Question

A particle describes a horizontal circle of radius $r$ on the smooth surface of an inverted cone as shown. The height of plane of circle above vertex is $h$. The speed of particle should be

• A. $\sqrt{rg}$
• B. $\sqrt{2rg}$
• C. $\sqrt{gh}$
• D. $\sqrt{2gh}$

Answer 1

Answer: (C)

$N\cos \left(90^{\circ }-\theta \right)=\frac{m{v}^2}{r}$
$N\sin \left(90^{\circ }-\theta \right)=mg$
$N=\frac{mg}{\cos \theta }$ ...(i)
and $N\sin \theta =\frac{m{v}^2}{r}$ ...(ii)
dividing (ii) and (i)
$mg\tan \theta =\frac{m{v}^2}{r}$
$g\left(\frac{h}{r}\right)=\frac{v^2}{r}$
$\Rightarrow v=\sqrt{gh}$