The correct answer is (d) 11.78 mm
The explanation: Explanation: \(\frac{Nflywheel}{Nfeed \, roller}=\frac{Nworm}{Nworm \, gear}=\frac{Tworm \, gear}{Tflywheel} \)
\(\frac{60}{Nr}=\frac{20}{2} \)
Nr = \(\frac{60*2}{20}\) = 6 rev/min
L*n*Nf = π*0.15*6
L = \(\frac{π*0.15*6}{4*60}\) = 11.78 mm.