Question

A particle rotates in 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)

Options

• `V^2/(2g)`

• `V/g`

• `V^2/g`

• `V/(2g)`

Answer 1

A particle rotates in 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 `bbunderline(V^2/g)`.

Explanation:

N sin θ = mg     ...(1)

N cos θ = `(mV^2)/R`     ...(2)

Divide equation (2) by equation (1):

`(N cos theta)/(N sin theta) = ((mV^2)/(R))/(mg) = cot theta = (V^2)/(Rg)`

cot θ = `h/R`

`h/R = (V^2)/(Rg)`

h = `(V^2)/g`