Q74 / 75 · JEE Main 2025 (24 Jan Shift 1) · Physics
The least count of a screw guage is 0.01 mm . If the pitch is increased by $75 \%$ and number of divisions on the circular scale is reduced by $50 \%$, the new least count will be _____ $\times 10^{-3} \mathrm{~mm}$
Correct answer: 35
Solution
$\begin{aligned} & \text { Given least count of Screw Gauge }=0.01 \mathrm{~mm} \\ & \text { L.C }=\frac{(\text { pitch })}{\text { No. of circular turn }}=\frac{\mathrm{P}}{\mathrm{N}}=0.01 \mathrm{~mm} \\ & \text { New pitch }=\frac{\mathrm{P}(1+0.75)}{\mathrm{N}(1-0.5)}=\frac{\mathrm{P}}{\mathrm{N}}\left[\frac{1.75}{0.5}\right] \\ & =(0.01) 3.5 \\ & =0.035 \mathrm{~mm} \\ & =35 \times 10^{-3} \mathrm{~mm}\end{aligned}$
Hence, the correct answer is 35.