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(C) The linear acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{mr^2}}$.For a

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$\frac{\sqrt{2} g}{3}$

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(C) The linear acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{mr^2}}$.For a uniform solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} mr^2$.Substituting this into the acceleration formula: $a = \frac{g \sin 	heta}{1 + \frac{1/2 mr^2}{mr^2}} = \frac{g \sin 	heta}{1 + 1/2} = \frac{g \sin 	heta}{3/2} = \frac{2}{3} g \sin 	heta$.Given the angle of inclination $	heta = 45^{\circ}$,we have $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$.Substituting the value of $\sin 45^{\circ}$ into the expression for $a$: $a = \frac{2}{3} g \left( \frac{1}{\sqrt{2}} ight) = \frac{\sqrt{2} g}{3}$.

$A$ 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 axis will be:

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