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

Finding the linear acceleration of a solid sphere rolling without slipping requires applying Newton’s second law to both translational and rotational motion. The gravitational force component along the incline is $Mg\sin\theta$, while friction $f_s$ acts up the incline.

The net force equation along the incline is:

$Mg\sin\theta - f_s = Ma$

For rotation about the center, the torque from friction gives:

$f_s R = I\alpha$

The no-slip condition relates linear and angular acceleration: $a = R\alpha$. For a solid sphere, $I = \frac{2}{5}MR^2$. Substituting yields:

$f_s R = \left(\frac{2}{5}MR^2\right)\left(\frac{a}{R}\right)$

which simplifies to $f_s = \frac{2}{5}Ma$.

Inserting this into the force equation:

$Mg\sin\theta - \frac{2}{5}Ma = Ma$

$Mg\sin\theta = \frac{7}{5}Ma$

Solving for acceleration:

$a = \frac{5}{7}g\sin\theta$

For $\theta = 30^\circ$, $\sin\theta = \frac{1}{2}$, so:

$a = \frac{5g}{14}$

The correct choice is B.