TOPIC: Error Analysis, Significant Figures, Vernier Calipers, and Screw Gauge Mechanics
IIT-JEE MAINS PYQ Arena — Set 10
Error Analysis, Significant Figures, Vernier Calipers & Screw Gauge
Set 10
Spectrometer Circular Vernier Scale
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. On the circular main scale of a Spectrometer, each degree is divided into 2 equal parts (so 1 MSD = 0.5° = 30′). If 30 divisions of the circular Vernier scale coincide with 29 divisions of the main scale, what is the angular Least Count of the Spectrometer?
✔ Optical Spectrometer LC Nailed! +4 Marks
✖ Recall: 1° = 60 arc-minutes (60′), so 1 MSD = 0.5° = 30′!
Concept & Solution:• Value of 1 Main Scale Division (1 MSD) = 0.5° = 30′ (30 arc-minutes).
• Since 30 VSD = 29 MSD ⇒ 1 VSD = (29 / 30) MSD.
• Least Count = 1 MSD − 1 VSD = (1 / 30) × 30′ = 1′ (1 arc-minute).
Question 1 of 15
Hollow Cylinder Volume & Vernier Error
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. Using the external and internal jaws of a Vernier Calipers of least count 0.1 mm, the external and internal diameters of a hollow metal pipe are measured as D₂ = (20.0 ± 0.1) mm and D₁ = (16.0 ± 0.1) mm. What is the maximum percentage error in the calculated wall thickness t = (D₂ − D₁) / 2 of the pipe?
✔ Wall Thickness Error Mastered! +4 Marks
✖ Notice that the factor of 1/2 cancels in relative error: Δt / t = (ΔD₂ + ΔD₁) / (D₂ − D₁)!
Concept & Solution:• Difference (D₂ − D₁) = 20.0 − 16.0 = 4.0 mm (so wall thickness t = 2.0 mm).
• Error in (D₂ − D₁) = ΔD₂ + ΔD₁ = 0.1 + 0.1 = 0.2 mm (so Δt = 0.1 mm).
• Maximum % Error in t = (Δt / t) × 100 = (0.1 mm / 2.0 mm) × 100 = 5.0%.
Screw Gauge Number of Circular Divisions
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. The linear distance moved by the screw of a Screw Gauge in 5 complete rotations of the circular cap is 5 mm. If the instrument is designed to have a least count of 0.005 mm, how many divisions must be engraved on its circular scale?
✔ Clean Pitch-to-Divisions Calculation! +4 Marks
✖ Pitch = 5 mm / 5 = 1 mm, so N = Pitch / LC = 1 mm / 0.005 mm = 200!
Concept & Solution:• Pitch (P) = Distance moved / Number of rotations = 5 mm / 5 = 1.0 mm.
• Since Least Count (LC) = Pitch / N:
• Number of circular scale divisions N = 1.0 mm / 0.005 mm = 200 divisions.
Significant Figures in Subtraction (Precision Loss)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A student subtracts two close length measurements: 18.45 cm − 18.39 cm. How many significant figures are present in each initial measurement and in the final subtracted result respectively?
✔ Classic Subtraction Precision-Drop Caught! +4 Marks
✖ 18.45 − 18.39 = 0.06 cm, and leading zeros in 0.06 are NOT significant, leaving only 1 sig fig!
Concept & Solution:• Both 18.45 cm and 18.39 cm have 4 significant figures and 2 decimal places.
• Difference = 18.45 − 18.39 = 0.06 cm (retaining 2 decimal places).
• Since leading zeros are never significant, 0.06 cm has only 1 significant figure.
Centripetal Force Error Propagation
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. A particle of mass m moves in a circle of radius r with speed v. If the percentage errors in the measurement of m, v, and r are 1%, 2%, and 1.5% respectively, what is the maximum percentage error in the calculated Centripetal Force (F = m v² / r)?
✔ Centripetal Error Nailed! +4 Marks
✖ Double the error in speed v (2 × 2% = 4%) and add the errors in m (1%) and r (1.5%)!
Concept & Solution:• From F = m v² / r:
• (ΔF / F) × 100 = (Δm / m)% + 2(Δv / v)% + (Δr / r)%
• Maximum % Error = 1% + 2(2%) + 1.5% = 1% + 4% + 1.5% = 6.5%.
Capillary Rise Surface Tension Error
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. In an experiment to determine the surface tension of water by capillary rise using S = (r h ρ g) / 2, the internal diameter of the capillary tube is measured as D = (1.25 ± 0.025) mm and the height of water rise is measured as h = (2.50 ± 0.025) cm. Assuming ρ and g are exact, what is the maximum percentage error in S?
✔ Capillary Rise Error Solved! +4 Marks
✖ Note that Δr / r = ΔD / D = 0.025 / 1.25 = 2%, and Δh / h = 0.025 / 2.50 = 1%!
Concept & Solution:• Since r = D / 2, relative error in radius equals relative error in diameter: (Δr / r)% = (0.025 / 1.25) × 100 = 2.0%.
• Relative error in height: (Δh / h)% = (0.025 / 2.50) × 100 = 1.0%.
• Maximum % Error in S = (Δr / r)% + (Δh / h)% = 2.0% + 1.0% = 3.0%.
Vernier Calipers Coinciding Mark Deduction
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A standard Vernier Calipers has 1 MSD = 1 mm and 10 VSD = 9 MSD (zero error = 0). When measuring a metal rod of true length 4.37 cm, the zero of the Vernier scale lies just past the 4.3 cm mark on the main scale. Which division of the Vernier scale coincides with a main scale division, and what is the distance of that coinciding main scale mark from the zero of the main scale?
✔ Deep Physical Insight into Vernier Geometry! +4 Marks
✖ Check the position of the 7th VSD: it is 7 × 0.09 cm = 0.63 cm ahead of the Vernier zero (4.37 cm), landing at 5.00 cm? Wait—let’s check!
Concept & Solution:• Since Reading = 4.30 cm + n × (0.01 cm) = 4.37 cm ⇒ n = 7th Vernier division.
• Note that 1 VSD = 0.9 mm = 0.09 cm.
• Position of the coinciding mark from Main Scale Zero = (Position of Vernier Zero) + 7 × (1 VSD) = 4.30 cm + 7 × 0.09 cm = 4.93 cm (or if Vernier zero is at 4.37 cm, 4.37 + 0.63 = 5.00 cm; relative to the 4.30 cm mark, 4.30 + 0.63 = 4.93 cm).
Series Capacitors Equivalent Capacitance Error
JEE Main 2022 (June Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. Two capacitors of capacitances C₁ = (6.0 ± 0.3) μF and C₂ = (3.0 ± 0.3) μF are connected in series. What is the equivalent capacitance (C_s) along with its maximum absolute error?
✔ Reciprocal Series Capacitance Error Solved! +4 Marks
✖ Use ΔC_s = C_s² (ΔC₁ / C₁² + ΔC₂ / C₂²) = 4 × (0.3/36 + 0.3/9)!
Concept & Solution:• Equivalent Capacitance: C_s = (6 × 3) / (6 + 3) = 18 / 9 = 2.0 μF.
• Differentiating 1/C_s = 1/C₁ + 1/C₂:
• ΔC_s = C_s² × (ΔC₁ / C₁² + ΔC₂ / C₂²)
• ΔC_s = (2.0)² × (0.3 / 36 + 0.3 / 9) = 4 × (1.5 / 36) = 6 / 36 ≈ 0.17 μF.
Screw Gauge Backlash Error Mechanics
JEE Main 2021 (Aug Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. In a Screw Gauge or Spherometer, Backlash Error occurs when the direction of rotation of the thimble is suddenly reversed. What is the primary physical cause of Backlash Error and how is it eliminated during measurement?
✔ Experimental Mechanics Concept Nailed! +4 Marks
✖ Backlash error is caused by loose play between worn screw threads and nut, and is avoided by always approaching the final reading from one direction!
Concept & Solution:• Due to wear and tear or loose fitting between the screw threads and the nut, reversing the rotation causes the circular scale to rotate slightly without any linear movement of the screw tip.
• It is completely avoided by always rotating the screw in one single direction while taking the final reading.
Fractional Power Error Propagation
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A physical quantity X is related to four measurable quantities a, b, c, and d as X = (a^(1/2) b²) / (c³ d^(1/3)). If the percentage errors in a, b, c, and d are 4%, 1.5%, 1%, and 6% respectively, what is the maximum percentage error in X?
✔ Fractional Exponents Handled Flawlessly! +4 Marks
✖ Multiply each percentage error by the magnitude of its power: (1/2)(4%) + 2(1.5%) + 3(1%) + (1/3)(6%)!
Concept & Solution:• (ΔX / X)% = (1/2)(Δa / a)% + 2(Δb / b)% + 3(Δc / c)% + (1/3)(Δd / d)%
• Maximum % Error = 0.5(4%) + 2(1.5%) + 3(1%) + (1/3)(6%) = 2% + 3% + 3% + 2% = 10.0%.
Unit Conversion & Significant Figures Invariance
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A length is measured as 0.07080 m. When this measurement is converted into centimeters as 7.080 cm and into millimeters as 70.80 mm, how many significant figures are present in the three representations respectively?
✔ Fundamental Rule: Unit Change Never Alters Significant Figures! +4 Marks
✖ Changing units (m ⇒ cm ⇒ mm) never changes the number of significant figures!
Concept & Solution:• A change of system of units does not change the number of significant figures in a measurement.
• In 0.07080 m, the leading zeros (0.0) are not significant, while 7, 0, 8, 0 are significant ⇒ 4, 4, 4.
Archimedes’ Principle & Specific Gravity Error
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. The weight of a solid body in air is measured by a spring balance as W₁ = (10.0 ± 0.1) N and its apparent weight when completely immersed in water is W₂ = (6.0 ± 0.1) N. What is the maximum percentage error in the calculated Relative Density (Specific Gravity) = W₁ / (W₁ − W₂)?
✔ Relative Density Error Mastered! +4 Marks
✖ Loss of weight ΔW = W₁ − W₂ = 4.0 N has maximum error ΔW₁ + ΔW₂ = 0.2 N (5%), plus 1% from W₁!
Concept & Solution:• Let Loss of weight L = W₁ − W₂ = 10.0 − 6.0 = 4.0 N.
• Maximum absolute error in L: ΔL = ΔW₁ + ΔW₂ = 0.1 + 0.1 = 0.2 N.
• Using standard independent-term error addition: (ΔRD / RD)% = (ΔW₁ / W₁)% + (ΔL / L)% = (0.1 / 10.0) × 100 + (0.2 / 4.0) × 100 = 1.0% + 5.0% = 6.0%.
Screw Gauge Thin Foil Stack Measurement
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. To measure the thickness of a very thin aluminum foil with high accuracy, a student stacks 50 identical sheets tightly together and measures the stack using a Screw Gauge of pitch 0.5 mm and 50 circular scale divisions (zero error = 0). If the main scale reads 1.5 mm and the 25th circular scale division coincides with the reference line, what is the thickness of a single foil sheet along with its least-count uncertainty?
✔ Stack Uncertainty Reduction by 50x! +4 Marks
✖ Divide both the total thickness (1.75 mm) and the Screw Gauge least count (0.01 mm) by 50!
Concept & Solution:• Least Count of Screw Gauge = 0.5 mm / 50 = 0.01 mm.
• Total thickness of 50 sheets = 1.5 mm + (25 × 0.01 mm) = (1.75 ± 0.01) mm.
• Thickness of 1 sheet = (1.75 / 50) ± (0.01 / 50) = (0.0350 ± 0.0002) mm.
Sonometer Wire Frequency Error
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. The fundamental frequency of a Sonometer wire is given by f = (1 / 2L) √(T / μ). If the percentage errors in measuring the vibrating length (L), tension (T), and linear mass density (μ) are 1%, 2%, and 4% respectively, what is the maximum percentage error in the calculated frequency (f)?
✔ Square-Root Exponent Halving Mastered! +4 Marks
✖ Both T and μ are inside the square root, so their percentage errors get multiplied by 1/2!
Concept & Solution:• Since f = (1 / 2) L⁻¹ T^(1/2) μ^(−1/2):
• (Δf / f)% = (ΔL / L)% + (1/2)(ΔT / T)% + (1/2)(Δμ / μ)%
• Maximum % Error = 1% + 0.5(2%) + 0.5(4%) = 1% + 1% + 2% = 4.0%.
Comparing Vernier Calipers vs. Screw Gauge
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. Instrument A is a Vernier Calipers in which 25 VSD coincide with 24 MSD (where 1 MSD = 0.5 mm). Instrument B is a Screw Gauge of pitch 0.5 mm having 100 circular scale divisions. What is the ratio of the Least Count of Instrument A to the Least Count of Instrument B (LC_A / LC_B)?
✔ Grand Finale 150th Question Conquered! +4 Marks
✖ Check LC_A = 0.5 mm / 25 = 0.02 mm, and LC_B = 0.5 mm / 100 = 0.005 mm!
Concept & Solution:• For Vernier Calipers (A): LC_A = 1 MSD / 25 = 0.5 mm / 25 = 0.020 mm.
• For Screw Gauge (B): LC_B = Pitch / 100 = 0.5 mm / 100 = 0.005 mm.
• Ratio LC_A / LC_B = 0.020 mm / 0.005 mm = 4 : 1.
150-PYQ Chapter 1 Mega-Series Completed!
Congratulations! With Set 10 complete, you now have a 150-question, zero-dependency JEE Main PYQ bank for Units, Dimensions & Experimental Physics.
👑 Set 10 Finale Tracker
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