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

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$\frac{5g}{14}$

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(B) The linear acceleration $a$ of a body rolling down an inclined plane without slipping 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 of mass 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}$ and $\sin 30^{\circ} = 0.5$,we have:$a = \frac{5}{7} g (0.5) = \frac{5}{7} g \left(\frac{1}{2}ight) = \frac{5g}{14}$.Therefore,the correct option is $B$.

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down an inclined plane from rest without slipping has a linear acceleration (where $g$ is the acceleration due to gravity and $\sin 30^{\circ} = 0.5$).

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