A particle describes a horizontal circle in a conical funne whoses inner surface is smooth with speed of `0.5m//s` . What is the height of the plane of circle from vertex the funnel?

Options

A. 0.25 cm
B. 2 cm
C. 4 cm
D. 2.5 cm

Explanation

To solve the problem of finding the height of the plane of the circle from the vertex of the conical funnel, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Particle:**
- The particle is moving in a horizontal circle inside a conical funnel. The forces acting on the particle are:
- Gravitational force (weight) acting downwards: \( mg \)
- Normal force \( N \) acting perpendicular to the surface of the funnel.

2. **Identifying the Geometry:**
- Let \( h \) be the height of the circular path from the vertex of the funnel, and \( r \) be the radius of the circular path.
- The angle between the radius of the funnel and the vertical line is \( \theta \).

3. **Using Trigonometric Relationships:**
- From the geometry of the funnel, we can relate the radius \( r \) and height \( h \) using the tangent function:
\[ \tan \theta = \frac{r}{h} \]

4. **Setting Up the Equations of Motion:**
- In the vertical direction, the vertical component of the normal force must balance the weight of the particle:
\[ N \cos \theta = mg \quad \text{(1)} \]
- In the horizontal direction, the horizontal component of the normal force provides the necessary centripetal force:
\[ N \sin \theta = \frac{mv^2}{r} \quad \text{(2)} \]

5. **Dividing the Two Equations:**
- Dividing equation (1) by equation (2):
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{\frac{mv^2}{r}}{mg} \]
- This simplifies to:
\[ \tan \theta = \frac{v^2}{g r} \]

6. **Substituting the Value of \( \tan \theta \):**
- From the earlier relationship, we have \( \tan \theta = \frac{r}{h} \). Therefore:
\[ \frac{r}{h} = \frac{v^2}{g r} \]
- Rearranging gives:
\[ r^2 = \frac{v^2 h}{g} \]

7. **Finding the Height \( h \):**
- Rearranging the equation gives:
\[ h = \frac{g r^2}{v^2} \]

8. **Substituting Known Values:**
- Given \( v = 0.5 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
- We need to find \( r \) in terms of \( h \). From the earlier equations, we can express \( r \) in terms of \( h \):
\[ h = \frac{v^2}{g} = \frac{(0.5)^2}{10} = \frac{0.25}{10} = 0.025 \, \text{m} \]
- Converting to centimeters:
\[ h = 0.025 \times 100 = 2.5 \, \text{cm} \]

### Final Answer:
The height of the plane of the circle from the vertex of the funnel is **2.5 cm**.