A particle rotates in horizontal circle of radius 'R' in a conical funnel, with speed 'V'. The inner surface…

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 $(\mathrm{g}=$ acceleration due to gravity $)$
  1. $\frac{\mathrm{V}^{2}}{2 \mathrm{~g}}$
  2. $\frac{\mathrm{V}}{\mathrm{g}}$
  3. $\frac{\mathrm{V}^{2}}{\mathrm{~g}}$
  4. $\frac{V}{2 g}$

Solution

$\mathrm{mg}=\mathrm{R} \sin \theta$ $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{R} \cos \theta$ $\tan \theta=\frac{\mathrm{rg}}{\mathrm{v}^{2}}$ $\tan \theta=\frac{\mathrm{r}}{\mathrm{h}}$ $\mathrm{h}=\frac{\mathrm{v}^{2}}{\mathrm{~g}}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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