Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A mass \(\pi\) slips along the wall of a semispherical surface of radius \(\mathcal{R}\) . The velocity at the bottom of the surface is
Text Solution
Verified by ExpertsThe correct answer is:
B
By applying law of conservation of energy
\(mgR - \frac{1}{2} mv^{2} = v - \sqrt{2Rg}\)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A stone falls from a balloon that is descending at a uniform rate of \(12 \quad 111 \quad 15\) . Th…
A body A is projected upwards with a velocity of \(28 \; m \; i \; s\) . The second body B is proje…
A ball is dropped on the floor from a height of 10 m . It rebounds to a height of 2.5 m . If the ba…
Two bodies of different masses mc and m_n are dropped from two different heights \hat{r} and \(\the…
A body falls freely from rest. It covers as much distance in the last second of its motion as cover…
A stone is dropped into water from a bridge 441 \(\pi\) above the water. Another stone is thrown ve…