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(D) Let $R$ be the radius of the semi-spherical vessel. When the ball is at point $Q$ at an angle $\alpha$ from the horizontal,its speed $v$ is given

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(D) Let $R$ be the radius of the semi-spherical vessel. When the ball is at point $Q$ at an angle $\alpha$ from the horizontal,its speed $v$ is given by conservation of energy: $\frac{1}{2}mv^2 = mg(R \sin \alpha)$,which gives $v^2 = 2gR \sin \alpha$.The centripetal force $F_c$ is $\frac{mv^2}{R} = 2mg \sin \alpha$.The forces acting on the ball in the radial direction are the normal reaction $N$ and the component of gravity $mg \sin \alpha$. The equation of motion is $N - mg \sin \alpha = \frac{mv^2}{R} = 2mg \sin \alpha$.Therefore,$N = 3mg \sin \alpha$.The ratio $A$ is defined as the ratio of centripetal force to normal reaction: $A = \frac{F_c}{N} = \frac{2mg \sin \alpha}{3mg \sin \alpha} = \frac{2}{3}$.Since $A = \frac{2}{3}$ is a constant value independent of $\alpha$,the graph of $A$ versus $\alpha$ is a horizontal straight line.

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