A solid sphere is released from rest at the top of an incline of angle $\theta$. If it rolls without slipping to the bottom, its linear acceleration down the incline is:

**Setup.** For pure rolling, $a = \dfrac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)}$. **Substitute.** For a solid sphere $I_{\text{cm}} = \dfrac{2}{5}MR^2$, so $\dfrac{I_{\text{cm}}}{MR^2} = \dfrac{2}{5}$. $a = \dfrac{g\sin\theta}{1 + 2/5} = \dfrac{g\sin\theta}{7/5} = \dfrac{5}{7}g\sin\theta$. **Why option A is tempting.** $g\sin\theta$ is the answer for a *sliding* (frictionless) object. Pure rolling diverts some gravitational PE into rotational KE, so the linear acceleration must be smaller — option A overstates it.