Import Question JSON

$\text{Hint: } a_c = \frac{mv^2}{r}$ $\text{Step 1: Find the velocity of the ball when the angular position is } \alpha.$ $mgh = \frac{1}{2}mv^2$ $\Rightarrow mg(r \sin \alpha) = \frac{1}{2}mv_Q^2 \quad [h = r \sin \alpha]$ $\Rightarrow v_Q^2 = 2gr \sin \alpha \quad \ldots (1)$ $\text{Step 2: Find the force at } Q.$ $\text{The force at } Q \text{ is given by:}$ $F_c = \frac{mv_Q^2}{r}$ $\Rightarrow F_c = \frac{m(2gr \sin \alpha)}{r}$ $\Rightarrow F_c = 2mg \sin \alpha \quad \ldots (2)$ $\text{Step 3: Find the normal force acting on the ball at } Q.$ $\text{The normal reaction at point } Q \text{ is given by:}$ $N - mg \sin \alpha = F_c$ $\Rightarrow N - mg \sin \alpha = 2mg \sin \alpha$ $\Rightarrow N = 3mg \sin \alpha \quad \ldots (3)$ $\text{Step 4: Find the value of } A \text{ and find the correct graph.}$ $\text{The value of } A \text{ is given as the ratio of centripetal force and the}$ $\text{normal reaction on the ball at point } Q \text{ i.e., } A = \frac{F}{N}$ $\Rightarrow A = \frac{2mg \sin \alpha}{3mg \sin \alpha} = \frac{2}{3}$ $\Rightarrow A = \frac{2}{3}$ $\text{Therefore, the value of } A \text{ is constant and independent of the angle } \alpha$ $\text{and the correct graph is shown below;}$ $\begin{array}{c} A \\ \alpha \end{array}$ $\text{Hence, option (3) is the correct answer.}$