Correct option is (b) 2g sinθ
Let the angular acceleration be a
torque equation
mgr = I x α
Where I = moment of inertia
\(mg \,sin\,θ \,r = \frac{mr^2}{2}\times \alpha\)
\(\alpha=\frac{2g\,sin\,\theta}{\pi}\)
for pure rolling a = α r
a \(=\frac{2g\,sin\,\theta}{r}.r\)
a = 2g sinθ