begin{aligned} t=\frac{1}{8} \times \frac{2 \pi}{\omega}=\frac{\pi}{4 \omega} & \\ x \text{-coordinate of } P &=\omega R T \\ &=\frac{\pi R}{4}>R \sin 45^{\circ} \end{aligned} To reach the unshaded part, particlePP needs to travel horizontal range >Rsin⁡45∘≃0.7R> R \sin 45^{\circ} \simeq 0.7 \mathrm{R} But its range is less than R/2\mathrm{R} / 2 so it will land on shaded part. QQ is near the origin, its velocity will be nearly along QR so it will land in unshaded part. Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O . The disc has a
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