Question

A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of $0.5\text{m/s}$. What is the height of the plane of circle from vertex of the funnel? (take $g=10{m/s}^2$)

• A. $0.25\text{cm}$
• B. $2\text{cm}$
• C. $4\text{cm}$
• D. $2.5\text{cm}$

Answer 1

The correct option is D $2.5\text{cm}$

The particle is moving in a circular path.



From the figure,
$mg=N\sin \theta$ ... (1)
$\frac{m{v}^2}{r}=N\cos \theta$ ... (2)
From Eqs. (1) and (2),
we get
$\tan \theta =\frac{rg}{v^2}$ but $\tan \theta =\frac{r}{h}$
$\therefore h=\frac{v^2}{g}=\frac{(0.5{)}^2}{10}=0.025\text{m}=2.5\text{cm}$