A thin uniform rod of mass ' m ' and length ' L ' is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above pivot and then released. The angular acceleration of the rod when it makes an angle ' ' with the vertical is [consider negligible friction at the pivot] ( g = acceleration due to gravity)

Step-by-step solution

For a uniform rod of length L and mass m pivoted at one end, the net torque about the pivot is due solely to gravity. The gravitational force mg acts at the center of mass, located at a distance L/2 from the pivot. When the rod makes an angle with the vertical, the lever arm is (L/2) , yielding a torque = mg(L/2) . The moment of inertia about the pivot is I = (1/3)mL^2 . Applying Newton's second law for rotation, = I , gives: mg(L/2) = (1/3)mL^2 Solving for : = mg(L/2) (1/3)mL^2 = 3g 2L This corresponds to option A