A body of m kg slides from rest along the curve of vertical circle from point A to B in friction less path. The velocity of the body at B is:
(given, R=14 m, g=10 m / s² and √2=1.4)

Solution

Physics sir · 4 steps

On a frictionless track only the height drop matters, so measure carefully how far A sits above B.

1. Height of A. A is on the circle at 45° above the horizontal radius, so it sits R sin 45° = R/√2 = 14/1.4 = 10 m above the centre.
2. Total drop. B is the lowest point, a full R = 14 m below the centre. The body therefore falls h = 10 + 14 = 24 m from A to B.
3. Energy conservation. The path is frictionless and the normal force does no work, so all the lost potential energy becomes kinetic energy. The body starts from rest.

   1/2mv² = mgh ⇒ v = √2gh = √(2 × 10 × 24) = √480

4. So. v = √480 ≈ 21.9 m s⁻¹, which is option (d).

The idea to keep. For any frictionless curved path, v = √2gh with h the vertical drop. Draw the centre line and add the heights above and below it.

Answer from the NTA final answer key. Solution by PrepWiser.