An inclined plane makes an angle of 30^{circ} | vedclass

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(D) For a body rolling down an inclined plane without slipping,the linear acceleration $a$ is given by the formula: $a = \frac{g \sin 	heta}{1 + \fra

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$\frac{5g}{14}$

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(D) For a body rolling down an inclined plane without slipping,the linear acceleration $a$ is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a solid sphere,the moment of inertia $I$ about its center is $\frac{2}{5} MR^2$.Substituting this into the formula: $a = \frac{g \sin 	heta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin 	heta}{1 + 2/5} = \frac{g \sin 	heta}{7/5} = \frac{5}{7} g \sin 	heta$.Given $	heta = 30^{\circ}$,we have $\sin 30^{\circ} = 0.5$.Therefore,$a = \frac{5}{7} g 	imes \frac{1}{2} = \frac{5g}{14}$.

An inclined plane makes an angle of $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to:

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