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?

Updated On: Apr 29, 2024
  • $ 0.25\,cm $
  • $ 2\,cm $
  • $ 4\,cm $
  • $ 2.5\,cm $
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The Correct Option is D

Solution and Explanation

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
$h =\frac{v^{2}}{g}$
$\therefore h=\frac{(0.5)^{2}}{10}=0.025 \,m$
$\therefore h=2.5\, cm$
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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration