A particle rotates in a horizontal circle of radius ' R ' in a conical funnel with constant speed ' V '. The…
A particle rotates in a horizontal circle of radius ' R ' in a conical funnel with constant 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)
$\frac{\mathrm{V}}{\mathrm{g}}$
$\frac{\mathrm{V}}{2 \mathrm{~g}}$
$\frac{\mathrm{V}^2}{2 \mathrm{~g}}$
$\frac{\mathrm{V}^2}{\mathrm{~g}}$
Solution
$\begin{array}{ll} & \text { From figure, } \\ & N \sin \theta=m g \\ & N \cos \theta=\frac{m V^2}{R} \\ \therefore \quad & \tan \theta=\frac{R g}{V^2} \\ \therefore \quad & \frac{R}{h}=\frac{R g}{V^2} \\ \therefore \quad & h=\frac{V^2}{g}\end{array}$
