Let's break down the question and solve it step by step: **Given:** - Initial diameter (\(D\)) = **6 cm** - Final diameter (\(d\)) = **5.4 cm** - Length (\(L\)) = **25 cm** - Spindle speed (\(N\)) = **900 rpm** - Axial speed (feed, \(f\)) = **30 cm/min** ### Step 1: **Calculate Depth of Cut** Depth of cut per pass, \( t \) : \[ t = \frac{D - d}{2} = \frac{6 - 5.4}{2} = 0.3 \text{ cm} \] ### Step 2: **Material Removal Rate (MRR) Formula for Turning** \[ MRR = \pi \times D_{avg} \times f \times d \] But for a single pass: \[ MRR = \pi \times D_{avg} \times \text{feed} \times \text{depth of cut} \] Where, - \(D_{avg}\) = average diameter during cut = \(\frac{D + d}{2}\) So, \[ D_{avg} = \frac{6 + 5.4}{2} = 5.7 \text{ cm} \] - Feed per minute = **30 cm/min** - Depth of cut = **0.3 cm** ### Step 3: **Calculate MRR** \[ MRR = \pi \times 5.7 \times 30 \times 0.3 \] \[ MRR = \pi \times 5.7 \times 9 \] \[ MRR = \pi \times 51.3 \] \[ MRR \approx 3.14 \times 51.3 \approx 161 \text{ cm}^3/\text{min} \] ### Step 4: **Match with Options** So, the correct answer is: **161 cm³/min** --- **Final Answer:** **161 cm³/min**