
Since the disc completes \(\frac{1}{8}\) of a rotation, the time for \(\frac{1}{8}\) rotation is \(\frac{T}{8}\), where T is the period of the disc.
The period T is given by \(T=\frac{2\pi}{\omega}\)
Therefore, the time for 1/8 rotation is \(t = \frac{T}{8} = \frac{2\pi}{8\omega} = \frac{\pi}{4\omega}\)
X- coordinate of P = ωRt
\(= \frac{πR}{4} \gt Rcos45\degree\)
Therefore, P and Q lands in the unshaded region.
So. the correct option is (D): both P and Q land in the shaded region
A cylindrical tube \(AB\) of length \(l\), closed at both ends, contains an ideal gas of \(1\) mol having molecular weight \(M\). The tube is rotated in a horizontal plane with constant angular velocity \(\omega\) about an axis perpendicular to \(AB\) and passing through the edge at end \(A\), as shown in the figure. If \(P_A\) and \(P_B\) are the pressures at \(A\) and \(B\) respectively, then (consider the temperature to be same at all points in the tube) 
As shown in the figure, radius of gyration about the axis shown in \(\sqrt{n}\) cm for a solid sphere. Find 'n'. 
When rod becomes horizontal find its angular velocity. It is pivoted at point A as shown. 
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?