MAT Exam 2023 Paper

Concept:The plane flies at constant height, so the change in angle of elevation gives the horizontal distance covered by the plane.Explanation:Let the height of the plane be $h = 3000$ m.When the angle of elevation is $45^\circ$, the horizontal distance from the point on the ground is:$$d_1 = \frac{3000}{\tan 45^\circ} = \frac{3000}{1} = 3000 \text{ m}$$When the angle becomes $30^\circ$, the horizontal distance is:$$d_2 = \frac{3000}{\tan 30^\circ} = \frac{3000}{1/\sqrt{3}} = 3000\sqrt{3} \text{ m}$$So, the distance travelled by the plane in $15$ seconds is:$$d = d_2 - d_1 = 3000\sqrt{3} - 3000 = 3000(\sqrt{3}-1) \text{ m}$$Using $\sqrt{3} \approx 1.732$:$$d = 3000(1.732-1) = 3000 \times 0.732 = 2196 \text{ m}$$Time taken $= 15$ seconds, so speed is:$$v = \frac{2196}{15} = 146.4 \text{ m/s}$$Convert to km/hr using $1 \text{ m/s} = 3.6 \text{ km/hr}$:$$v = 146.4 \times 3.6 = 527.04 \text{ km/hr} \approx 527.0 \text{ km/hr}$$Answer:The speed of the plane is approximately $527.0$ km/hr, which is Option D.