MHTCET 20 Apr 2026 Shift 2 Paper

Concept:The problem uses related rates and the Pythagoras theorem to connect horizontal and slant distances.Explanation:Let the observer be at point $O$ on the ground.The aeroplane flies horizontally at a constant height of $1$ km.At any instant, let its horizontal distance from $O$ be $x$ km and its slant distance be $s$ km.By Pythagoras theorem:$$s^{2}=x^{2}+1^{2}$$The horizontal speed is:$$\frac{dx}{dt}=-600\ \mathrm{km/hr}$$The negative sign shows that $x$ is decreasing as the plane approaches the observer.Given that the plane is $1250$ m away from the observer:$$1250\ \mathrm{m}=1.25\ \mathrm{km}$$So, $s=1.25$ km.Now find $x$:$$(1.25)^{2}=x^{2}+1$$$$1.5625=x^{2}+1$$$$x^{2}=0.5625$$$$x=0.75\ \mathrm{km}$$Differentiate $s^{2}=x^{2}+1$ with respect to time $t$:$$2s\frac{ds}{dt}=2x\frac{dx}{dt}$$$$\frac{ds}{dt}=\frac{x}{s}\cdot\frac{dx}{dt}$$Substitute the values:$$\frac{ds}{dt}=\frac{0.75}{1.25}\times(-600)$$$$\frac{ds}{dt}=\frac{3}{5}\times(-600)$$$$\frac{ds}{dt}=-360\ \mathrm{km/hr}$$The negative sign indicates that the distance is decreasing.Therefore, the rate at which the plane is approaching the observer is $360$ km/hr.Answer:Option A: $360\ \mathrm{km/hr}$