UPSC NDA I 2023 General Ability Solved Paper

Concept:Average acceleration is the total change in velocity divided by the total time taken.For half a circle, the velocity vector reverses direction, so the magnitude of change in velocity is $2v$.The time taken is half the period: $T/2 = \pi R / v$.Thus, average acceleration $\bar{a} = 2v^2 / (\pi R)$.Explanation:The particle moves with constant speed $v$ in a circle of radius $R$.Its instantaneous centripetal acceleration is $v^2/R$, directed toward the centre.However, average acceleration over a finite path depends only on the initial and final velocities.After moving half the circle, the velocity direction is exactly opposite to the starting direction.If the initial velocity is $\vec{v}$, the final velocity is $-\vec{v}$.Change in velocity $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = (-\vec{v}) - \vec{v} = -2\vec{v}$.Magnitude $|\Delta v| = 2v$.Time taken to cover half the circle: distance = $\pi R$, speed = $v$, so $\Delta t = \pi R / v$.Average acceleration $\bar{a} = |\Delta v| / \Delta t = 2v / (\pi R / v) = 2v^2 / (\pi R)$.Answer:$\displaystyle \frac{2v^2}{\pi R}$ (Option C).