A solid sphere ($I = \tfrac{2}{5}MR^2$) rolls without slipping down a plane inclined at $\theta$. Its **linear acceleration** is:
- $g\sin\theta$
- $\tfrac{2}{7} g\sin\theta$
- $\tfrac{5}{7} g\sin\theta$
- $\tfrac{1}{2} g\sin\theta$
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Correct answer: C. $\tfrac{5}{7} g\sin\theta$
$a = g\sin\theta /(1 + I/MR^2) = g\sin\theta /(1 + 2/5) = (5/7) g\sin\theta$.