$[5]$

$7.$ Derive the expression for the smallest pinion tooth count that will run with a rack instead of a gear without any interference. Find its value for $=20.$

$[10]$

For interference free operation between pinion and gear following relationships are valid for smallest number of teeth on pinion and largest number of teeth on gear:

$N_P=\frac{2k}{(1+2m)si{n}^2}(m+\sqrt[\phantom{2}]{m{m}^2+(1+2m)si{n}^2}),N_G=\frac{N_p^{2}si{n}^2-4k^2}{4k-2N_{p}si{n}^2}$

$m=\frac{N_G}{N_P}$;$N_P$ and $N_G$ are number of teeth on pinion and gear, respectively and $k=1$ for full depth teeth.

$2$