MHTCET 16 Apr 2026 Shift 1 Paper

Concept:In a smooth conical funnel, the normal reaction's horizontal component provides the centripetal force, while its vertical component balances the weight.Explanation:Let the particle move in a horizontal circle of radius $r$ at a height $h$ from the vertex of the funnel.Let $\alpha$ be the semi-vertical angle of the cone.The forces acting on the particle are its weight $mg$ and the normal reaction $N$.From the geometry of the cone, $\tan \alpha = \frac{r}{h}$.The horizontal component of $N$ is $N \cos \alpha$, and the vertical component is $N \sin \alpha$.Thus, $\frac{N \cos \alpha}{N \sin \alpha} = \frac{h}{r}$.For horizontal circular motion:$$N \cos \alpha = \frac{mv^2}{r}$$For vertical equilibrium:$$N \sin \alpha = mg$$Dividing these two equations:$$\frac{v^2}{rg} = \frac{h}{r}$$Cancelling $r$ from both sides:$$h = \frac{v^2}{g}$$Substituting $v = 0.5 \, \mathrm{m/s}$ and $g = 10 \, \mathrm{m/s^2}$:$$h = \frac{(0.5)^2}{10} = \frac{0.25}{10} = 0.025 \, \mathrm{m}$$Converting into centimetres:$$h = 0.025 \, \mathrm{m} = 2.5 \, \mathrm{cm}$$Answer:The height of the plane of the circle from the vertex is $2.5 \, \mathrm{cm}$.Correct option: B.