The diagram illustrates a right circular cone-shaped hypothetical mountain, of apex ‘O’. If you build a shortest distance track for a sightseeing train around the mountain, in which the track starts at point A and ends at point B, the track will first go uphill, but then it will go downhill. AB=10km, radius of cone=20 km. Slant height of the cone=60km. The speed of train (consider it a particle) is set at a constant 20 kmph. Mark the CORRECT statements(s).

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The velocity of the train is always perpendicular to the line joining ‘O’ to the train

The journey time of the train is $\frac{\sqrt{91}}{2} hr$

The length of the downhill part of the journey is $\frac{200}{\sqrt{91}} km$

At the instant of shortest distance from ‘O’ the angular velocity of train about ‘O’ is maximum

answer is A, B.

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Detailed Solution

By the Pythagorean Theorem:

$\frac{\begin{array}{l} {(10 \sqrt{91} - x)}^{2} + + h^{2} = 60^{2} \\ x^{2} + h^{2} = 50^{2} \end{array}}{\begin{array}{l} 9100 - 2 (10 \sqrt{91}) x = 60^{2} - 50^{2} \\ (20 \sqrt{91}) x = 8000 \\ x = \frac{400}{\sqrt{91}} \end{array}}$

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