JEE Main 2025 — 24 January, Morning Shift — Question 68

Given least count of Screw Gauge $=0.01 \mathrm{~mm}$

L.C $=\frac{(\text { pitch })}{\text { No. of circular turn }}=\frac{\mathrm{P}}{\mathrm{N}}=0.01 \mathrm{~mm}$

New pitch $=\frac{\mathrm{P}(1+0.75)}{\mathrm{N}(1-0.5)}=\frac{\mathrm{P}}{\mathrm{N}}\left[\frac{1.75}{0.5}\right]$

$=(0.01) 3.5$

$=0.035 \mathrm{~mm}$

$=35 \times 10^{-3} \mathrm{~mm}$

$\therefore$ Ans. is 35