Let's break down the problem: **Given:** - Length of rod, \( L = 25 \) cm - Initial diameter, \( D_1 = 6 \) cm - Final diameter, \( D_2 = 5.4 \) cm - Spindle speed, \( N = 900 \) rpm - Axial (feed) speed, \( f = 30 \) cm/min You are asked to **calculate the material removal rate (MRR)**. ### Step 1: Volume of Material Removed per Revolution The volume removed in one revolution is the volume of the cylindrical shell removed: \[ \text{Volume per revolution} = \text{Length} \times \left[ \frac{\pi}{4} (D_1^2 - D_2^2) \right] \] ### Step 2: Volume Removed per Minute (MRR) Since the tool moves at 30 cm/min axially, in 1 minute, the tool covers 30 cm. But the length of the rod is 25 cm, so we use the feed rate per minute. \[ \text{MRR} = \text{Feed rate} \times \left[ \frac{\pi}{4} (D_1^2 - D_2^2) \right] \] Plug in the values: \[ \text{MRR} = 30 \times \frac{\pi}{4} \left[(6)^2 - (5.4)^2\right] \] \[ = 30 \times \frac{\pi}{4} (36 - 29.16) \] \[ = 30 \times \frac{\pi}{4} \times 6.84 \] \[ = 30 \times 1.71 \times 6.84 \] \[ = 30 \times 11.7084 \] \[ = 351.252 \text{ cm}^3/\text{min} \] But let's check the answer choices. The calculation seems too large, so let's try the step-by-step approach: #### Let's recalculate: \[ \frac{\pi}{4} = 0.7854 \] \[ 0.7854 \times 6.84 = 5.372 \] \[ 30 \times 5.372 = 161.16 \] **So, the correct answer is:** \[ \boxed{161\ \text{cm}^3/\text{min}} \] **Option:** 161 cm³/min If you need the explanation for the formula, let me know!