NDA General Ability Question (2026) — Solution

Step-by-step solution

Using the mirror formula: 1 v + 1 u = 1 f Differentiating with respect to time t, we get the velocity of the image v_i in terms of the velocity of the object v_o: - 1 v^2 dv dt - 1 u^2 du dt = 0 v_i = - ( v u )^2 v_o = - f^2 (u-f)^2 v_o Since the object is moving towards the mirror, u is increasing (becoming less negative), so v_o > 0. This makes v_i To find the acceleration of the image a_i, we differentiate v_i with respect to t: a_i = dv_i dt = - f^2 (u-f)^2 a_o + 2f^2 (u-f)^3 v_o^2 Here, a_o > 0 because the object is accelerating towards the mirror. Since the object is placed in front of the concave mirror, u and f are negative, and |u| > |f|, which means (u-f) Therefore, both terms in the expression for a_i are negative, yielding a_i Since both the velocity v_i and the acceleration a_i of the image are negative, the speed of the image is increasing in the direction away from the mirror. Hence, the image accelerates towards the centre of curvature. Answer: The image will accelerate towards the centre of curvature.