# Experimental Physics for JEE > Vernier calipers, screw gauge least count, zero error and graph plotting — JEE Physics experimental skills that convert instrument readings into marks. - Canonical URL: https://prepelephant.com/topics/jee/physics/experimental-physics-jee - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Experimental Physics for JEE", PrepElephant, https://prepelephant.com/topics/jee/physics/experimental-physics-jee ## Direct answer A JEE Physics measurement is complete only after the least count and the zero error have been applied. For a vernier caliper, least count = 1 MSD − 1 VSD, which is 0.1 mm on the standard instrument where 10 vernier divisions span 9 main-scale divisions; for a screw gauge, least count = pitch divided by the number of circular-scale divisions, typically 0.01 mm for a 1 mm pitch with 100 divisions. The corrected reading equals the observed reading minus the zero error, sign included. Most experimental questions then reduce to graph sense: rearrange the relation so a straight line results, and pull the physical constant out of the slope rather than from any single reading. ## What you must remember - **Vernier caliper:** LC = 1 MSD − 1 VSD; reading = main-scale division just before the zero mark + (coincident vernier division × LC); outside jaws for thickness, upper jaws for internal diameter, tail strip for depth. - **Screw gauge:** LC = pitch/number of circular divisions; reading = pitch-scale reading + (coincident circular division × LC); with the stud closed, whatever the instrument shows is the zero error, and you subtract it with its sign. - **Standard least counts to memorise:** vernier caliper 0.1 mm (0.02 mm on a 50-division vernier), screw gauge 0.01 mm, metre scale 1 mm, stopwatch 0.1 s, ordinary thermometer 1 degree Celsius. - **Sign discipline:** if the closed instrument reads +0.04 mm, every measurement loses 0.04 mm; a closed reading of 98 divisions on a 100-division screw gauge means a zero error of −0.02 mm, which adds to observed readings. - **Linearisation pairs worth knowing:** T² versus l for a pendulum (slope 4π²/g), v versus t for uniform acceleration, 1/v versus 1/u for a lens, V versus I for Ohm's law. - **Graph hygiene:** choose scales so the plot covers more than half the graph paper, draw the best-fit straight line (never join dots), and compute the slope from two widely separated points on the line, not from the data table. - **Practical habits examiners reward:** rotate a screw gauge in one direction only to avoid backlash error; measure a wire's diameter along two perpendicular directions and average; take several timing trials and average. ## How a reading comes together Take a screw gauge with a 1 mm pitch, 100 circular divisions, and a closed-jaw reading of +0.04 mm. A wire held between the stud and spindle shows the pitch scale past the 0.5 mm mark, with the 27th circular division on the reference line. The observed diameter is 0.5 mm + 27 × 0.01 mm = 0.77 mm, and the corrected diameter is 0.77 − 0.04 = 0.73 mm. Rotate the wire through 90 degrees and repeat, because drawn wire is rarely perfectly round; the accepted value is the mean of the two. Now the graph half of the skill. For a simple pendulum you could compute g from one reading using T = 2π√(l/g), but T versus l gives a parabola and hides systematic error. Squaring gives T² = (4π²/g) l, a straight line through the origin; plotting measured T² against l and taking the slope m gives g = 4π²/m. Random timing scatter averages into the best-fit line, and a non-zero intercept exposes a fixed error such as a wrongly located suspension point. This is exactly the reasoning JEE Advanced rewards when it asks which plot yields a given slope. ## How the exam frames it JEE Main keeps matters computational: find the least count of a vernier whose 20 divisions coincide with 19 mm, or correct a reading for a stated zero error, and the options differ by exactly the kind of sign slip the paper is fishing for. JEE Advanced leans toward design questions — identify which graph linearises a relation, decide what the slope and intercept mean, or explain why the diameter is measured at several places along the wire. The classic trap is the zero-error direction: students add when they should subtract because they treat the closed-jaw reading as a correction instead of an error. Anchor the habit: corrected equals observed minus zero error, and the answer must make physical sense — a corrected zero must give zero. ## Frequently asked questions ### What is the least count of a standard vernier caliper? 0.1 mm, because ten vernier scale divisions coincide with nine main-scale divisions, so 1 VSD = 0.9 mm and LC = 1 − 0.9 = 0.1 mm; 50-division verniers achieve 0.02 mm. ### How is zero error applied to a measurement? The corrected reading equals the observed reading minus the zero error with its sign, so a gauge that already reads +0.04 mm with closed jaws has that amount removed from every subsequent reading. ### Why plot T squared against l instead of T against l? Because T = 2π√(l/g) gives a curve, whereas T² = (4π²/g) l is a straight line through the origin whose slope 4π²/g delivers g cleanly and averages out random error. ### Which instrument suits a wire's diameter versus a tube's internal diameter? A screw gauge (0.01 mm least count) for the wire, and the internal jaws of a vernier caliper for the tube, since a screw gauge cannot reach inside a bore. ### Why must a screw gauge be rotated in one direction only? Reversing the rotation lets play in the worn screw thread appear as a false reading — the backlash error — so all final adjustments are made turning the same way.