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: Energy-free approach:
Take the net force on the particle: the normal reaction's horizontal part provides the centripetal force and its vertical part balances weight.

Step 2: Geometry:
Normal reaction is at angle $\theta$ to the vertical, where $\tan\theta = h/r$ is the slope of the cone wall. Then $\dfrac{mv^2/r}{mg} = \tan\theta = \dfrac hr$.

Step 3: Evaluate:
$v^2 = gh$, so $h = 0.25/10 = 0.025$ m $= 2.5$ cm, option (B).

Final Answer:
The height is v squared over g, equal to 2.5 cm. \[ \boxed{\text{(B) }2.5\ \text{cm}} \]