Question

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. $2.5\,cm$

Answer 1

Answer: (D)

$2.5\,cm$

Answer 2

Equating the vertical forces, we have
$N \sin \theta=m g...$ (i)
and $\frac{m v^{2}}{R}=N \cos \theta ...$ (ii)
$\therefore \tan \theta=\frac{R g}{v^{2}} , . .$ (iii)
Also, in triangle OAB,
$\tan \theta=\frac{R}{h} ...$ (iv)
Equating Eqs. (iii) and (iv), we get
$h =\frac{v^{2}}{g}$
$\therefore h=\frac{(0.5)^{2}}{10}=0.025 \,m$
$\therefore h=2.5\, cm$