An aircraft executes a horizontal loop with a speed of 150 m - s with it-s wings banked at an angle of 12- - The radius of the loop is g -10 m - s 2- tan 12-0-21A- 10-7 kmB- 9-6 kmC- 7-4 kmD- 5-8 km

The correct option is A 10.7 kmUsing the relation for the radius (r) of loop tanθ = v^2rg or tan 12^∘ = (150)^2r × 10 or r = 22500.21 = 10.7 × 10^3 m = 10.7 k ...

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Standard XII

Physics

Banking of Roads

An aircraft e...

Question

An aircraft executes a horizontal loop with a speed of $150 m/s$ with it's wings banked at an angle of $12^{\circ}$ . The radius of the loop is $(g = 10 {m/s}^{2}, tan 12^{\circ} = 0.21)$

10.7 km

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9.6 km

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7.4 km

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5.8 km

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Solution

The correct option is A $10.7 km$
Using the relation for the radius $(r)$ of loop
$tan θ = \frac{v^{2}}{r g}$
$or tan 12^{\circ} = \frac{(150)^{2}}{r \times 10}$
or $r = \frac{2250}{0.21} = 10.7 \times 10^{3} m = 10.7 km .$

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An aircraft executes a horizontal loop with a speed of 150 m/s with it's wings banked at an angle of 12∘. The radius of the loop is (g=10 m/s2,tan12∘=0.21)

The correct option is A 10.7 kmUsing the relation for the radius (r) of loop tanθ=v2rg or tan12∘=(150)2r×10 or r=22500.21=10.7×103 m=10.7 km.

An aircraft executes a horizontal loop with a speed of $150 m/s$ with it's wings banked at an angle of $12^{\circ}$ . The radius of the loop is $(g = 10 {m/s}^{2}, tan 12^{\circ} = 0.21)$

The correct option is A $10.7 km$
Using the relation for the radius $(r)$ of loop
$tan θ = \frac{v^{2}}{r g}$
$or tan 12^{\circ} = \frac{(150)^{2}}{r \times 10}$
or $r = \frac{2250}{0.21} = 10.7 \times 10^{3} m = 10.7 km .$