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Question

For having central angle `alpha`, the area of cross-section of sewers running partially full, is


Options:
A .  ` A = D^2/2 [ (pi alpha)/(180^circ) - sin alpha /2]`
B .  ` A = D^2/4 [ (pi alpha)/(180^circ) - sin alpha /2]`
C .  ` A = D^2/4[ (pi alpha)/(180^circ) - cos alpha /2]`
D .  ` A = D^2/2 [ (pi alpha)/(180^circ) - cos alpha /2]`
Answer: Option B



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