COMEDK 2025 Evening Solved Paper

Concept:Use trigonometry to relate the angle of elevation to the horizontal distance, then differentiate with respect to time to find the rate of change of the angle.Explanation:Let $x$ be the horizontal distance of the man from the foot of the tower.The effective height of the tower above the man's eye level is $41.6 - 1.6 = 40$ m.So, $\tan \theta = \frac{40}{x}$, where $\theta$ is the angle of elevation.Differentiate both sides with respect to time $t$:$\sec^2 \theta \cdot \frac{d\theta}{dt} = -\frac{40}{x^2} \cdot \frac{dx}{dt}$.We know $\frac{dx}{dt} = 2$ m/s (moving away, so positive).Using $\sec^2 \theta = 1 + \tan^2 \theta = 1 + \left(\frac{40}{x}\right)^2 = \frac{x^2+1600}{x^2}$.Substitute: $\frac{x^2+1600}{x^2} \cdot \frac{d\theta}{dt} = -\frac{40}{x^2} \cdot 2$.Simplify: $\frac{d\theta}{dt} = -\frac{80}{x^2+1600}$.At $x = 30$ m: $\frac{d\theta}{dt} = -\frac{80}{30^2+1600} = -\frac{80}{900+1600} = -\frac{80}{2500} = -\frac{4}{125}$ rad/s.The negative sign indicates that the angle is decreasing as the man moves away.Answer:$-\frac{4}{125}$ rad/s, which corresponds to Option A.