A vertical disc plough with 5 discs is operated at a depth of 0.15 m. The disc angle and disc diameter are 40⁰ and 0.6 m, respectively. If overlap between two consecutive discs is 0.12 m at 0.15 m depth of cut, the total width of cut at the specified depth in m is
Options:
(A) 1.19
(B) 1.55
(C) 2.11
(D) 2.36
Official Answer Key:
A
Step-by-Step Solution & Explanation
At a working depth d on a disc of diameter D, only the lower segment of the disc actually engages the soil. The effective (working) chord-diameter of this engaged segment is D_e = 2√[d(D−d)] = 2√[0.15×(0.6−0.15)] = 2√(0.15×0.45) = 2√0.0675 = 0.5196 m.
The horizontal width cut by a single disc = D_e×sin(disc angle) = 0.5196×sin40° = 0.5196×0.6428 = 0.3339 m.
With n=5 discs, each overlapping the previous by 0.12 m: total width = n×(width per disc) − (n−1)×overlap = 5×0.3339 − 4×0.12 = 1.6693 − 0.48 = 1.19 m. Answer: (A). *Distractor note:* (C) 2.11 results from omitting the depth-based effective-diameter correction and using full D sinα instead.
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