Question:

An object is sliding from the top of the smooth inclined plane of height \( h \) from rest and it just completes a vertical circle of diameter 20 cm. Then the minimum height \( h \) of smooth inclined plane is:

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For any object to complete a vertical circle, the required release height on a smooth incline is always \( 2.5 \) times the radius of the circle.
Updated On: Jun 9, 2026
  • \( 0.25 \) m
  • \( 0.2 \) m
  • \( 0.5 \) m
  • \( 2.5 \) m
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The Correct Option is A

Solution and Explanation

Concept: For an object sliding down a smooth inclined plane and entering a loop-the-loop to complete a vertical circle, the conservation of energy dictates the minimum height. To barely complete the circle, the velocity at the top of the loop must be \( v_{top} = \sqrt{gr} \).

Step 1: State the energy conservation condition.
Using conservation of energy from the release height \(h\) to the top of the circle \(2r\): $$ mgh = mg(2r) + \frac{1}{2}m v_{top}^2 $$ $$ mgh = 2mgr + \frac{1}{2}m(gr) = 2.5mgr $$ $$ h = 2.5r = \frac{5}{2}r $$

Step 2: Convert diameter to radius.
Diameter \( D = 20 \) cm, so radius \( r = 10 \) cm \(= 0.1\) m.

Step 3: Calculate height.
$$ h = 2.5 \times 0.1 \text{ m} = 0.25 \text{ m} $$ $$\boxed{0.25 \text{ m}}$$
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