Subtopic: Mechanics and energy requirement in size reduction of agriculture produce
Question Statement:
A ball mill of 200 cm diameter grinds solid materials while operating with 10 cm size balls. If the same ball mill is used for wet grinding, charged with 20 cm diameter balls, the change in the operating speed in rpm is ______. [round off to 2 decimal places] (Take π = 3.14 and g = 9.81 m·s⁻²)
Official Answer Key:
3.60 to 4.20
Step-by-Step Solution & Explanation
Critical speed of a ball mill: N_c = (1/2π)√[g/(R−r)] rev/s, where R = mill radius, r = ball radius. Mill radius R = 1.00 m. Case 1 (10 cm balls, r₁=0.05 m): R−r₁ = 0.95 m; N_c1 = 9.549×√(9.81/0.95) = 9.549×3.213 = 30.68 rpm. Case 2 (20 cm balls, r₂=0.10 m): R−r₂ = 0.90 m; N_c2 = 9.549×√(9.81/0.90) = 9.549×3.302 = 31.53 rpm. Change = N_c2 − N_c1 = ≈0.85 rpm by this standard critical-speed relation — noticeably smaller than the officially accepted range of 3.60–4.20 rpm. This gap could not be reconciled from the given data alone (a matching value would require an additional assumption, such as operating speed being a specified fraction of critical speed that differs between dry and wet grinding, which is not stated in the question); the official key value is adopted for scoring purposes.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.