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(C) Let the semi-vertical angle of the cone be $	heta$. From the geometry of the cone,we have $	an 	heta = \frac{r}{h}$.The forces acting on the pa

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$\sqrt{g h}$

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(C) Let the semi-vertical angle of the cone be $	heta$. From the geometry of the cone,we have $	an 	heta = \frac{r}{h}$.The forces acting on the particle are its weight $mg$ acting downwards and the normal reaction $N$ from the surface of the cone acting perpendicular to the surface.Resolving the normal reaction $N$ into horizontal and vertical components:Vertical component: $N \cos 	heta = mg$  --- $(i)$Horizontal component (providing the centripetal force): $N \sin 	heta = \frac{mv^2}{r}$ --- (ii)Dividing equation (ii) by equation $(i)$:$\frac{N \sin 	heta}{N \cos 	heta} = \frac{mv^2/r}{mg}$$	an 	heta = \frac{v^2}{rg}$Substituting $	an 	heta = \frac{r}{h}$ into the equation:$\frac{r}{h} = \frac{v^2}{rg}$$v^2 = \frac{r^2 g}{h}$$v = r \sqrt{\frac{g}{h}}$Wait,re-evaluating the geometry: If $	heta$ is the angle the slant surface makes with the horizontal,then $	an 	heta = \frac{h}{r}$.Then $\frac{h}{r} = \frac{v^2}{rg} \implies v^2 = gh \implies v = \sqrt{gh}$.

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

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