Q.Which of the following time measuring devices is most precise?
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The Problem: How Fine Can You Measure?
Imagine you have a standard 15 cm ruler. You look at the markings — there's a line for every centimetre, and between those, smaller lines for every millimetre. Now try to measure the thickness of a single sheet of paper. You place it against the ruler. Does it line up exactly with a millimetre mark? Almost certainly not. It falls somewhere between two millimetre lines.
What do you report? You can't say "2.3 mm" because your ruler doesn't show tenths of a millimetre. The best you can honestly say is "about 2 mm" or "between 2 and 3 mm". That limitation — the smallest change in the quantity that your instrument can reliably detect and display — is its least count.
Least count is not about how good you are at estimating. It is a property of the instrument itself. Even a perfect observer cannot extract more detail than the instrument's scale allows.
The Precise Definition
Least count is the smallest value of a physical quantity that can be measured accurately using a given instrument. It is the resolution of the measuring device.
For a simple scale (like a ruler, voltmeter, or thermometer) with equally spaced markings, the least count is:
Least Count=Number of divisions on the vernier (or sub-scale)Value of one main scale division
For instruments without a vernier (like a plain ruler), the least count is simply the value of the smallest division on the scale.
Examples you already know:
| Instrument | Smallest Division | Least Count |
|---|---|---|
| Metre ruler | 1 mm | 1 mm |
| Stopwatch (digital) | 0.01 s | 0.01 s |
| Laboratory thermometer | 1 °C | 1 °C |
| Screw gauge | 0.01 mm | 0.01 mm |
Why This Matters in Exams
When you record a measurement, you must write it to the correct number of decimal places — the least count determines that. If a ruler has a least count of 1 mm, you cannot report a length as 12.35 cm. The correct recording is 12.3 cm or 12.4 cm (the last digit is uncertain, but it must be at the place of the least count).
A common mistake: writing 2.50 cm when your ruler only shows millimetres. That implies you measured to 0.01 cm (0.1 mm), which you didn't. Write 2.5 cm — the last digit is at the tenths place, matching your least count of 0.1 cm.
The Core Intuition …
Why this formula?
Least Count & Precision: Why the Formula Holds
Let's build this from first principles — understanding why the least count formula works, not just memorizing it.
1. What is Least Count?
Least Count (LC) is the smallest measurement an instrument can reliably indicate.
Think of a ruler with 1 cm marks but no mm marks — you can't measure 0.5 cm precisely. The least count is 1 cm.
2. The Core Formula
For a scale-type instrument (ruler, vernier caliper, micrometer):
Least Count=Number of divisions on the vernier/circular scaleValue of 1 main scale division
Why this formula?
Reasoning step-by-step:
- The main scale has fixed divisions (e.g., 1 mm each).
- The vernier scale has N divisions that exactly span (N−1) main scale divisions.
- So, 1 vernier division = NN−1 main scale divisions.
The difference between 1 main scale division and 1 vernier division is:
LC=1 MSD−1 VSD=1−NN−1=N1 MSD
That's exactly the formula above.
3. Example: Vernier Caliper
- Main scale: 1 mm per division
- Vernier scale: 10 divisions covering 9 mm
Then:
LC=101 mm=0.1 mm
Why 0.1 mm? Because the 10th vernier mark aligns with the 9th main scale mark — the smallest shift you can detect is 0.1 mm.
4. For Circular Scales (Micrometer Screw Gauge)
Same logic, different geometry:
LC=Number of circular scale divisionsPitch
Pitch = distance moved by spindle in one full rotation.
Why? One full rotation moves the spindle by the pitch. If the circular scale has N divisions, each division corresponds to Npitch linear movement.
5. Precision vs. Least Count
Precision is half the least count (or sometimes ± LC/2).
Precision=±2LC
Why half? …
Precision depends on an instrument's least count (smallest interval it can detect):
- Wall clock: ≈1 s
- Stopwatch: ≈0.1–0.01 s
- Digital watch: ≈0.01 s …
Precision is set by an instrument's least count — the smallest interval it can reliably detect. Among a wall clock, stopwatch, digital watch, and atomic clock, the atomic clock has by far the smallest least count (about 1 part in 1013, an uncertainty of order 10−13 s), so (d) the atomic clock is the most precise.
Comparing the four devices by least count
- Wall clock — typically has only a second hand; least count ≈1 s. Cannot resolve fractions of a second.
- Stop watch — mechanical or digital, least count typically 0.1 s or 0.01 s.
- Digital watch — usually displays to 0.01 s, better than a mechanical stopwatch but still far from atomic precision.
- Atomic clock — uses the fixed frequency of a specific atomic transition (e.g. the cesium-133 hyperfine transition, which defines the SI second) as its "tick." This makes it accurate to about 1 part in 1013 — an uncertainty of roughly 10−13 s, meaning it would gain or lose only a few microseconds over an entire year. …
Method: Least Count Comparison
Method name: Least Count (or Resolution) Comparison
Concept: The precision of a measuring device is determined by its least count — the smallest change in the measured quantity that the device can detect. A smaller least count means higher precision.
Steps to solve:
-
Identify the least count of each device
- Wall clock: typically measures up to 1 second (least count = 1 s)
- Stopwatch: usually measures up to 0.1 s or 0.01 s (least count = 0.1 s or 0.01 s)
- Digital watch: commonly shows up to 0.01 s (least count = 0.01 s)
- Atomic clock: accurate to about 1 part in 1013 — an uncertainty of order 10−13 s (from the fixed cesium-133 hyperfine transition frequency that defines the SI second)
-
Compare the least counts
Smaller least count → more precise device.
-
Rank them in increasing order of least count
- Atomic clock: ∼10−13 s (smallest) …
Here are the common mistakes students make on this question, along with how to avoid each one.
Mistake 1: Confusing "Precision" with "Accuracy"
- What students do: They think a wall clock is "precise" because it shows the correct time every day, or they pick a stopwatch because it "seems scientific."
- Why it's wrong: Precision is about the smallest change a device can detect (least count), not whether it tells the correct time. A wall clock has a least count of 1 second (or more), while an atomic clock is accurate to about 1 part in 1013 — an uncertainty of order 10−13 seconds or smaller.
- How to avoid: Always define least count first. The most precise device is the one with the smallest least count.
Key rule: Precision = Least Count. Smaller least count = more precise.
Mistake 2: Ignoring the "Least Count" of a Digital Watch
- What students do: They pick a digital watch because it shows digits (e.g., 10:05:32) and assume it's very precise.
- Why it's wrong: A typical digital watch (wristwatch) has a least count of 1 second — same as a wall clock. It just displays seconds, but it cannot measure fractions of a second.
- How to avoid: Check the smallest division or smallest digit change the device can reliably measure. For a digital watch, that is 1 s.
Example: A digital watch showing 10:05:32 cannot show 10:05:32.5 — so its least count is 1 s.
Mistake 3: Forgetting the Atomic Clock's Least Count
- What students do: They think an atomic clock is "too complicated" or "not a common clock," so they ignore it.
- Why it's wrong: An atomic clock uses vibrations of atoms (e.g., cesium-133) and is accurate to about 1 part in 1013 — an uncertainty of order 10−13 s. This is far smaller than any mechanical or digital watch.
- How to avoid: Memorise the order of magnitude of least counts:
- Wall clock: 1 s
- Stopwatch: 0.1 s or 0.01 s
- Digital watch: 1 s
- Atomic clock: ∼10−13 s (1 part in 1013)
Correct answer: (d) An atomic clock — because its least count is the smallest.
--- …
- KCET 2023Set A-31 markMCQQ.The true length of a wire is 3.678cm. When the length of this wire is measured using instrument A, the length of the wire is 3.5cm. When the length of the wire is measured using instrument B, it is found to have length 3.38cm. Then the (A) measurement with A is more accurate while measurement with B is more precise. (B) measurement with B is more accurate and precise. (C) measurement with A is more precise while measurement with B is more accurate. (D) measurement with A is more accurate and precise.
›Reveal solutionSolution
Compare absolute errors for accuracy (A wins) and least counts / significant figures for precision (B wins).
Step 1 — The two distinct ideas
- Accuracy — how close a measured value is to the true value. Measured by the magnitude of the error.
- Precision — the resolution (least count) of the instrument, reflected in the number of decimal places/significant figures it can report. It says nothing about closeness to truth.
Step 2 — Accuracy
True length L=3.678 cm.
ΔA=∣3.678−3.50∣=0.178 cm
ΔB=∣3.678−3.38∣=0.298 cm
Since ΔA<ΔB, instrument A is the more accurate one.
Step 3 — Precision
- A reports 3.5 cm → resolves to 0.1 cm (2 significant figures). …
- KCET 2022Set B-31 markMCQQ.The Vernier scale of a travelling microscope has 50 divisions which coincides with 49 main scale divisions. If each main scale division is 0.5 mm, then the lease count of the microscope is (A) 0.01 mm (B) 0.5 cm (C) 0.01 cm (D) 0.5 mm
›Reveal solutionSolution
Least count of a vernier =number of vernier divisionsvalue of 1 main scale division=500.5 mm=0.01 mm.
Step 1 — Understand what the vernier construction tells us.
We are told 50 vernier scale divisions (VSD) coincide with 49 main scale divisions (MSD):
50 VSD=49 MSD⟹1 VSD=5049 MSD
Step 2 — Least count from first principles.
The least count is the difference between one main-scale division and one vernier division — that is the smallest length the instrument can resolve:
L.C.=1 MSD−1 VSD=1 MSD−5049 MSD=501 MSD
This is exactly the standard shortcut formula:
L.C.=number of divisions on the vernier scalevalue of 1 MSD
Step 3 — Substitute the given values. …
- KCET 2020Set A-11 markMCQQ.Two poles are separated by a distance of 3.14 m. The resolving power of human eye is 1 minute of an arc. The maximum distance from which he can identify the two poles distinctly is (A) 10.8 km (B) 5.4 km (C) 188 m (D) 376 m
›Reveal solutionSolution
The key idea is to relate the angular resolution of the eye (1 minute of arc) to the linear separation of the poles (3.14 m) using the small-angle approximation. The maximum distance is 10.8 km, which corresponds to option (A).
The resolving power of the human eye is given as 1 minute of arc. This means that two points can be seen as distinct only if the angle they subtend at the eye is at least 1 minute of arc. If the angle is smaller, the eye cannot resolve them — they appear as a single point.
The problem gives the actual separation between the two poles (3.14 m) and asks for the maximum distance at which this separation still subtends an angle of exactly 1 minute of arc. Beyond that distance, the angle would be smaller, and the poles would blur together.
The natural tool here is the small-angle approximation: for a small angle θ (in radians), the arc length s and the distance r are related by s≈rθ, where s is the linear separation perpendicular to the line of sight. This works because the angle is tiny — 1 minute of arc is only about 2.91×10−4 radians.
Watch outA common mistake is to forget to convert minutes of arc to radians. The formula s=rθ requires θ in radians, not degrees or minutes. Always convert first.
Let’s work through it step by step.
- Convert the angular resolution to radians. 1 minute of arc = 601 degree. Since 1∘=180π radians,
1′=601×180π=10800π radians.
Numerically, π≈3.1416, so
θ=108003.1416≈2.909×10−4 rad.
- Apply the small-angle relation. The linear separation s=3.14 m is the arc length subtended at distance r. So …
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