MCQ
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 ........ $cm$
- A$0.25$
- B$2 $
- C$4$
- ✓$2.5 $
✓
Answer
Correct option: D.
$2.5 $
d
(d) The particle is moving in circular path
(d) The particle is moving in circular path
From the figure, $mg = R\sin \theta $ …$(i)$
$\;\frac{{m{v^2}}}{r} = R\cos \theta $ …$(ii)$
From equation $(i)$ and $(ii)$ we get
$\tan \theta = \frac{{rg}}{{{v^2}}}$ but $\tan \theta = \frac{r}{h}$
$h = \frac{{{v^2}}}{g} = \frac{{{{(0.5)}^2}}}{{10}} = 0.025m = 2.5\,cm$
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