MHTCET 15 Apr 2026 Shift 1 Paper

Concept:A small sphere moving inside a smooth watch glass executes simple harmonic motion along the curved surface.The effective length of the equivalent pendulum is the radius of curvature $R$ of the watch glass.So the time period is $T = 2\pi\sqrt{\frac{R}{g}}$.Explanation:Let the sphere be displaced through a small angle $\theta$ from the mean position.The displacement along the arc is $x = R\theta$, so $\theta = \frac{x}{R}$.The restoring acceleration is the tangential component of gravity: $a = -g\sin\theta$.For small oscillations, $\sin\theta \approx \theta$.Therefore, $a = -g\theta = -g\frac{x}{R}$. This is of the form $a = -\omega^{2}x$, where $\omega^{2} = \frac{g}{R}$.Hence, the time period is $T = 2\pi\sqrt{\frac{R}{g}}$.Substitute $R = 1.6\ \mathrm{m}$ and $g = 10\ \mathrm{m/s^{2}}$:$$T = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi\sqrt{0.16} = 2\pi(0.4) = 0.8\pi\ \mathrm{s}.$$Answer:The period of oscillation of the sphere is $0.8\pi\ \mathrm{s}$, which corresponds to Option C.