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A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of $0.5 m s^{- 1}$ . The height (in $c m$ ) of the plane of circle from vertex of the funnel is

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$3 c m$

$1.5 c m$

$2 c m$

$2.5 c m$

answer is C.

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Detailed Solution

$N \sin θ = \frac{m v^{2}}{r}$ and $N \cos θ = m g$

$∴ \tan θ = \frac{v^{2}}{r g}$

or $\frac{h}{r} = \frac{v^{2}}{r g}$

$∴ h = \frac{v^{2}}{g} = \frac{0 . 5^{2}}{10} = 2.5 c m$

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