Question

A body slides down a smooth inclined plane of inclination \(θ\) and reaches the bottom with velocity 'V'. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of the plane is

• A. \(\frac{V}{\sqrt{2}}\)
• B. \(\frac{V}{2}\)
• C. \(V\)
• D. \(2V\)

Hint:

Compare energy conservation for sliding with energy conservation for rolling, where rotation takes half the energy for a ring.

Answer 1

The Correct Option is A

Step 1: Use the rolling formula.
For rolling without slipping: $v^2 = \dfrac{2gh}{1 + k^2/R^2}$, where $k$ is the radius of gyration.

Step 2: For a ring.
$k = R$, so $v^2 = \dfrac{2gh}{2} = gh$.

Step 3: Compare with sliding.
Sliding gives $V^2 = 2gh$. So $v^2 = V^2/2$ and $v = V/\sqrt{2}$.

Final Answer:
Option (A). \[ \boxed{\frac{V}{\sqrt{2}}} \]