• $summary_{type}$: concept
• $word_{count}$: 170

The vernier caliper measures to 0.01 cm (0.1 mm). Principle: n vernier divisions = (n-1) main scale divisions. LC = 1 MSD - 1 VSD = $\frac{MSD}{n}$. Standard: 10 VSD = 9 MSD, LC = 0.1 mm. Reading = MSR + VSR x LC. Three jaws: external (lengths), internal (bore), depth gauge. The screw gauge measures to 0.001 cm (0.01 mm). LC = $\frac{pitch}{N_circular}$. Standard: pitch = 0.5 mm, N = 50, LC = 0.01 mm. Reading = MSR + CSR x LC. Both instruments require zero error correction. Vernier: positive if zero of VS is right of MS zero (subtract correction), negative if left (add correction). Screw gauge: positive if zero of CS is below reference line (subtract), negative if above (add). Correction = +(n x LC) for negative, -(n x LC) for positive. JEE questions almost always include zero error — computing the corrected reading is the most tested skill. Always state the measurement with the correct number of significant figures matching the least count.