Prerequisites
Before studying this topic, make sure you understand:
- Vernier Caliper - Basic precision measurement concepts
- Units and Measurements - Errors and significant figures
- Circular scale reading
The Hook: How Do You Measure Hair Thickness?
Human hair thickness: ~0.07 mm = 70 micrometers
Can you measure this with:
- Ruler? No (precision only 1 mm)
- Vernier caliper? Barely (precision 0.1 mm = 100 µm)
- Screw gauge? Perfect! (precision 0.01 mm = 10 µm)
Screw gauge (micrometer) achieves 100× better precision than ruler!
Real measurements:
- Paper thickness: 0.08 mm
- Wire diameter: 0.34 mm
- Metal sheet: 1.25 mm
- Microscope slide: 1.03 mm
Fun fact: The micrometer screw gauge enabled the precision engineering revolution - allowing mass production of interchangeable parts. Without it, no modern manufacturing!
How does a simple screw achieve such incredible precision? Let’s unlock the secret!
Interactive Demo
Practice reading screw gauge measurements:
The Core Concept: Screw Principle
Screw Gauge Diagram
Basic Parts
1. U-Frame:
- Fixed frame holding components
2. Anvil (Fixed End):
- Fixed reference point
3. Spindle (Moving End):
- Moves via screw rotation
- Touches object being measured
4. Sleeve (Barrel):
- Fixed graduated scale
- Linear (pitch) scale
5. Thimble:
- Rotating graduated scale
- Circular scale (100 divisions)
6. Ratchet:
- Ensures constant pressure
- Click when proper tightness reached
The Screw Mechanism
Key principle: When screw rotates 1 complete turn, it advances linearly by pitch distance
Pitch (p) = Distance moved by screw in one complete rotation
Standard values:
- Metric: 0.5 mm or 1 mm
- Typically 0.5 mm for precision instruments
Magic of screw gauge:
1 complete rotation = Spindle moves 0.5 mm (pitch)
Thimble has 100 divisions
Each division corresponds to:
$$\text{LC} = \frac{\text{Pitch}}{100} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}$$But wait! We can only read to 0.01 mm precision (2 divisions)
Think of it like:
- Vernier caliper: Uses sliding scale for precision
- Screw gauge: Uses rotation for precision
- Advantage: Rotation is smoother and more precise than sliding!
Your smartphone camera focus ring works the same way - tiny rotation = precise focus adjustment!
Least Count Calculation
Definition
Least Count (LC) = Smallest measurement possible
Formula
$$\boxed{\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}}$$Common Values
Standard Screw Gauge:
- Pitch = 0.5 mm
- Circular scale divisions = 100
- LC = 0.5/100 = 0.005 mm = 0.0005 cm
But practical reading: 0.01 mm = 0.001 cm (due to human eye limitation)
High-precision Screw Gauge:
- Pitch = 0.5 mm
- Circular scale divisions = 50
- LC = 0.01 mm = 0.001 cm
Pitch Calculation
Pitch = Distance moved in 1 complete rotation
Method: Rotate thimble by known number of rotations, measure linear distance
$$\boxed{\text{Pitch} = \frac{\text{Linear distance moved}}{\text{Number of rotations}}}$$Example: 5 rotations move spindle by 2.5 mm
$$\text{Pitch} = \frac{2.5}{5} = 0.5 \text{ mm}$$Reading the Screw Gauge
Step-by-Step Method
Step 1: Linear Scale Reading (LSR) or Pitch Scale Reading (PSR)
Read the last visible division on sleeve scale (pitch scale)
Note: Some screw gauges have 0.5 mm divisions, some have 1 mm
Example: If last visible mark is 5.0 mm:
$$\text{LSR} = 5.0 \text{ mm}$$Step 2: Circular Scale Reading (CSR)
Find division on thimble (circular scale) that coincides with reference line on sleeve
Example: 28th division coincides:
$$\text{CSR} = 28$$Step 3: Calculate Reading
$$\boxed{\text{Reading} = \text{LSR} + \text{CSR} \times \text{LC}}$$Example: LSR = 5.0 mm, CSR = 28, LC = 0.01 mm
$$\text{Reading} = 5.0 + 28 \times 0.01 = 5.0 + 0.28 = 5.28 \text{ mm}$$Alternative Notation
$$\boxed{\text{Measurement} = \text{PSR} + \text{CSR} \times \text{LC}}$$where PSR = Pitch Scale Reading (same as LSR)
Zero Error and Corrections
Types of Zero Error
When spindle and anvil just touch (no object), circular scale zero should align with reference line.
1. Positive Zero Error
Condition: Zero of circular scale is below (beyond) reference line
Reading shows: Object appears thicker than actual
Example: When closed, CSR shows +3
Zero error: $+3 \times 0.01 = +0.03$ mm
Correction:
$$\boxed{\text{Actual} = \text{Observed} - \text{(+Error)}}$$Example:
- Observed: 2.45 mm
- Zero error: +0.03 mm
- Actual: 2.45 - 0.03 = 2.42 mm
2. Negative Zero Error
Condition: Zero of circular scale is above reference line
How to read: Count divisions from zero in backward direction
Example: 97th division coincides (3 divisions before completing 100)
Zero error: $-(100 - 97) \times 0.01 = -3 \times 0.01 = -0.03$ mm
Correction:
$$\boxed{\text{Actual} = \text{Observed} - (-\text{Error}) = \text{Observed} + |\text{Error}|}$$Example:
- Observed: 1.58 mm
- Zero error: -0.03 mm
- Actual: 1.58 + 0.03 = 1.61 mm
Zero Error Summary
| Type | CSR Position | Example | Correction |
|---|---|---|---|
| Positive | Below line | +3 → +0.03 mm | Subtract |
| Negative | Above line | 97 → -0.03 mm | Add |error| |
Backlash error = Play/looseness in screw threads
Cause:
- Wear and tear
- Poor manufacturing
- Loose screw fitting
Effect:
- Spindle position changes when rotation direction reversed
- Inconsistent readings
How to avoid:
- Always rotate thimble in ONE direction (typically clockwise)
- If you overshoot, rotate back more and approach from same direction
- Use ratchet to ensure consistent pressure
In exam: If asked about backlash - explain it’s avoided by rotating in single direction!
Important Formulas Summary
Pitch
$$\boxed{\text{Pitch} = \frac{\text{Distance moved}}{\text{Number of rotations}}}$$Least Count
$$\boxed{\text{LC} = \frac{\text{Pitch}}{\text{Number of CSR divisions}}}$$Standard: LC = 0.01 mm = 0.001 cm
Reading
$$\boxed{\text{Reading} = \text{LSR} + \text{CSR} \times \text{LC}}$$Zero Error Correction
$$\boxed{\text{Actual} = \text{Observed} - \text{Zero Error}}$$Positive error: Subtract Negative error: Add magnitude
Memory Tricks & Patterns
Mnemonic for Reading
“Linear First, Circular Counts”
- Linear scale reading (LSR/PSR)
- Circular scale coincidence (CSR)
- Multiply CSR by LC
- Add to LSR
Same as vernier caliper!
Pitch Memory
“Half-mm Pitch is Standard”
- Standard pitch = 0.5 mm
- Each rotation moves 0.5 mm
Least Count Quick Value
“Point-Zero-One mm” → 0.01 mm
- Standard screw gauge LC = 0.01 mm = 0.001 cm
- 10× better than vernier caliper!
Zero Error Direction
“Below Positive, Above Negative”
- CSR zero below line → Positive error
- CSR zero above line → Negative error
Same logic as vernier caliper!
Backlash Prevention
“One Direction Only”
- Rotate in one direction (clockwise) always
- Never reverse during measurement
Pattern Recognition
LC is always:
- 0.01 mm (standard)
- 0.005 mm (high precision)
- Never bigger!
Final reading format:
- If LC = 0.01 mm: Reading ends in 0.00, 0.01, 0.02, … 0.99
- Example: 3.47 mm ✓, 3.475 mm ✗ (too precise!)
Pitch × Rotations = Distance:
- 1 rotation → 0.5 mm
- 10 rotations → 5.0 mm
- Linear relationship!
When to Use This
Use screw gauge when:
- Object thickness < 5 mm
- Need precision of 0.01 mm
- Measuring wire diameter
- Measuring sheet thickness
- Measuring spherical objects (diameter)
Screw gauge vs Vernier caliper:
| Property | Screw Gauge | Vernier Caliper |
|---|---|---|
| Precision | 0.01 mm | 0.1 mm |
| Range | 0-25 mm | 0-15 cm |
| Best for | Thin objects | Thicker objects |
| Measures | Thickness, diameter | Length, depth, internal/external |
For JEE problems:
Given pitch and divisions → Find LC:
$$\text{LC} = \frac{\text{Pitch}}{\text{Divisions}}$$Given LSR and CSR → Find reading:
$$\text{Reading} = \text{LSR} + \text{CSR} \times \text{LC}$$Apply zero error correction:
$$\text{Actual} = \text{Observed} - \text{Error}$$Common Mistakes to Avoid
Wrong: “Pitch = Least Count = 0.5 mm”
Correct:
- Pitch: Distance per full rotation (0.5 mm)
- Least Count: Distance per one division (0.01 mm)
- LC = Pitch/100 (for 100-division thimble)
Example:
- Pitch = 0.5 mm
- Divisions = 100
- LC = 0.5/100 = 0.005 mm (theoretical)
- Practical LC = 0.01 mm (2 divisions minimum)
JEE trap: “Find least count given pitch = 1 mm, 50 divisions” → LC = 1/50 = 0.02 mm (not 0.01 mm!)
Wrong: “Last visible division is 5.5, so LSR = 5.5”
Correct: Check if divisions are in 0.5 mm or 1 mm intervals
Sleeve scale types:
- Type 1: 1, 2, 3, 4, 5 (1 mm divisions) → LSR = 5.0 mm
- Type 2: 0, 0.5, 1.0, 1.5, 2.0 (0.5 mm shown) → LSR = 5.5 mm
Always check graduation!
Safest: Count how many 0.5 mm divisions are visible
Wrong:
- 95th division coincides when closed
- Zero error = 95 × 0.01 = 0.95 mm ✗
Correct:
- 95th means 5 divisions before completing circle
- Zero error = (100 - 95) × 0.01 = -0.05 mm ✓
- Negative because zero is above line!
Quick rule: If CSR > 50 when closed → Negative error
- Error = -(100 - CSR) × LC
Wrong: Rotating thimble back and forth freely
Correct:
- Rotate in one direction only
- If overshoot, go further back and approach from same side
- Use ratchet for consistent pressure
JEE conceptual question: “Why should screw gauge be rotated in one direction?” Answer: To avoid backlash error due to loose screw threads!
Practice Problems
Level 1: Foundation (NCERT/Basic)
A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. Find the least count.
Solution:
$$\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}}$$ $$= \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}$$Answer: LC = 0.01 mm = 0.001 cm
This is the standard screw gauge precision!
In a measurement, LSR = 3.0 mm, CSR = 42, LC = 0.01 mm. Find the reading.
Solution:
$$\text{Reading} = \text{LSR} + \text{CSR} \times \text{LC}$$ $$= 3.0 + 42 \times 0.01$$ $$= 3.0 + 0.42 = 3.42 \text{ mm}$$Answer: 3.42 mm
Level 2: JEE Main
A screw advances 2.5 mm when rotated 5 complete turns. If circular scale has 100 divisions, find: (a) Pitch (b) Least count
Solution:
(a) Pitch:
$$\text{Pitch} = \frac{\text{Distance}}{\text{Rotations}} = \frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm}$$(b) Least count:
$$\text{LC} = \frac{\text{Pitch}}{\text{Divisions}} = \frac{0.5}{100} = 0.005 \text{ mm}$$Practical LC = 0.01 mm (2 divisions)
Answer:
- (a) Pitch = 0.5 mm
- (b) LC = 0.005 mm (theoretical), 0.01 mm (practical)
A screw gauge has positive zero error of 0.04 mm. Reading for a wire is 1.36 mm. Find actual diameter.
Solution:
Zero error: +0.04 mm (positive)
$$\text{Actual} = \text{Observed} - \text{Zero Error}$$ $$= 1.36 - 0.04 = 1.32 \text{ mm}$$Answer: Actual diameter = 1.32 mm
Insight: Positive error makes reading appear larger!
When anvil and spindle are in contact, 96th division of circular scale coincides with reference line. If LC = 0.01 mm, find: (a) Zero error (b) How to correct measurements
Solution:
(a) Zero error:
CSR = 96 (> 50) → Negative zero error
$$\text{Error} = -(100 - 96) \times 0.01 = -4 \times 0.01 = -0.04 \text{ mm}$$(b) Correction:
$$\text{Actual} = \text{Observed} - (-0.04) = \text{Observed} + 0.04$$Answer:
- (a) Zero error = -0.04 mm
- (b) Add 0.04 mm to all readings
Example: Observed = 2.45 mm → Actual = 2.45 + 0.04 = 2.49 mm
Level 3: JEE Advanced
A student measures wire diameter 5 times with screw gauge (LC = 0.01 mm, zero error = +0.02 mm):
Readings: 0.54, 0.53, 0.55, 0.54, 0.54 mm
Find: (a) Actual diameters, (b) Mean, (c) Mean absolute error, (d) Result with error
Solution:
(a) Actual diameters:
Zero error = +0.02 mm → Subtract from each
| Observed | Actual |
|---|---|
| 0.54 | 0.52 |
| 0.53 | 0.51 |
| 0.55 | 0.53 |
| 0.54 | 0.52 |
| 0.54 | 0.52 |
(b) Mean:
$$\text{Mean} = \frac{0.52 + 0.51 + 0.53 + 0.52 + 0.52}{5}$$ $$= \frac{2.60}{5} = 0.52 \text{ mm}$$(c) Mean absolute error:
$$\overline{|\Delta d|} = \frac{|0.52-0.52| + |0.51-0.52| + |0.53-0.52| + |0.52-0.52| + |0.52-0.52|}{5}$$ $$= \frac{0 + 0.01 + 0.01 + 0 + 0}{5} = \frac{0.02}{5} = 0.004 \text{ mm}$$Round to LC: 0.01 mm
(d) Final result:
$$d = (0.52 \pm 0.01) \text{ mm}$$Answer:
- (a) See table
- (b) Mean = 0.52 mm
- (c) Error = 0.01 mm
- (d) d = (0.52 ± 0.01) mm
Insight: Random error ~ LC, showing good measurement technique!
Explain why screw gauge is more precise than vernier caliper. Calculate the factor of improvement.
Solution:
Vernier Caliper:
- Least count = 0.1 mm
Screw Gauge:
- Least count = 0.01 mm
Improvement factor:
$$\frac{\text{LC}_{\text{vernier}}}{\text{LC}_{\text{screw}}} = \frac{0.1}{0.01} = 10$$Screw gauge is 10× more precise!
Why?
1. Screw mechanism advantage:
- 1 full rotation = only 0.5 mm linear movement
- 100 divisions on thimble
- Each division = 0.005 mm (very small!)
2. Amplification:
- Small rotation → easily visible on large circumference
- But small linear movement
- Circular motion amplifies precision
3. Elimination of sliding friction:
- Vernier uses sliding scale (friction, wear)
- Screw uses rotation (smoother, more precise)
Analogy:
- Steering wheel: Small turn → precise wheel angle
- Screw gauge: Small rotation → precise linear measurement
Answer: Screw gauge is 10× more precise due to screw mechanism that converts rotation to small linear motion with amplified reading on circular scale.
Quick Revision Box
| Parameter | Formula | Standard Value |
|---|---|---|
| Pitch | Distance / Rotations | 0.5 mm |
| Least Count | Pitch / Divisions | 0.01 mm |
| Reading | LSR + CSR × LC | - |
| Positive zero error | CSR below line | Subtract |
| Negative zero error | CSR above line (100-CSR) × LC | Add |error| |
| Actual Reading | Observed - Zero Error | - |
Key advantage: 10× more precise than vernier caliper!
JEE Strategy: High-Yield Points
Pitch calculation - Very common!
- Pitch = Distance / Number of rotations
- Standard: 0.5 mm
Least count calculation:
- LC = Pitch / Divisions
- Standard: 0.01 mm (50 or 100 divisions)
- Don’t confuse with pitch!
Reading calculation:
- Reading = LSR + CSR × LC
- LSR from sleeve, CSR from thimble
- Same pattern as vernier!
Zero error correction - Appears every year!
- Positive: CSR below (beyond) line → Subtract
- Negative: CSR above line → Add |error|
- Negative: Use (100 - CSR) × LC
Backlash error: Conceptual favorite
- Caused by loose screw threads
- Avoided by rotating in one direction only
Comparison with vernier:
- Screw gauge: 10× more precise (0.01 vs 0.1 mm)
- But smaller range (0-25mm vs 0-150mm)
Time-saving tricks:
- Standard pitch = 0.5 mm (memorize!)
- Standard LC = 0.01 mm (memorize!)
- For negative error: If CSR = 97 → Error = -(100-97)×0.01 = -0.03 mm
- Quick check: Final reading must be multiple of LC
Common numerical values:
- Pitch: 0.5 mm or 1 mm
- Divisions: 50 or 100
- LC: 0.01 mm (most common) or 0.02 mm
Related Topics
Within Experimental Skills
- Vernier Caliper - Less precise but longer range
- Common Experiments - Where to use screw gauge
Connected Chapters
- Units and Measurements - Errors, significant figures
- Rotational Motion - Screw as inclined plane
Real-world Applications
- Manufacturing - Precision quality control
- Mechanical engineering - Bearing, shaft measurements
- Metallurgy - Sheet thickness verification
- Textiles - Fabric thickness
- Electronics - Wire gauge measurement
- Research labs - Microscopic measurements
Teacher’s Summary
Screw gauge achieves 0.01 mm precision - 10× better than vernier caliper (0.1 mm)
Screw principle: Rotation → Linear motion
- Pitch: Distance moved in 1 rotation (0.5 mm standard)
- Amplifies precision through circular scale
Least Count = Pitch / Number of divisions
- Standard: 0.5 mm / 50 = 0.01 mm
- Don’t confuse with pitch!
Reading = LSR + CSR × LC
- LSR: Linear (pitch) scale reading from sleeve
- CSR: Circular scale reading from thimble
- Same logic as vernier caliper
Zero error correction:
- Positive (CSR below line): Subtract error
- Negative (CSR above, use 100-CSR): Add |error|
- Formula: Actual = Observed - Error (with sign)
Backlash error avoided by rotating in one direction only
Applications: Measuring anything thin (< 5mm)
- Wire diameter, sheet thickness, paper, hair!
- Essential in precision manufacturing
“The screw gauge converts circular motion into amplified linear precision - enabling measurements down to 10 micrometers, essential for modern manufacturing from smartphones to spacecraft!”