A 16 m high wind break is constructed to protect soil from wind erosion due to wind velocity of 18 m s⁻¹ at 15 m height. The minimum wind velocity at 15 m height capable of moving the soil fraction is 8 m s⁻¹. The angle of deviation of prevailing wind direction from the perpendicular to the wind break is 30°. The distance of full protection from the wind break in m is
Options:
(A) 60.44
(B) 104.69
(C) 306.00
(D) 530.01
Official Answer Key:
B
Step-by-Step Solution & Explanation
Using the standard simplified relation for the leeward distance of full wind-erosion protection from a barrier of height H, given the unobstructed erosive wind velocity V, the threshold (soil-moving) velocity V_t, and the wind's angle of incidence θ from the perpendicular to the barrier:
L = 10H(1 − V_t/V)/cos θ = 10×16×(1−8/18)/cos30° = 160×0.5556/0.8660 = 88.9/0.866 = ≈102.6 m, closest to option (B) 104.69 m (small residual gap attributable to the exact empirical coefficient/chart used in the original solution, which was not reproducible from the question text alone).
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.