Accuracy, Precision of Measuring Instruments, Significant Figures and Error in Measurements: NEET MCQs with Answers

15 free questions on Accuracy, Precision of Measuring Instruments, Significant Figures and Error in Measurements for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
A screw gauge has a least count of 0.01 mm and there are 50 divisions on its circular scale. The pitch of the screw gauge is [NEET 2020]
  1. 0.25 mm
  2. 0.5 mm
  3. 1.0 mm
  4. 0.01 mm
Show answer
Answer: (B) 0.5 mm
Explanation: Since least count = pitch / number of circular scale divisions, pitch = LC × n = 0.01 × 50 = 0.5 mm. The pitch is the linear distance moved by the screw in one complete rotation.

Question 2

Previous-year question
Taking into account the significant figures, what is the value of 9.99 m − 0.0099 m? [NEET 2020]
  1. 9.98 m
  2. 9.980 m
  3. 9.9 m
  4. 9.9801 m
Show answer
Answer: (A) 9.98 m
Explanation: The arithmetic difference is 9.9801 m. In subtraction the result is rounded to the least number of decimal places among the data; 9.99 m has two decimal places, so the answer is 9.98 m.

Question 3

Previous-year question
In an experiment, the percentage errors in the measurement of the physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage error in the measurement of X, where X = A²B¹ᐟ²/(C¹ᐟ³D³), will be [NEET 2019]
  1. (3/13)%
  2. 16%
  3. −10%
  4. 10%
Show answer
Answer: (B) 16%
Explanation: ΔX/X = 2(ΔA/A) + ½(ΔB/B) + ⅓(ΔC/C) + 3(ΔD/D) = 2(1) + ½(2) + ⅓(3) + 3(4) = 2 + 1 + 1 + 12 = 16%. Every exponent contributes as a positive multiplying factor.

Question 4

Previous-year question
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 the zero of the circular scale division 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 [NEET 2018]
  1. 0.521 cm
  2. 0.525 cm
  3. 0.529 cm
  4. 0.053 cm
Show answer
Answer: (C) 0.529 cm
Explanation: Observed reading = MSR + CSR × LC = 0.5 cm + 25 × 0.001 cm = 0.525 cm. Corrected reading = observed reading − zero error = 0.525 − (−0.004) = 0.529 cm. A negative zero error is always added back to the observed reading.

Question 5

Previous-year question
A steel ball's diameter is measured with a screw gauge of least count 0.001 cm. The main scale reads 5 mm and the 0 of the circular scale is 25 divisions above the reference line. With a zero error of −0.004 cm, the correct diameter is [NEET 2018]
  1. 0.053 cm
  2. 0.525 cm
  3. 0.521 cm
  4. 0.529 cm
Show answer
Answer: (D) 0.529 cm
Explanation: Reading = 0.5 cm + 25(0.001) = 0.525 cm; corrected = 0.525 − (−0.004) = 0.529 cm.

Question 6

Previous-year question
A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e₁ and e₂ respectively, the percentage error in the estimation of g is [2010]
  1. e₂ − e₁
  2. e₁ + 2e₂
  3. e₁ + e₂
  4. e₁ − 2e₂
Show answer
Answer: (B) e₁ + 2e₂
Explanation: For free fall from rest s = ½gt² ⇒ g = 2s/t². Hence %g = %s + 2(%t) = e₁ + 2e₂.

Question 7

Previous-year question
Taking into account the significant figures, what is the value of 9.99 m − 0.0099 m? [2020]
  1. 9.9801 m
  2. 9.98 m
  3. 9.9 m
  4. 9.980 m
Show answer
Answer: (B) 9.98 m
Explanation: 9.99 − 0.0099 = 9.9801. In addition/subtraction the result keeps the least number of decimal places among the data — here 2 (from 9.99) — so it is rounded to 9.98 m.
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Question 8

Previous-year question
The area of a rectangular field (in m²) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is [2019]
  1. 138 × 10¹
  2. 1382
  3. 1382.5
  4. 14 × 10²
Show answer
Answer: (D) 14 × 10²
Explanation: Area = 55.3 × 25 = 1382.5 m². In multiplication the result carries the least number of significant figures of the data — here 2 (from 25) — so it is rounded to 1.4 × 10³ = 14 × 10² m².

Question 9

Previous-year question
Each side of a metallic cube of mass 5.580 kg is measured to be 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X × 10³ kg m⁻³. The value of X is [2019]
  1. 7.654
  2. 7.7
  3. 7.65
  4. 7.6
Show answer
Answer: (B) 7.7
Explanation: ρ = m/a³ = 5.580 / (0.090)³ = 5.580 / 7.29×10⁻⁴ = 7654 kg m⁻³ = 7.654 × 10³. The side (9.0 cm) has only 2 significant figures, so the result is rounded to 2 significant figures ⇒ 7.7 × 10³ kg m⁻³, i.e. X = 7.7.

Question 10

Previous-year question
In an experiment, the percentage error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage error in the measurement of X, where X = A²B^(1/2) / (C^(1/3)D³), will be [2023]
  1. 10%
  2. (3/13)%
  3. 16%
  4. −10%
Show answer
Answer: (C) 16%
Explanation: %X = 2(%A) + ½(%B) + ⅓(%C) + 3(%D) = 2(1) + ½(2) + ⅓(3) + 3(4) = 2 + 1 + 1 + 12 = 16%.

Question 11

Previous-year question
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are [2023]
  1. Least count errors
  2. Random errors
  3. Instrumental errors
  4. Personal errors
Show answer
Answer: (B) Random errors
Explanation: Random errors occur irregularly and arise from unpredictable fluctuations in experimental conditions such as temperature and supply voltage. Their effect is reduced by taking many readings and averaging them.

Question 12

Previous-year question
A physical quantity P is related to four observations a, b, c and d as follows: P = a³b² / (√c · d). The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. The percentage error in the quantity P is [2025]
  1. 13%
  2. 15%
  3. 10%
  4. 2%
Show answer
Answer: (A) 13%
Explanation: For P = a³b²c^(−1/2)d^(−1), the fractional error adds with the magnitude of each power: %P = 3(%a) + 2(%b) + ½(%c) + 1(%d) = 3(1) + 2(3) + ½(4) + 1(2) = 3 + 6 + 2 + 2 = 13%.

Question 13

Which one of the following statements is incorrect?
  1. The parsec is a unit of distance.
  2. Absolute error is a pure number that carries no unit.
  3. An exact count, such as the number of students in a class, has an unlimited number of significant figures.
  4. The absolute error in a single reading may be positive, negative or zero.
Show answer
Answer: (B) Absolute error is a pure number that carries no unit.
Explanation: Absolute error is the difference between a measured value and the true value, so it has the same unit and dimensions as the quantity measured. It is the relative error that has no unit. The other statements are correct.

Question 14

The radius of a sphere is measured with a percentage error of 1.5%. The percentage error in the surface area of the sphere calculated from this radius is:
  1. 4.5%
  2. 3%
  3. 2.25%
  4. 1.5%
Show answer
Answer: (B) 3%
Explanation: Surface area S = 4πr², so the fractional error in S is twice the fractional error in r. The percentage error in S is 2 × 1.5% = 3%.

Question 15

A glass prism has a refracting angle of 60°, taken as exact. Its angle of minimum deviation is measured as 40° with a maximum error of 2°. Using μ = sin[(A + δ)/2]/sin(A/2), within what range does the experimental value of the refractive index lie? (Take sin 50° = 0.766, cot 50° = 0.84 and 2° = 0.035 rad.)
  1. 1.43 ≤ μ ≤ 1.63
  2. 1.51 ≤ μ ≤ 1.55
  3. 1.49 ≤ μ ≤ 1.58
  4. 1.52 ≤ μ ≤ 1.54
Show answer
Answer: (B) 1.51 ≤ μ ≤ 1.55
Explanation: The central value is μ = sin 50°/sin 30° = 0.766/0.5 = 1.532. With A fixed, Δμ/μ = ½ cot[(A + δ)/2] Δδ, with Δδ in radians, which gives ½ × 0.84 × 0.035 ≈ 0.0147. Hence Δμ ≈ 1.532 × 0.0147 ≈ 0.022, so μ lies between about 1.51 and 1.55.

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