Q65 / 75 · JEE Main 2025 (24 Jan Shift 1) · Physics

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

  • A
    $\frac{1}{\sqrt{2}} \mathrm{~g}$
  • B
    $\frac{1}{3 \sqrt{2}} \mathrm{~g}$
  • Correct answer
    C
    $\frac{\sqrt{2} \mathrm{~g}}{3}$
  • D
    $\sqrt{2} g$

Solution

\(\begin{aligned} & a=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^2}} \\ & a=\frac{\frac{g}{\sqrt{2}}}{1+\frac{1}{2}}=\frac{2 \frac{g}{\sqrt{2}}}{3}=\frac{\sqrt{2} g}{3}\end{aligned}\)

Hence, the correct answer is option: 'C'.

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