A uniform solid cylinder of mass ' $m$ ' and radius ' $r$ ' rolls along an inclined rough plane of…
- $\frac{1}{\sqrt{2}} \mathrm{~g}$
- $\frac{1}{3 \sqrt{2}} \mathrm{~g}$
- $\frac{\sqrt{2} \mathrm{~g}}{3}$
- $\sqrt{2} g$
Solution

$\Rightarrow \quad a=\frac{2 g}{3} \sin \theta=\frac{2 g}{3 \sqrt{2}}=\frac{\sqrt{2}}{3} g$
Asked in: JEE Main 2025 (24 Jan Shift 1)