A 20 mm diameter Aluminium rod is turned to 19 mm diameter in a single pass. If the axial speed of the tool is 200 mm/min and the spindle rotate at 600 rpm, the material removal rate is nearly

Correct option is A

​$$\begin{aligned}&\text{In turning, the \textbf{Material Removal Rate (MRR)} is given by the formula:} \\&\text{MRR} = \pi \cdot d_{\text{avg}} \cdot d_c \cdot f \\[1em]&\text{Where:} \\&d_{\text{avg}} = \frac{d_{\text{initial}} + d_{\text{final}}}{2} = \frac{20 + 19}{2} = 19.5\, \text{mm} \\&d_c = \text{depth of cut} = \frac{20 - 19}{2} = 0.5\, \text{mm} \\&f = \text{feed rate (mm/min)} = 200\, \text{mm/min} \\[1em]&\textbf{Step-by-step calculation:} \\&\text{MRR} = \pi \cdot 19.5 \cdot 0.5 \cdot 200 \\&\quad = 3.1416 \cdot 19.5 \cdot 0.5 \cdot 200 \\&\quad = 3.1416 \cdot 9.75 \cdot 200 = 3.1416 \cdot 1950 \approx \boxed{6124.1\, \text{mm}^3/\text{min}}\end{aligned}$$​​