MHTCET 23 Apr 2025 Shift 2 Paper

Concept:For a particle moving in a horizontal circle inside a smooth cone, the normal reaction provides the necessary centripetal force, and the vertical force balance links the speed to the height above the vertex.Explanation:Let the semi-vertical angle of the cone be $\theta$ and the radius of the horizontal circle be $R$.The height of the circle above the vertex is $h = 10\ \text{cm} = 0.1\ \text{m}$.From the force balance on the particle:Vertical component: $N \sin\theta = mg$Horizontal component: $N \cos\theta = \dfrac{mv^2}{R}$Dividing these two equations gives:$\tan\theta = \dfrac{Rg}{v^2}$From the geometry of the cone, $\tan\theta = \dfrac{R}{h}$.Therefore, $\dfrac{R}{h} = \dfrac{Rg}{v^2}$Simplifying, $v^2 = gh$Substitute $g = 10\ \text{m/s}^2$ and $h = 0.1\ \text{m}$:$v = \sqrt{gh} = \sqrt{10 \times 0.1} = \sqrt{1} = 1\ \text{m/s}$Answer:The speed of the particle is $1\ \text{m/s}$, which corresponds to option C.