Question

A particle describes a horizontal circle on the smooth inner surface of a conical funnel whose vertex angle is ${90^ \circ }$. If the height of the plane of the circle above the vertex is $9.8\,\,cm$, the speed of the particle is:
(A) $\sqrt {9.8} \,m{s^{ - 1}}$
(B) $0.98\,\,m{s^{ - 1}}$
(C) $19.6\,\,m{s^{ - 1}}$
(D) $14.7\,\,m{s^{ - 1}}$

Answer 1

Hint Horizontal surface replicates the motion of a mass on a stiff rod that is moving along a horizontally oriented circular path, this known as the horizontal circular motion. It also explores the relationship between the inward force acting on an object travelling in uniform circular motion.

Useful formula
The formula for the speed of the particle is given as;
$\dfrac{N}{{\sqrt 2 }} = mg$
Where, $N$ denotes the force of the particle, $m$ denotes the mass of the particle, $g$ denotes the gravitational force acting on the particle.

Complete step by step solution
The data given in the problem are as follows;
The height of the plane of the circle is, $r = 9.8\,\,cm$.
Vertex angle of the conical flask is, ${90^ \circ }$.
Acceleration due to gravity on the particle is given as, $g = 9.8\,\,m{s^{ - 2}}$.

The formula for the speed of the particle is given as;
\[
  \dfrac{N}{{\sqrt 2 }} = mg \\
  N = \sqrt 2 mg \\
 \]
We know that;
$\dfrac{N}{{\sqrt 2 }} = \dfrac{{m{V^2}}}{r}$
Since it is found that;
$
  \Rightarrow mg = \dfrac{{m{V^2}}}{r} \\
  \Rightarrow {V^2} = \dfrac{{r\,mg}}{m} \\
  \Rightarrow {V^2} = rg \\
 $
Since we only the speed of the particle, by taking square root on both side we get;
$V = \sqrt {rg} $
Substitute the values of acceleration due to gravity and the height of the plane of the circle above the vertex in the above equation;
Convert the height of the plane of the circle from $cm$ to $m$.
\[
  \Rightarrow V = \sqrt {\dfrac{{\left( {9.8} \right) \times \left( {9.8} \right)}}{{100}}} \\
  \Rightarrow V = \sqrt {\dfrac{{96.04}}{{100}}} \\
  \Rightarrow V = \sqrt {0.9604} \\
 \]
By taking square root for the given value we get;
$V = 0.98\,\,m{s^{ - 1}}$

Therefore, the speed of the particle is calculated as $V = 0.98\,\,m{s^{ - 1}}$.

Hence the option (B) $V = 0.98\,\,m{s^{ - 1}}$ is the correct answer.

Note: The velocity of the particle is the speed of a particle in a medium as it transfers a wave. The SI unit of particle velocity is the metre per second $\left( {m{s^{ - 1}}} \right)$. The wave travels moderately fast, while the particles vibrate around their original position with a moderately small particle velocity.