A uniform solid cylinder of mass ' $m$ ' and radius ' $r$ ' rolls along an inclined rough plane of…

A uniform solid cylinder of mass ' $m$ ' and radius ' $r$ ' rolls along an inclined rough plane of inclination $45^{\circ}$. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be
  1. $\frac{1}{\sqrt{2}} \mathrm{~g}$
  2. $\frac{1}{3 \sqrt{2}} \mathrm{~g}$
  3. $\frac{\sqrt{2} \mathrm{~g}}{3}$
  4. $\sqrt{2} g$

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)

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