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Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre $\mathrm{O}$. 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 $P$ and $Q$ are simultaneously projected at an angle towards $R$. The velocity of projection is in the $y$ - $z$ 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 $1 / 8$ rotation, (ii) their range is less than half the disc radius, and (iii) $\omega$ remains constant throughout. Then
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The correct answer is:
Both $P$ and $Q$ land in the unshaded region.
For $\frac{1}{8}$ of rotation of disc,
$\begin{aligned}
t=\frac{1}{8} \times \frac{2 \pi}{\omega}=\frac{\pi}{4 \omega} & \\
x \text {-coordinate of } P &=\omega R T \\
&=\frac{\pi R}{4}>R \sin 45^{\circ}
\end{aligned}$
To reach the unshaded part, particle$P$ needs to travel horizontal range $>$ $R \sin 45^{\circ} \simeq 0.7 \mathrm{R}$
But its range is less than $\mathrm{R} / 2$ so it will land on shaded part. $Q$ is near the origin, its velocity will be nearly along QR so it will land in unshaded part.
$\begin{aligned}
t=\frac{1}{8} \times \frac{2 \pi}{\omega}=\frac{\pi}{4 \omega} & \\
x \text {-coordinate of } P &=\omega R T \\
&=\frac{\pi R}{4}>R \sin 45^{\circ}
\end{aligned}$
To reach the unshaded part, particle$P$ needs to travel horizontal range $>$ $R \sin 45^{\circ} \simeq 0.7 \mathrm{R}$
But its range is less than $\mathrm{R} / 2$ so it will land on shaded part. $Q$ is near the origin, its velocity will be nearly along QR so it will land in unshaded part.
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