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An air craft executes a horizontal loop at a speed at 540 km per hour with its wings banked at 17°. The diameter of the loop is $(given g = 10 {ms}^{- 2} & \tan^{- 1} (0 .3) = 17^{0})$

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9 Km

15 Km

12 Km

13 Km

answer is A.

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Detailed Solution

$\tan θ = \frac{v^{2}}{rg} ∴ r = \frac{v^{2}}{gtan θ} = \frac{150 \times 150}{10 x 0 .3} = 7500 m = 7 .5 km Diameter of the loop = 2 r = 15 KM$

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