d
A force \(f\) acts on the sphere towards left, so the particle experience pseudo force \(f =\) ma towards right.

Initial kinetic energy of the particle is zero.

Let the speed of the particle at point \(C\) be \(v\).

From work-energy theorem, \(W _{ g }+ W _{ f }=\Delta K \cdot E = K \cdot E _{ f }\)

where \(W _{ g }\) is work done by gravity and \(W _{ f }\) is work done by pseudo force.

From figure, we get \(AB = R - R \cos \theta= R (1-\cos \theta)\)

Also \(AC = R \sin \theta\)

\(\therefore mg ( AB )+ f ( AC )=\frac{1}{2} mv ^2\)

Or \(m g R(1-\cos \theta)+(m a)(R \sin \theta)=\frac{1}{2}{m v^2}^2\)

Or \(m g R(1+\sin \theta-\cos \theta)=\frac{1}{2}{m v^2}^2 \quad(\because a=g)\)

\(\Rightarrow v =\sqrt{2 gR (1+\sin \theta-\cos \theta)}\)