Question

A particle describes a horizontal circle of radius 'r' in conical funnel with smooth inner surface with a speed of \(0.5\) m/s. The height of the plane of the circle from the vertex of the funnel is (acceleration due to gravity, \(g = 10\,\text{m/s}^2\))

• A. \(1.5\) cm
• B. \(2.5\) cm
• C. \(3.5\) cm
• D. \(0.5\) cm

Hint:

Resolve the normal force: v squared = g h for a smooth cone.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The particle moves in a horizontal circle on the smooth inner surface of a cone. The forces are gravity \(mg\) and the normal reaction \(N\), perpendicular to the surface. Let the surface make angle \(\theta\) with the horizontal.

Step 2: Force balance:
Vertical: \(N\cos\theta = mg\). Horizontal (centripetal): \(N\sin\theta = \dfrac{mv^2}{r}\).
Dividing: \(\tan\theta = \dfrac{v^2}{rg}\).

Step 3: Relate to the height:
The height of the circle above the vertex is \(h = r\tan\theta\). So
\[ v^2 = g r\tan\theta = g h \Rightarrow h = \frac{v^2}{g} \]

Step 4: Compute:
\[ h = \frac{(0.5)^2}{10} = 0.025\ \text{m} = 2.5\ \text{cm} \]
Option (B). The other values (1.5, 3.5, 0.5 cm) do not satisfy \(h = v^2/g\).

Final Answer:
For a smooth cone, v squared equals g h, so h = 2.5 cm. \[ \boxed{\text{(B) }2.5\ \text{cm}} \]