A small disc is on the top of a smooth hemisphere of radius ' R '. The smallest horizontal velocity ' $V$ '…
- $V=\sqrt{g^2 R}$
- $V=\sqrt{2 g R}$
- $V=\sqrt{g R}$
- $V=\sqrt{g / R}$
Solution

At the top of hemisphere, $\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv}^2}{\mathrm{R}}$
For the disc leaves the hemisphere, $\mathrm{N}=0 \quad \therefore \mathrm{mg}=\frac{\mathrm{mv}^2}{\mathrm{R}} \Rightarrow \mathrm{v}=\sqrt{\mathrm{gR}}$
Asked in: AP EAMCET 2024 (20 May Shift 2)