Question

A particle rotates in a horizontal circle of radius \( R \) in a conical funnel, with speed \( V \). The inner surface of the funnel is smooth. The height of the plane of the circle from the vertex of the funnel is (g = acceleration due to gravity)

• A. \( \frac{v^2}{2g} \)
• B. \( \frac{v}{g} \)
• C. \( \frac{v^2}{g} \)
• D. \( \frac{V}{2g} \)

Hint:

For a particle moving in a conical funnel, the height of the circle is determined by the balance between the centripetal force and the gravitational force component.

Answer 1

The Correct Option is C

Step 1: Understanding the forces involved.
In this problem, the particle rotates in a horizontal circle inside a conical funnel. The centripetal force is provided by the component of the gravitational force acting along the surface of the funnel. This force is balanced by the vertical component of the tension in the string. The height \( h \) of the circle from the vertex of the funnel can be related to the velocity \( V \) and the radius \( R \) using the equation: \[ h = \frac{v^2}{g} \] where \( v \) is the speed of the particle and \( g \) is the acceleration due to gravity.
Step 2: Conclusion.
Thus, the height of the circle from the vertex of the funnel is \( \frac{v^2}{g} \), which is option (C).