Q. A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of $\ 0.5\ m/s$. What is the height of the plane of circle from vertex of the funnel
A) $0.25\ cm$
B) $2\ cm$
C) $4\ cm$
D) $2.5\ cm$
Explanation
(d) The particle is moving in circular path
From the figure, $mg = R\sin\theta\ \ \ \ \ \ \ldots(i)$
$\frac{mv^{2}}{r} = \ \ R\cos\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From equation (i) and (ii) we get
$\tan\theta = \frac{rg}{v^{2}}$ but $\tan\theta = \frac{r}{h}$
$\therefore h = \frac{v^{2}}{g} = \frac{(0.5)^{2}}{10} = 0.025m = 2.5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.