A particle describes a horizontal circle on the smooth inner surface of conical funnel whose vertex angle is `90^(@)`. If the height of the plane of the circle above the vertex is 9.8cm, the speed of the particle is

Options

A. `sqrt(9.8)m//sec`
B. `0.98m//sec`
C. `19.6 m//sec`
D. `14.7 m//sec`

Answer

To find the speed of a particle moving in a horizontal circle on the smooth inner surface of a conical funnel with a vertex angle of 90 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Height (h) above the vertex: \( h = 9.8 \, \text{cm} = 0.098 \, \text{m} \)
- Vertex angle of the cone: \( 90^\circ \)

2. **Determine the Angle of the Cone:**
- Since the vertex angle is \( 90^\circ \), the angle \( \theta \) at which the particle moves in the circular path is \( 45^\circ \) (because the half-angle of the cone is \( 45^\circ \)).

3. **Use the Relationship Between Forces:**
- The normal force \( N \) acting on the particle can be resolved into two components: one acting vertically (balancing the weight) and the other providing the centripetal force.
- The vertical component of the normal force is given by:
\[ N \sin(45^\circ) = mg \]
- Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we can write:
\[ N \cdot \frac{1}{\sqrt{2}} = mg \quad \Rightarrow \quad N = mg\sqrt{2} \quad \text{(Equation 1)} \]

4. **Centripetal Force:**
- The centripetal force required for circular motion is given by:
\[ mg = \frac{mv^2}{r} \]
- Here, \( r \) is the radius of the circular path. From the geometry of the cone, we can relate the radius \( r \) to the height \( h \):
\[ r = h \tan(45^\circ) = h \quad \text{(since } \tan(45^\circ) = 1\text{)} \]
- Thus, \( r = 0.098 \, \text{m} \).

5. **Substituting for Centripetal Force:**
- From the centripetal force equation:
\[ mg = \frac{mv^2}{r} \quad \Rightarrow \quad g = \frac{v^2}{r} \]
- Rearranging gives:
\[ v^2 = g \cdot r \]

6. **Calculate the Speed:**
- Substituting the known values \( g = 9.8 \, \text{m/s}^2 \) and \( r = 0.098 \, \text{m} \):
\[ v^2 = 9.8 \cdot 0.098 \]
- Calculate \( v^2 \):
\[ v^2 = 0.9604 \quad \Rightarrow \quad v = \sqrt{0.9604} \approx 0.98 \, \text{m/s} \]

### Final Answer:
The speed of the particle is approximately \( 0.98 \, \text{m/s} \).