JEE Main 2024 — 8 April, Shift 2 — Question 45

Least count of vernier calipers $=\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$

$\because$ Least count $=1 \mathrm{MSD}-1 \mathrm{VSD}$

let $x$ no. of divisions of main scale coincides with N division of vernier scale, then

$1 \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}}$

$\therefore \frac{1}{20 \mathrm{~N}} \mathrm{~cm}=1 \mathrm{~mm}-\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}}$

$\frac{1}{2 \mathrm{~N}} \mathrm{~mm}=1 \mathrm{~mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm}$ $x=\left(1-\frac{1}{2 \mathrm{~N}}\right) \mathrm{N}$

$\mathrm{x}=\frac{2 \mathrm{~N}-1}{2}$