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1. A student repeats a measurement five times and gets values very close to each other, but all are far from the accepted value. Which statement best describes the result?
Explanation: Reproducibility of results means high precision . Being far from the true value means low accuracy .
2. Which of the following is a characteristic of a systematic error?
Explanation: Systematic errors are consistent and unidirectional; averaging does not eliminate them. Random errors scatter and can be reduced by averaging.
3. The least count of a measuring instrument is:
Explanation: Least count (LC) is the smallest division that can be read directly without estimation.
4. A vernier caliper has 10 divisions on the vernier scale that coincide with 9 main scale divisions of 1 mm each. What is its least count?
Explanation: LC = 1 MSD − 1 VSD = 1 − 9/10 = 0.1 mm.
5. A screw gauge has a pitch of 0.5 mm and 50 circular scale divisions. Its least count is:
Explanation: LC = pitch / circular scale divisions = 0.5 / 50 = 0.01 mm.
6. How many significant figures are in the number 0.00720?
Explanation: Leading zeros are not significant. 7, 2, and the trailing 0 after the decimal point are significant — giving 3 significant figures.
7. The result of 3.400 × 2.0 should be reported with how many significant figures?
Explanation: In multiplication, the result has as many significant figures as the factor with the fewest — here 2.0 has 2 significant figures.
8. If the true value of a quantity is 50 units and a measured value is 48 units, the absolute error is:
Explanation: Absolute error = |measured − true| = |48 − 50| = 2 units (always taken as positive).
9. If absolute error in a measurement is 0.2 cm and the measured value is 4.0 cm, the relative error is:
Explanation: Relative error = Δx / x = 0.2 / 4.0 = 0.05.
10. Percentage error is related to relative error by:
Explanation: Percentage error = (Δx / x) × 100.
11. A screw gauge shows a reading of +0.03 mm when jaws are fully closed. This is called:
Explanation: When the instrument reads a positive value with nothing between the jaws, it has a positive zero error, and the correction (−0.03 mm) must be applied to readings.
12. If A = (5.0 ± 0.1) cm and B = (3.0 ± 0.2) cm, then A + B =
Explanation: For addition/subtraction, absolute errors add: ΔC = ΔA + ΔB = 0.1 + 0.2 = 0.3 cm.
13. If A = (10.0 ± 0.2) m and B = (4.0 ± 0.1) m, then A − B =
Explanation: In subtraction, absolute errors still add: Δ(A−B) = ΔA + ΔB = 0.2 + 0.1 = 0.3 m.
14. If percentage error in length is 2% and in breadth is 3%, the percentage error in area (length × breadth) is:
Explanation: For a product, relative (percentage) errors add: %error in A = %error in L + %error in B = 2 + 3 = 5%.
15. If Q = R³ and percentage error in R is 2%, the percentage error in Q is:
Explanation: For a power, the percentage error multiplies: %error in Q = 3 × %error in R = 3 × 2 = 6%.
16. Readings of a measurement are 2.50, 2.53, 2.47, 2.51, 2.49 (in cm). The mean value is:
Explanation: Mean = (2.50 + 2.53 + 2.47 + 2.51 + 2.49) / 5 = 12.50 / 5 = 2.50 cm.
17. Using the readings above (2.50, 2.53, 2.47, 2.51, 2.49 cm; mean = 2.50 cm), the mean absolute error is:
Explanation: Deviations: 0.00, 0.03, 0.03, 0.01, 0.01. Mean absolute error = (0+0.03+0.03+0.01+0.01)/5 = 0.08/5 = 0.016 ≈ 0.02 cm.
18. 13.20 + 1.5 + 0.126 reported to correct significant figures is:
Explanation: In addition, result is rounded to the least number of decimal places — here 1.5 has 1 decimal place, so the answer is 14.8.
19. Which instrument has a smaller least count — a metre rule (1 mm graduations) or a vernier caliper (LC = 0.1 mm)?
Explanation: A vernier caliper has LC = 0.1 mm, which is smaller than the metre rule's 1 mm, making it more precise.
20. The period T of a pendulum is measured as T = (2.00 ± 0.02) s. If g is calculated as g = 4π²L/T², and %error in L is 1%, the %error in g is:
Explanation: %error in g = %error in L + 2×(%error in T) = 1 + 2×1 = 3%.
21. In a vernier caliper, the main scale reads 2.3 cm and the 4th vernier division coincides with a main scale division (LC = 0.01 cm). The reading is:
Explanation: Reading = MSR + (VSD × LC) = 2.3 + 4 × 0.01 = 2.34 cm.
22. A screw gauge (LC = 0.01 mm) has MSR = 5 mm and circular scale reading = 25. The diameter is:
Explanation: Reading = MSR + CSR × LC = 5 + 25 × 0.01 = 5.25 mm.
23. A screw gauge shows −0.04 mm when jaws are closed. If a diameter reads 3.42 mm, the corrected diameter is:
Explanation: For negative zero error (instrument reads less than true), corrected reading = observed + |zero error| = 3.42 + 0.04 = 3.46 mm.
24. How many significant figures are in 5.060 × 10³?
Explanation: 5, 0, 6, 0 are all significant in 5.060 — trailing zeros after the decimal in scientific notation count. Total = 4.
25. 2.400 ÷ 1.2 should be reported as:
Explanation: 2.400 has 4 significant figures, 1.2 has 2 — the result takes the smaller count. So 2.0 (2 sig figs).
26. If %error in P is 4% and %error in Q is 2%, the %error in P/Q is:
Explanation: For a quotient, percentage errors add: %error in (P/Q) = %error in P + %error in Q = 4 + 2 = 6%.
27. The density ρ = M/V. If %error in M is 2% and %error in V is 3%, the %error in ρ is:
Explanation: %error in ρ = %error in M + %error in V = 2 + 3 = 5%.
28. The kinetic energy E = ½mv². If %error in m is 2% and %error in v is 3%, the %error in E is:
Explanation: %error in E = %error in m + 2 × %error in v = 2 + 2×3 = 8%.
29. A vernier caliper has a positive zero error of 0.04 cm. A reading shows MSR = 3.2 cm and 6th VSD coincides (LC = 0.01 cm). Corrected reading is:
Explanation: Observed = 3.2 + 6×0.01 = 3.26 cm. Corrected = 3.26 − 0.04 = 3.22 cm (subtract positive zero error).
30. Five readings of a length are 4.40, 4.42, 4.38, 4.41, 4.39 cm. The mean absolute error is:
Explanation: Mean = 4.40. Deviations: 0.00, 0.02, 0.02, 0.01, 0.01. Mean absolute error = (0+0.02+0.02+0.01+0.01)/5 = 0.06/5 ≈ 0.01 cm.
31. Which technique helps reduce random errors but NOT systematic errors?
Explanation: Averaging reduces random (statistical) errors. Systematic errors are directional and constant — they require calibration or correction to reduce.
32. The least count error is:
Explanation: Every measurement has an inherent uncertainty equal to at least the least count of the instrument.
33. Express 57,864 rounded to 3 significant figures:
Explanation: The first 3 significant figures are 5, 7, 8. The next digit is 6 (≥5), so round up: 57,900.
34. A measured value is 9.8 m/s² and the accepted value is 9.81 m/s². The percentage error is approximately:
Explanation: %error = |9.8 − 9.81|/9.81 × 100 = 0.01/9.81 × 100 ≈ 0.1%.
35. For X = A²B/C³, the percentage error in X when %errors in A, B, C are 1%, 2%, 3% respectively is:
Explanation: %error in X = 2×(1%) + 1×(2%) + 3×(3%) = 2 + 2 + 9 = 13%.
36. An instrument has a least count of 0.001 cm. A student records 2.340 cm. The number of significant figures in this reading is:
Explanation: 2, 3, 4, 0 — all four digits are significant. The trailing zero after the decimal is significant because it was measured.
37. Backlash error in a screw gauge occurs due to:
Explanation: Backlash arises when there is play (looseness) between the screw and nut, so reversing direction doesn't immediately move the thimble. Always rotate in one direction during measurement.
38. If X = A − B + C and ΔA = 0.2, ΔB = 0.3, ΔC = 0.1 (all in cm), then ΔX is:
Explanation: For any algebraic sum/difference, absolute errors always add: ΔX = ΔA + ΔB + ΔC = 0.2 + 0.3 + 0.1 = 0.6 cm.
39. Two measurements: A = 100 ± 1 and B = 1 ± 0.01. Which has smaller relative error?
Explanation: Relative error of A = 1/100 = 0.01. Relative error of B = 0.01/1 = 0.01. Both have the same relative (fractional) error.
40. A result is reported as (25.4 ± 0.2) cm. This means:
Explanation: The ± notation gives the confidence interval: the true value is most likely in [25.4−0.2, 25.4+0.2] = [25.2, 25.6] cm.
41. Resistance R = V/I. If V = (10.0 ± 0.1) V and I = (2.00 ± 0.02) A, the percentage error in R is:
Explanation: %error in R = %error in V + %error in I = (0.1/10.0)×100 + (0.02/2.00)×100 = 1% + 1% = 2%.
42. In an experiment, L = 1.000 ± 0.001 m and T = 2.00 ± 0.01 s. The percentage error in g = 4π²L/T² is:
Explanation: %error in g = %error in L + 2×(%error in T) = (0.001/1.000)×100 + 2×(0.01/2.00)×100 = 0.1% + 1.0% = 1.1%.
43. The mass of a sphere is 4.237 g and its radius is 1.54 cm. Density = m/(4πr³/3). How many significant figures should the answer have?
Explanation: The radius 1.54 cm has 3 significant figures — the fewest among the given data — so the final answer should have 3 significant figures.
44. A vernier caliper (LC = 0.01 cm) shows MSR = 1.2 cm and 8th VSD coincides. Zero error is +0.02 cm. Corrected length is:
Explanation: Observed = 1.2 + 8×0.01 = 1.28 cm. Corrected = 1.28 − 0.02 = 1.26 cm.
45. A screw gauge (LC = 0.01 mm, pitch = 1 mm) shows zero error = −0.05 mm. For a wire, MSR = 3 mm and CSR = 42. Corrected diameter is:
Explanation: Observed = 3 + 42×0.01 = 3.42 mm. Correction for negative zero error: +0.05 mm. Corrected = 3.42 + 0.05 = 3.47 mm.
46. If Y = √(A/B) and %errors in A and B are 4% and 2% respectively, the %error in Y is:
Explanation: Y = A^(1/2) / B^(1/2). %error in Y = (1/2)(4%) + (1/2)(2%) = 2% + 1% = 3%.
47. How many significant figures does 4500 have if written without a decimal point?
Explanation: Without a decimal point or additional indication, trailing zeros in an integer like 4500 are ambiguous — they may or may not be significant. Scientific notation removes ambiguity.
48. Two students measure a 1 m rod. Student A: (1.01 ± 0.01) m; Student B: (0.99 ± 0.01) m. Whose measurement is more precise?
Explanation: Precision is determined by the absolute error (uncertainty), which is 0.01 m for both. They are equally precise.
49. Which of the following reduces random error in an experiment?
Explanation: Random errors follow a statistical distribution; averaging over many trials reduces the standard error of the mean.
50. Kinetic energy E = p²/(2m). If %error in momentum p is 3% and in mass m is 2%, the %error in E is:
Explanation: %error in E = 2×(%error in p) + %error in m = 2×3 + 2 = 8%.
51. If the zero mark of a vernier scale is behind the 0 on the main scale when closed (vernier reads −2 divisions), this instrument has:
Explanation: When the vernier zero is to the left (behind) the main scale zero when closed, the instrument shows less than the true value — this is negative zero error. Correction: add the magnitude to all readings.
52. A student uses a stopwatch with LC = 0.1 s to time 10 oscillations of a pendulum, getting 15.3 s. The time period is (15.3 ± ?) s. What is the error in the time period?
Explanation: Error in total time = 0.1 s (LC). Time period T = 15.3/10 = 1.53 s. Error in T = 0.1/10 = 0.01 s.
53. Which expression has the largest percentage error? Given %errors in A, B, C are each 1%.
Explanation: For a sum, %errors don't simply add like products do. For A³B²C, %error = 3(1)+2(1)+1(1) = 6%. That is the largest.
54. The diameter of a wire is measured 5 times: 1.20, 1.22, 1.20, 1.19, 1.19 mm. Which of the following best represents the result?
Explanation: Mean = (1.20+1.22+1.20+1.19+1.19)/5 = 6.00/5 = 1.200 mm. Mean absolute error ≈ 0.01 mm. Result: 1.20 ± 0.01 mm.
55. Calculate: (3.0 × 10⁴) × (2.50 × 10²) and report with correct significant figures.
Explanation: 3.0 has 2 sig figs (fewer), so the answer is 7.5 × 10⁶ (2 significant figures).
56. A measurement is (50 ± 1) kg. A second measurement is (5 ± 1) kg. The measurement with smaller relative error is:
Explanation: Relative error of 50 kg: 1/50 = 2%. Relative error of 5 kg: 1/5 = 20%. The 50 kg measurement has much smaller relative error.
57. X = A − B. If A = 50.0 ± 0.1 and B = 49.5 ± 0.1, then X and its relative error are:
Explanation: X = 0.5, ΔX = 0.1+0.1 = 0.2. Relative error = 0.2/0.5 × 100 = 40%. This shows why subtracting nearly equal quantities amplifies relative error enormously.
58. To measure the thickness of a thin metal sheet (~0.5 mm) most accurately, the best instrument is:
Explanation: For thin sheets where measurements are in fractions of a millimetre, the screw gauge with LC = 0.01 mm provides the best precision.
59. The formula for the time period of a simple pendulum is T = 2π√(L/g). Which error contributes more to %error in T — a 2% error in L or a 1% error in g?
Explanation: %error in T = (1/2)×(%error in L) = 1%, and (1/2)×(%error in g) = 0.5%. Actually 2% in L contributes 1% and 1% in g contributes 0.5%, so 2% in L contributes more.
60. When rounding 2.45 to 2 significant figures, the result is:
Explanation: The third significant figure is 5 — when the digit to round is 5 and there's nothing after it, round to even (banker's rounding): 2.4 (even) is preferred. In many JEE contexts, rounding 2.45 → 2.4 or 2.5 are both accepted; the standard rule gives 2.4.
61. Instrumental error is best reduced by:
Explanation: Instrumental errors arise from the instrument itself (manufacturing defects, wear). Calibration against a known standard corrects these errors.
62. A vernier scale has 20 divisions equal to 19 main scale divisions of 1 mm each. The least count is:
Explanation: 1 VSD = 19/20 mm. LC = 1 MSD − 1 VSD = 1 − 19/20 = 1/20 = 0.05 mm.
63. A screw gauge has pitch 0.5 mm and 50 circular divisions. What is the LC? If the thimble reads 48 and MSR = 2.5 mm, the reading is:
Explanation: LC = 0.5/50 = 0.01 mm. Reading = 2.5 + 48×0.01 = 2.5 + 0.48 = 2.98 mm.
64. In Kater's pendulum experiment, g is measured to be 9.8 m/s² with T measured to 0.1% accuracy and L to 0.2% accuracy. The %error in g is:
Explanation: Since g ∝ L/T², %error in g = %error in L + 2×%error in T = 0.2 + 2×0.1 = 0.4%.
65. 1.0 × 10⁻² has how many significant figures?
Explanation: 1.0 has 2 significant figures (the zero after the decimal is significant). The exponent doesn't count.
66. The frequency of a tuning fork is found using f = n/t where n = 500 (exact count) and t = 10.0 ± 0.1 s. The %error in f is:
Explanation: Since n is exact (no error), %error in f = %error in t = (0.1/10.0)×100 = 1%.
67. A screw gauge has zero error +0.03 mm. The observed reading for a wire diameter is 0.78 mm. The actual diameter is:
Explanation: Positive zero error means the instrument reads more than true. Actual = observed − zero error = 0.78 − 0.03 = 0.75 mm.
68. Which measurement has higher accuracy: (a) 50.00 ± 0.01 cm or (b) 2.00 ± 0.01 cm?
Explanation: Accuracy is judged by relative error: (a) 0.01/50.00 = 0.02%; (b) 0.01/2.00 = 0.5%. (a) has smaller relative error, hence higher accuracy.
69. Surface area of a sphere is S = 4πR². If %error in radius R is 1.5%, the %error in S is:
Explanation: S ∝ R². %error in S = 2 × %error in R = 2 × 1.5 = 3.0%.
70. Repeated measurement of a length gives (all in cm): 5.01, 4.99, 5.02, 4.98, 5.00. The result should be reported as:
Explanation: Mean = 5.00. Deviations: 0.01, 0.01, 0.02, 0.02, 0.00. Mean absolute error = 0.06/5 = 0.012 ≈ 0.01 cm. Result: 5.00 ± 0.01 cm.
71. In a series of measurements, the deviation of each reading from the mean is called:
Explanation: The absolute error of each reading is its deviation from the mean of all readings.
72. Express (8.765 − 8.23) with correct significant figures:
Explanation: In subtraction, result has fewest decimal places: 8.23 has 2, so result = 0.54 (2 decimal places = 2 sig figs here).
73. A stop clock has a LC of 1 second and a stopwatch has LC of 0.01 s. The stopwatch is:
Explanation: Smaller LC means the instrument can resolve smaller time intervals, making the stopwatch more precise.
74. If f = A^m B^n / C^p, the maximum percentage error in f is:
Explanation: Maximum percentage error = m×(%error in A) + n×(%error in B) + p×(%error in C). All terms are added.
75. Four measurements of a 10 Ω resistor: 9.9, 10.1, 9.8, 10.2 Ω. The mean is 10.0 Ω. This set is:
Explanation: The mean (10.0 Ω) equals the true value — so the set is accurate. But the spread (max deviation 0.2 Ω) is large — so it is not very precise.
76. A vernier (LC = 0.02 mm) has zero error −0.04 mm. If MSR = 12 mm and 8th VSD coincides, the corrected reading is:
Explanation: Observed = 12 + 8×0.02 = 12.16 mm. Negative zero error → correction = +0.04 mm. Corrected = 12.16 + 0.04 = 12.20 mm.
77. Young's modulus Y = (FL)/(AΔl). If %errors in F, L, A, Δl are 1, 2, 3, 4% respectively, the %error in Y is:
Explanation: %error in Y = %F + %L + %A + %Δl = 1 + 2 + 3 + 4 = 10%.
78. To measure a very small length (like wavelength of light ≈ 500 nm), which method is appropriate?
Explanation: Wavelengths of light (hundreds of nanometres) are far below the LC of mechanical instruments. Interferometry uses wave interference to measure sub-micron lengths.
79. Write 0.00309 in scientific notation with correct significant figures:
Explanation: 0.00309 has 3 significant figures (3, 0, 9). In scientific notation: 3.09 × 10⁻³.
80. A student measures the diameter of a marble 6 times and takes the average. This is done to minimise:
Explanation: Random errors fluctuate around the mean and are reduced by averaging. Systematic errors are not reduced by repeating measurements.
81. A physical quantity P = a³b²/(c√d). If %errors in a, b, c, d are α, β, γ, δ respectively, the %error in P is:
Explanation: %error in P = 3α + 2β + γ + (1/2)δ. All terms add (regardless of whether quantities appear in numerator or denominator, the power determines the multiplier).
82. In a vernier caliper, if 49 main scale divisions = 50 vernier scale divisions, and 1 MSD = 1 mm, the LC is:
Explanation: 1 VSD = 49/50 mm. LC = 1 − 49/50 = 1/50 = 0.02 mm.
83. If refractive index μ = sin((A+D)/2)/sin(A/2) where A = 60° ± 0.1° and D = 30° ± 0.2°, which angle dominates the error in μ?
Explanation: D appears only once but its error (0.2°) is larger than A's contribution per degree. Differentiation shows the numerator angle (A+D)/2 changes more with D changes.
84. 10⁰ has how many significant figures?
Explanation: 10⁰ = 1 exactly (a mathematical identity), so it has infinite significant figures — it is a pure number, not a measurement.
85. A student times 20 oscillations with a stopwatch (LC = 0.01 s) and gets 40.06 s. The time period T and its error are:
Explanation: T = 40.06/20 = 2.003 s. Error in total time = 0.01 s (LC). Error in T = 0.01/20 = 0.0005 s. Report: 2.003 ± 0.0005 s.
86. A screw gauge has 0.5 mm pitch and 50 circular divisions (LC = 0.01 mm). Positive zero error = 0.03 mm. For a wire: MSR shows 1 complete division past 0, and CSR = 32. The correct diameter is:
Explanation: MSR: 1 division × 0.5 mm = 0.5 mm. CSR: 32 × 0.01 = 0.32 mm. Observed = 0.5 + 0.32 = 0.82 mm. Corrected = 0.82 − 0.03 = 0.79 mm.
87. Why is subtraction of two nearly equal measured quantities considered unreliable?
Explanation: If A ≈ B, then A − B ≈ 0, making the relative error (ΔA+ΔB)/(A−B) enormous even if absolute errors are small.
88. A student consistently gets a measured value of gravity as 9.75 m/s² when the accepted value is 9.81 m/s². This is most likely due to:
Explanation: A consistent offset from the true value in one direction indicates a systematic error — likely miscalibrated length or stopwatch.
89. Surface tension T is found using T = Fθ/(2L). If %error in F = 2%, in θ = 1%, in L = 3%, then %error in T is:
Explanation: %error in T = %error in F + %error in θ + %error in L = 2 + 1 + 3 = 6%.
90. The pitch of a screw gauge is doubled while keeping the number of circular divisions the same. The new least count is:
Explanation: LC = pitch / circular divisions. If pitch doubles and divisions stay fixed, LC doubles.
91. If a result is (4.0 ± 0.2) and another is (3.0 ± 0.1), then their product and percentage error are:
Explanation: %error in product = (0.2/4.0 + 0.1/3.0) × 100 = 5% + 3.33% = 8.33% ≈ 8.3%.
92. For measuring an angle of a crystal (~ 1 minute of arc precision required), the appropriate instrument is:
Explanation: A spectrometer with a vernier on the circular scale can measure angles to arc-minute precision, needed for crystal angle and prism experiments.
93. A measurement is made with a device whose LC = 0.1 cm. The measured value is 5.6 cm. The true value could reasonably be anywhere in the range:
Explanation: With LC = 0.1 cm, the reading 5.6 cm means the true value is within ±0.05 cm (half LC): range = [5.55, 5.65] cm.
94. In Young's modulus experiment, stress = F/A and strain = ΔL/L. If %errors in F, A, ΔL, L are 0.5, 1.0, 2.0, 0.5 respectively, the %error in Y is:
Explanation: Y = stress/strain = FL/(AΔL). %error in Y = %F + %A + %L + %ΔL = 0.5 + 1.0 + 0.5 + 2.0 = 4.0%.
95. A student calculates a speed as 12.345678 m/s, where both distance and time were measured to 3 significant figures. The correctly reported speed is:
Explanation: The inputs have 3 significant figures, so the output must also be limited to 3: 12.3 m/s.
96. In a graph-based experiment, the slope is found to be 2.0 with an uncertainty of ±0.1. If a physical quantity Q = slope², its value and %error are:
Explanation: Q = slope². %error in Q = 2 × %error in slope = 2 × (0.1/2.0 × 100) = 2 × 5% = 10%. Q = 4.0 ± 10%.
97. A thermometer is calibrated by dipping it in ice water (reads 2°C instead of 0°C) and boiling water (reads 98°C instead of 100°C). When it reads 52°C, the true temperature is:
Explanation: Scale maps [2,98] → [0,100]. Correction: True = (Reading−2)×(100/96) = 50×(100/96) = 52.08 ≈ 53.06°C. Each 1°C on thermometer = 100/96 ≈ 1.0417 true degrees.
98. In an experiment, a student increases the number of observations from 4 to 16. The standard error of the mean changes by a factor of:
Explanation: Standard error ∝ 1/√n. Going from n=4 to n=16: factor = √4/√16 = 2/4 = 1/2. The error is halved.