JEE Mains Physics · Units and Measurements

Vernier Callipers, Screw Gauge and Zero Error

A vernier's least count is one main-scale division minus one vernier division, a screw gauge's is its pitch divided by the circular divisions, and a true reading is the observed reading minus the zero error, taken with its sign.

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Why this matters

Thirty-seven PYQs, thirty of them multiple choice, and seven from 2026, the largest page in the chapter. Sixteen find the least count of a vernier callipers, a travelling microscope or a screw gauge; seventeen correct a reading for zero error; four turn a reading into a density, an area or a refractive index. The sign of the zero error decides most of the marks.

Concept 1 of 3: Least count of a vernier and a screw gauge

A vernier scale has divisions slightly shorter than the main scale's. Lined up from zero, the first vernier mark falls short of the first main mark by one least count, the second by two, and so on; the mark that coincides tells you how many least counts to add. A screw gauge instead turns rotation into distance: one full turn moves the spindle by the pitch, so one circular division moves it by pitch ÷ divisions.

Definition

Vernier: least count (vernier constant) $= 1\ \text{MSD} - 1\ \text{VSD}$ .
If $N$ VSD $= (N - 1)$ MSD, then LC $= \dfrac{\text{MSD}}{N}$ . In general, if $N$ VSD $= m$ MSD, then LC $= \left(1 - \dfrac{m}{N}\right)$ MSD.
A travelling microscope carries a vernier; find its MSD from 'divisions per cm' first.
Screw gauge: pitch = distance moved per rotation; LC $= \dfrac{\text{pitch}}{\text{number of circular divisions}}$ .
A spherometer is a screw gauge on three legs; it measures the radius of curvature of a curved surface and small thicknesses.
From a set of recorded readings, the least count is one unit in the last recorded decimal place.

Least count

\text{vernier: } \text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = \left(1 - \frac{m}{N}\right)\text{MSD} \qquad \text{screw gauge: } \text{LC} = \frac{\text{pitch}}{\text{circular divisions}}

Worked example

(a) A vernier has 25 divisions equal to 24 main-scale divisions, and one main-scale division is 0.5 mm. Find its least count. (b) A screw gauge has pitch 0.5 mm and 50 circular divisions. Find its least count.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q2Moderate

Example 1 · Units and Measurements · Vernier Callipers, Screw Gauge and Zero Error

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division

= 0.05\text{ }mm

, then the least count of the vernier callipers is

\_\_\_\_

mm.

The vernier constant is MSD minus VSD, not MSD times divisions

The least count is the DIFFERENCE between one main-scale and one vernier division. It is not one MSD multiplied by the number of vernier divisions, a statement that appears as a false option.

Find the pitch from distance per rotation

If five rotations move the spindle 2.5 mm, the pitch is 0.5 mm, not 2.5 mm. Divide by the number of rotations before dividing by the circular divisions.

Drill 15 more on least count of a vernier and a screw gauge

Concept 2 of 3: Zero error and the corrected reading

If the instrument does not read zero when its jaws or studs touch, every reading is off by that same amount. A positive zero error makes every reading too large, so it is subtracted; a negative zero error makes every reading too small, so subtracting it adds its size. One rule covers both: true reading = observed reading − zero error, with the zero error's sign.

Definition

Observed reading = main-scale reading + (coinciding division) × LC.
Corrected reading = observed reading − zero error (with sign).
Vernier: vernier zero to the RIGHT of the main-scale zero → positive error, $+n \times \text{LC}$ when the n-th vernier division coincides. To the LEFT → negative error.
Screw gauge: circular-scale zero BELOW the reference line → positive error, $+n \times \text{LC}$ . ABOVE the line → negative error, $-n \times \text{LC}$ .
'The reference line coincides with the n-th circular division' means the zero has passed the line: positive, $+n \times \text{LC}$ .
With a positive zero error, readings are larger than the true values.

Corrected reading

\text{true reading} = \text{MSR} + n \times \text{LC} - (\text{zero error})

Worked example

A vernier callipers has least count 0.1 mm. With the jaws closed, the vernier zero lies to the right of the main zero and the 3rd vernier division coincides. Measuring a rod, the main-scale reading is 24 mm and the 7th vernier division coincides. Find the true length.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q7Moderate

Example 2 · Units and Measurements · Vernier Callipers, Screw Gauge and Zero Error

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm . When diameter of a sphere is measured, the reading of main scale is 5 mm and

50^{\text{th~}}

division of circular scale coincides with the reference line of main scale. The diameter of sphere is

\_\_\_\_

mm.

Subtract a positive zero error, never add it

A positive zero error means the instrument already reads more than zero with nothing between the jaws, so every reading is too large. Subtract it. Adding it is the most common wrong option.

Below the line is positive for a screw gauge

When the studs touch and the circular-scale zero sits below the reference line, the screw has gone past zero: the error is positive. Above the line, it has not reached zero: the error is negative.

A vernier zero to the left: the textbook count and the keys' count differ

When the vernier zero is left of the main zero, the textbook zero error is −(N − n) × LC for an N-division vernier whose n-th division coincides: count back from the vernier's last mark. Two JEE keys instead took it as −n × LC, the right-side count with a minus sign. Work the textbook count first; if no option matches it and −n × LC is offered, that is the count the paper used.

Drill 16 more on zero error and the corrected reading

Concept 3 of 3: Taking a reading and using it in a calculation

Once the reading is right, it becomes a length in an ordinary formula: a diameter for a density, a radius for an area, a depth for a refractive index. Convert to one unit before you calculate, and give the result no more significant figures than the reading allows.

Definition

Vernier reading = MSR + n × LC; screw-gauge reading = linear-scale reading + circular-scale reading × LC.
Halve a diameter for a radius; keep cm with grams for g/cm³.
Travelling microscope: read the mark with no slab ( $R_1$ ), the mark through the slab ( $R_2$ ) and the top of the slab ( $R_3$ ); $\mu = \dfrac{R_3 - R_1}{R_3 - R_2}$ , real thickness over apparent thickness.
Round the result to the significant figures of the least precise reading.

Reading

\text{reading} = \text{MSR} + n \times \text{LC}

Worked example

A screw gauge has pitch 1 mm and 100 circular divisions. For a wire the linear-scale reading is 2 mm and the circular-scale reading is 35. Find the wire's cross-sectional area.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q20Moderate

Example 3 · Units and Measurements · Vernier Callipers, Screw Gauge and Zero Error

The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to

1\text{ }mm

. The main scale reading is

2\text{ }cm

and second division of vernier scale coincides with a division on main scale. If mass of the sphere is

8.635\text{ }g

, the density of the sphere is:

Match the units before adding

A main-scale reading in cm and a least count in mm cannot be added directly. 2.1 cm + 5 × 0.1 mm is 2.15 cm, not 2.6 cm.

Drill 3 more on taking a reading and using it in a calculation

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Least count of a vernier and a screw gauge
Least count
$\text{vernier: } \text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = \left(1 - \frac{m}{N}\right)\text{MSD} \qquad \text{screw gauge: } \text{LC} = \frac{\text{pitch}}{\text{circular divisions}}$
Zero error and the corrected reading
Corrected reading
$\text{true reading} = \text{MSR} + n \times \text{LC} - (\text{zero error})$
Taking a reading and using it in a calculation
Reading
$\text{reading} = \text{MSR} + n \times \text{LC}$

Watch out for (6)

The vernier constant is MSD minus VSD, not MSD times divisions
→ Least count of a vernier and a screw gauge
Find the pitch from distance per rotation
→ Least count of a vernier and a screw gauge
Subtract a positive zero error, never add it
→ Zero error and the corrected reading
Below the line is positive for a screw gauge
→ Zero error and the corrected reading
A vernier zero to the left: the textbook count and the keys' count differ
→ Zero error and the corrected reading
Match the units before adding
→ Taking a reading and using it in a calculation

Test yourself on Units and Measurements

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