A block of mass m, slides down along the surface of a bowl (radius R) from the rim to the…
Physics · JEE Advanced · NTA Exams — Rotational Motion
A block of mass m, slides down along the surface of a bowl (radius R) from the rim to the bottom. The velocity of the block at the bottom will be :
- \(\sqrt{\pi \mathrm{Rg}}\)
- \(2 \sqrt{\pi \mathrm{Rg}}\)
- \(\sqrt{2 R g}\)
- \(\sqrt{\mathrm{gR}}\)
Answer
(C) 2 R g
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