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This question was previously asked in

AAI JE (Technical) Official Paper 2020

- 16.755
- 167.55
- 1675.5
- 16755

Option 3 : 1675.5

Free

ST 2: Strength of materials

2026

15 Questions
15 Marks
15 Mins

** Concept**:

Material removal rate, MRR = V × f × d

Where V = cutting speed (mm/s), f = feed (mm/rev), d = depth of cut (mm)

Cutting speed V = \(\frac{{\pi DN}}{{60}}\)

Where D = diameter of shaft (mm), N = RPM

__Calculation:__

__Given:__

D = 100 mm, f = 0.50 mm/rev, d = 2 mm, N = 320 RPM

∴ V = \(\frac{{\pi DN}}{{60}}\)= \(\frac{{\pi \ \times\ 100\ \times \ 320}}{{60}}\)

= 1675.516 mm/s

MRR = V × f × d = 1675.516 × 0.50 × 2

= **1675.516 mm ^{3}/s**

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