Screw Gauge: Pitch, Least Count and Reading

Physics · Units And Measurements · NEET

A screw gauge measures very small lengths (like wire diameter). Its Pitch = distance the screw moves in one full rotation of the circular scale, and its Least Count (LC) = Pitch / (number of circular scale divisions). The final Reading = Main Scale Reading (MSR) + (Circular Scale Reading x LC). Memory hook: "Pitch is one full spin, LC is that spin cut into 50 (or 100) tiny slices."
Screw Gauge: Main Scale (mm) + Circular Scale01234Main scale (mm) MSR = 5 mmref line0CSR = 25Circular scale: 50 divisions, LC = Pitch/50 = 0.5/50 = 0.01 mmReading = MSR + CSR x LC
A screw gauge reading = Main Scale Reading (straight mm scale) plus Circular Scale Reading times Least Count. Here MSR = 5 mm and CSR = 25 divisions with LC = 0.01 mm, giving 5 + 0.25 = 5.25 mm before any zero-error correction.

Your doubts, answered

Are pitch and least count the same thing?

No. Pitch is the bigger number: it is how far the screw moves forward in ONE full turn (usually 0.5 mm or 1 mm). Least count is much smaller: it is the pitch divided by the number of divisions on the circular scale. Example: pitch = 0.5 mm and 50 divisions gives LC = 0.5/50 = 0.01 mm. So LC is always smaller than pitch.

How do I find pitch if only least count is given?

Turn the formula around. LC = Pitch / N, so Pitch = LC x N, where N is the number of circular scale divisions. If LC = 0.01 mm and N = 50, then Pitch = 0.01 x 50 = 0.5 mm. NEET 2020 asked exactly this, so learn both directions of the formula.

What is MSR and CSR when reading a screw gauge?

MSR (Main Scale Reading) is the value on the straight sleeve scale you can see just before the thimble edge, read in mm. CSR (Circular Scale Reading) is the division number on the round thimble that lines up with the main scale reference line. Final reading = MSR + (CSR x LC).

Why does a screw gauge measure smaller lengths than a vernier caliper?

A vernier caliper has LC = 0.01 cm = 0.1 mm. A screw gauge has LC = 0.001 cm = 0.01 mm, which is 10 times finer. This is because the screw turns many times over a small distance, magnifying tiny movements onto a large circular scale. So use a screw gauge for wire or sheet thickness, and a vernier for slightly bigger objects.

Do I add the zero error inside this reading?

Not in the basic reading. First get MSR + CSR x LC. Then correct for zero error separately: corrected reading = observed reading minus zero error (keep the sign). Zero error is a whole separate topic covered next; for pitch and least count, focus only on MSR + CSR x LC.

⚠️ The NEET trap
Pitch = Least Count x number of divisions, so students multiply when the question actually wants them to divide, or they mix up mm and cm and answer 0.52 cm instead of 0.052 cm.
LC = Pitch / N (divide to get LC). Pitch = LC x N (multiply to get pitch). And always convert carefully: 0.52 mm = 0.052 cm, not 0.52 cm.
🧠 Read the question direction: 'find LC' means divide, 'find pitch' means multiply. Then check your unit: mm to cm means divide by 10.

Real NEET questions

2020

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

A · 0.5 mm
B · 1.0 mm
C · 0.01 mm
D · 0.25 mm
Solution: Pitch = Least Count x number of circular divisions. Step 1: LC = 0.01 mm, N = 50. Step 2: Pitch = LC x N = 0.01 x 50 = 0.5 mm. Answer: 0.5 mm (option A). Trap: do not divide here; the question gives LC and asks for pitch, so multiply.
2021

A screw gauge gives the following readings when measuring the diameter of a wire: Main scale reading = 0 mm; Circular scale reading = 52 divisions. 1 mm on the main scale corresponds to 100 divisions on the circular scale. The diameter of the wire is:

A · 0.26 cm
B · 0.052 cm
C · 0.52 cm
D · 0.026 cm
Solution: Step 1: Find LC. Here 1 mm = 100 circular divisions, so LC = 1/100 = 0.01 mm. Step 2: Reading = MSR + CSR x LC = 0 + 52 x 0.01 = 0.52 mm. Step 3: Convert to cm = 0.52/10 = 0.052 cm. Answer: 0.052 cm (option B). Trap: option C (0.52 cm) is wrong by a factor of 10 from a unit mistake.
2018

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of the circular scale coincides with 25 divisions above the reference level. If the screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is:

A · 0.053 cm
B · 0.525 cm
C · 0.521 cm
D · 0.529 cm
Solution: Step 1: MSR = 5 mm = 0.5 cm. Step 2: CSR x LC = 25 x 0.001 = 0.025 cm. Step 3: Observed reading = 0.5 + 0.025 = 0.525 cm. Step 4: Correct for zero error. Corrected = observed minus zero error = 0.525 - (-0.004) = 0.525 + 0.004 = 0.529 cm. Answer: 0.529 cm (option D). Trap: subtracting a negative zero error means you add it, so 0.525 becomes 0.529, not 0.521.

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Frequently asked

What is the formula for least count of a screw gauge?

Least Count = Pitch / (number of divisions on the circular scale). Example: pitch 0.5 mm with 50 divisions gives LC = 0.5/50 = 0.01 mm = 0.001 cm.

What is the pitch of a screw gauge?

Pitch is the distance the screw (thimble) advances along the main scale in one complete rotation. For a standard NEET screw gauge it is usually 0.5 mm or 1 mm. Pitch = distance moved / number of full rotations.

How do you read a screw gauge?

Reading = Main Scale Reading (MSR, in mm) + (Circular Scale Reading x Least Count). Then subtract the zero error, keeping its sign, to get the true measurement.

What is the usual least count of a screw gauge in NEET?

Most NEET problems use LC = 0.01 mm = 0.001 cm. This is 10 times finer than a vernier caliper (0.1 mm), which is why the screw gauge is used for very thin wires and sheets.

Why is a screw gauge more precise than a vernier caliper?

The screw converts a small forward motion into a large rotation of the circular scale, so a tiny movement spreads across many divisions. This gives LC = 0.01 mm versus 0.1 mm for a vernier, making the screw gauge ten times more precise.