Measurements

Q: What is the least count of a vernier caliper?

The least count of a standard vernier caliper is 0.01 cm (0.1 mm). It is calculated as LC = 1 MSD - 1 VSD, or equivalently as the value of one main scale division divided by the total number of vernier divisions.

Q: What is the least count of a screw gauge?

The least count of a typical screw gauge is 0.01 mm (0.001 cm). It is calculated as LC = Pitch / N, where the pitch is the distance moved by the screw in one full rotation and N is the number of divisions on the circular scale.

Q: What is the difference between vernier caliper and screw gauge?

A vernier caliper measures lengths from a few mm to about 15 cm with a least count of 0.01 cm. A screw gauge measures very small lengths (up to a few mm) with a higher precision of 0.01 mm - ten times finer than a vernier. Verniers are used for external and internal dimensions; screw gauges for very thin wires, sheets, and small diameters.

PhysicsUnits And MeasurementFor NEET aspirants

Measurements in physics deal with the practical techniques and instruments used to determine physical quantities accurately - essential for both JEE and NEET aspirants. Ordinary metre scales measure only up to $0.1$ cm, so precision instruments like the vernier caliper (least count $0.01$ cm) and the screw gauge (least count $0.01$ mm) are used to measure smaller lengths. Alongside these instruments, mastering unit conversions - between SI, CGS, and practical units like electron volts, atmospheres and light years - is critical for solving numerical problems in JEE Main, JEE Advanced and NEET.

Key Formulas - Quick Reference

Least count of vernier caliper: $LC = 1 MSD - 1 VSD = \frac{value of 1 MSD}{number of vernier divisions}$
Least count of screw gauge: $LC = \frac{Pitch}{Number of circular scale divisions}$
Pitch of a screw gauge: distance moved on pitch scale per one full rotation of thimble.
Total reading: $R = Main scale reading (MSR) + Circular/Vernier scale reading \times LC$
Corrected reading: $R_{true} = R_{observed} - (zero error, with sign)$
Positive zero error is subtracted; negative zero error is added.
$1 eV = 1.6 \times 1 0^{- 19} J$ ; $1 atm = 1.013 \times 1 0^{5} Pa$ ; $1 A.U. = 1.496 \times 1 0^{11} m$ .

1. Need for Precise Measurement

A metre scale can measure lengths only up to about

0.1

cm. Its smallest accurate reading is called its least count. For measuring smaller lengths, such as the diameter of a wire or the thickness of a coin, we use more precise instruments:

Vernier callipers - least count typically $0.01$ cm.
Screw gauge (micrometer) - least count typically $0.001$ cm ( $0.01$ mm).

Least count: the smallest measurement that can be taken accurately with an instrument.

2. Vernier Callipers

A vernier calliper has two scales: a fixed main scale (marked in cm and mm) and a sliding vernier scale below it. The vernier scale is graduated so that its total length is one division shorter than an integer number of main scale divisions. This mismatch allows precise reading of fractional divisions.

Figure 1: Vernier caliper - fixed jaw is attached to the main scale, movable jaw slides with the vernier scale. Object to be measured sits between the jaws.

2.1 Least Count of Vernier Calipers

The least count (also called vernier constant) is:

LC = \frac{value of 1 main scale division}{number of divisions on vernier scale}

Equivalently:

LC = 1 MSD - 1 VSD

For a standard vernier caliper:

1 MSD = 0.1

cm and there are

10

vernier divisions, so

LC = \frac{0.1}{10} = 0.01

cm.

2.2 Zero Error of Vernier Calipers

When the jaws are closed, the zero of the vernier scale should coincide with the zero of the main scale. If it does not, the instrument has a zero error:

Positive zero error: the zero of the vernier scale is to the right of the main scale zero. The correction is subtracted.
Negative zero error: the zero of the vernier scale is to the left of the main scale zero. The correction is added.

Figure 5: Zero error - if zero of vernier or circular scale does not coincide with main scale zero when jaws are closed, the reading needs correction.

True reading = Observed reading - (zero error, with sign)

Solved Example 1

N

divisions of the vernier scale coincide with

(N - 1)

divisions of the main scale, and the value of 1 main scale division is

a

units, find the least count of the vernier.

Solution:

Length of $N$ vernier divisions = length of $(N - 1)$ main scale divisions.

So $1 VSD = \frac{N - 1}{N} \times 1 MSD = \frac{( N - 1 ) a}{N}$ .

$LC = 1 MSD - 1 VSD = a - \frac{( N - 1 ) a}{N} = \frac{a}{N}$

Therefore, least count $= \frac{a}{N}$ units.

Solved Example 2

A vernier caliper has a main scale graduated in mm and 10 vernier divisions coincide with 9 main scale divisions. Find its least count.

Solution:

Here $N = 10$ and $a = 1$ mm.

$LC = \frac{a}{N} = \frac{1 mm}{10} = 0.1$ mm $= 0.01$ cm.

3. Screw Gauge (Micrometer)

A screw gauge measures very small lengths (thickness of paper, diameter of wire) to an accuracy of

0.01

mm. It works on the principle of a screw: when the thimble is rotated, the screw moves linearly. The distance moved per full rotation is called the pitch.

Figure 2: Screw gauge (micrometer) - U-frame holds the fixed anvil; the spindle screws in from the right. Object sits between anvil and spindle. Main scale is on the sleeve, circular scale is on the thimble.

3.1 Pitch and Least Count

Pitch: the linear distance moved by the screw along the pitch scale in one complete rotation of the thimble.

Least count = \frac{Pitch}{Number of divisions on circular scale}

For a typical screw gauge: pitch

= 1

mm, circular scale has

100

divisions, so

LC = \frac{1}{100} = 0.01

mm.

3.2 Zero Error of Screw Gauge

When the studs are brought in contact, the zero of the circular scale should coincide with the reference line of the pitch scale. If not:

Positive zero error: the zero of the circular scale is below the reference line. Correction is subtracted.
Negative zero error: the zero of the circular scale is above the reference line. Correction is added.

3.3 Reading a Screw Gauge

Reading = Main scale reading (MSR) + Circular scale division (CSD) \times LC

Solved Example 3

A screw gauge has main scale reading

3

mm and the

45

th division of the circular scale coincides with the reference line. Least count is

0.01

mm. Find the reading.

Solution:

Reading = MSR + CSD $\times$ LC

$= 3 + 45 \times 0.01 = 3 + 0.45 = 3.45$ mm.

Figure 3: Screw gauge zoom - thimble edge past 3 mm on sleeve, 45th division on the centre line. Reading = 3 + 45(0.01) = 3.45 mm.

Solved Example 4

A screw gauge shows main scale reading

5.5

mm (half-mm mark visible) and the

16

th division of the circular scale coincides with the reference line. Least count is

0.01

mm. Find the reading.

Solution:

Reading = MSR + CSD $\times$ LC

$= 5.5 + 16 \times 0.01 = 5.5 + 0.16 = 5.66$ mm.

Figure 4: Screw gauge zoom - thimble edge past 5.5 mm (half-mm mark visible), 16th division on the centre line. Reading = 5.5 + 16(0.01) = 5.66 mm.

Solved Example 5

The pitch of a screw gauge is

0.5

mm and there are

50

divisions on its circular scale. Find its least count.

Solution:

$LC = \frac{Pitch}{N} = \frac{0.5}{50} = 0.01$ mm.

4. Important Unit Conversions

Memorising these conversions saves crucial time in JEE and NEET numerical problems. They come up directly in questions and indirectly as tools while solving mechanics, thermodynamics and modern physics problems.

4.1 Length

Quantity	SI equivalent
1 astronomical unit (A.U.)	$1.496 \times 1 0^{11}$ m
1 light year	$9.46 \times 1 0^{15}$ m
1 parsec	$3.08 \times 1 0^{16}$ m
1 angstrom (Å)	$1 0^{- 10}$ m
1 X-ray unit	$1 0^{- 13}$ m
1 fermi (fm)	$1 0^{- 15}$ m
1 foot	$30.48$ cm
1 inch	$2.54$ cm

4.2 Mass

Quantity	SI equivalent
1 metric ton (tonne)	$1000$ kg
1 quintal	$100$ kg
1 pound (lb)	$0.4536$ kg
1 atomic mass unit (a.m.u.)	$1.66 \times 1 0^{- 27}$ kg
1 carat	$200$ mg $= 2 \times 1 0^{- 4}$ kg
1 Chandrasekhar limit	$1.4$ times the mass of the Sun

4.3 Time

Quantity	SI equivalent
1 shake	$1 0^{- 8}$ s
1 year	$365.25$ days $= 3.156 \times 1 0^{7}$ s

4.4 Pressure

Quantity	SI equivalent
1 bar	$0.1$ MPa $= 1 0^{5}$ Pa
1 standard atmosphere	$1.013 \times 1 0^{5}$ Pa (or $N/m^{2}$ )
1 mm of Hg (torr)	$133.3 N/m^{2}$

4.5 Energy and Power

Quantity	SI equivalent
1 electron volt (eV)	$1.6 \times 1 0^{- 19}$ J
1 calorie	$4.19$ J
1 watt-hour	$3.6 \times 1 0^{3}$ J
1 horse power (hp)	$746$ W
Gas constant $R$	$8.314 J mol^{- 1} K^{- 1}$ $= 2 cal mol^{- 1} K^{- 1}$

4.6 Radioactivity and Radiation

Quantity	SI equivalent
1 curie (Ci)	$3.7 \times 1 0^{10}$ disintegrations/s $= 3.7 \times 1 0^{10} s^{- 1}$
1 roentgen (R)	$2.58 \times 1 0^{- 4}$ C/kg
1 barn (nuclear cross-section)	$1 0^{- 28} m^{2}$

4.7 Magnetic Units

Quantity	SI equivalent
1 weber (Wb)	$1 0^{8}$ maxwell
1 tesla (T)	$1 Wb/m^{2} = 1 0^{4}$ gauss
1 oersted (Oe)	$\approx 79.58 A/m$

5. Solved Examples on Measurements

Solved Example 6

The density of water at

4 °C

is approximately equal to what in SI units?

Solution:

The density of water depends on temperature. It is maximum at $4 °C$ and equals:

$ρ_{water} = 1000 kg/m^{3} = 1 0^{3} kg m^{- 3}$ .

In CGS, this is $1 g/cm^{3}$ .

Solved Example 7

Convert 1 standard atmospheric pressure into

N/m^{2}

and dyne/cm

^{2}

Solution:

1 atmosphere = 76 cm of Hg column.

Pressure $= h ρ g = 76 \times 13.6 \times 981$ dyne/cm $^{2}$

$= 1.013 \times 1 0^{6}$ dyne/cm $^{2}$

$= 1.013 \times 1 0^{5} N/m^{2}$ (since $1 N/m^{2} = 10$ dyne/cm $^{2}$ ).

Solved Example 8

C

is the capacitance and

R

is the resistance, find the dimensional formula of

\frac{1}{C R}

Solution:

$C R$ is the time constant of an RC circuit, which has dimensions of time.

$[C R] = [T]$ , therefore $[\frac{1}{C R}] = [T^{- 1}]$ .

SI unit: $s^{- 1}$ (same as frequency).

Solved Example 9

In a vernier caliper,

10

divisions of vernier coincide with

9

divisions of main scale. Main scale is graduated in mm. The reading is: main scale =

2.1

cm,

4

th vernier division coincides with a main scale line. Find the length of the object.

Solution:

Least count $= \frac{1 mm}{10} = 0.01$ cm.

Length = MSR + VSD $\times$ LC = $2.1 + 4 \times 0.01 = 2.14$ cm.

Solved Example 10

A screw gauge has a positive zero error of

0.05

mm. When measuring a wire, the observed reading is

2.35

mm. Find the corrected diameter of the wire.

Solution:

Since the zero error is positive, it is subtracted from the observed reading.

True reading $= 2.35 - 0.05 = 2.30$ mm.

Solved Example 11

A star is at a distance of

4

light years from Earth. Convert this distance into metres.

Solution:

$1$ light year $= 9.46 \times 1 0^{15}$ m.

Distance $= 4 \times 9.46 \times 1 0^{15} = 3.784 \times 1 0^{16}$ m.

Common Mistakes to Avoid

Watch out

Confusing "shake" with kilograms. $1$ shake is a unit of time ( $1 0^{- 8}$ s), not mass.
Forgetting the half-mm mark on screw gauge. When the half-mm line past the last full mm is visible, add $0.5$ mm to the main scale reading.
Getting the sign of zero error wrong. Positive zero error is subtracted, negative is added. A useful check: if closed jaws show a reading, that reading is the zero error.
Using wrong pitch value. Pitch is the distance moved per full rotation, not per division on the circular scale.
Mixing dyne and newton. $1 N = 1 0^{5}$ dyne, and $1 N/m^{2} = 10$ dyne/cm $^{2}$ .
Assuming $1$ oersted equals $1$ A/m. It does not: $1$ Oe $\approx 79.58$ A/m.
Forgetting to convert 76 cm of Hg to metres when computing pressure in SI units. Use $ρ = 13600$ kg/m $^{3}$ and $g = 9.8$ m/s $^{2}$ for the SI computation.

Frequently Asked Questions

Q1. What is the least count of a vernier caliper?

The least count of a standard vernier caliper is $0.01$ cm ( $0.1$ mm). It is calculated as $LC = 1 MSD - 1 VSD$ , or equivalently as the value of one main scale division divided by the total number of vernier divisions.

Q2. What is the least count of a screw gauge?

The least count of a typical screw gauge is $0.01$ mm ( $0.001$ cm). It is calculated as $LC = Pitch / N$ , where the pitch is the distance moved by the screw in one full rotation and $N$ is the number of divisions on the circular scale.

Q3. What is the difference between vernier caliper and screw gauge?

A vernier caliper measures lengths from a few mm to about $15$ cm with a least count of $0.01$ cm. A screw gauge measures very small lengths (up to a few mm) with a higher precision of $0.01$ mm - ten times finer than a vernier. Verniers are used for external and internal dimensions; screw gauges for very thin wires, sheets, and small diameters.

Q4. How do you correct for zero error in a screw gauge?

Bring the studs together and note the reading. If the circular zero lies below the reference line, the zero error is positive - subtract it from the observed reading. If it lies above, the error is negative - add its magnitude. The formula is: true reading = observed reading - (zero error with sign).

Q5. Are vernier caliper and screw gauge questions asked in NEET?

Yes. NEET frequently asks numerical questions on least count, zero error correction, and reading calculation for both instruments. These are direct-scoring questions that require careful reading of the instrument diagram and correct application of the reading formula.

Q6. How does JEE Main test measurement topics?

JEE Main typically asks 1-2 questions on measurement, focusing on: least count calculations, reading with zero-error correction, and unit conversions (especially electron volt, atmospheric pressure and astronomical units). Occasionally JEE Advanced combines measurement with error analysis and significant figures.

Q7. What is one astronomical unit (A.U.)?

One astronomical unit is the mean distance from the Earth to the Sun, approximately $1.496 \times 1 0^{11}$ m. It is used to measure distances within the solar system. For interstellar distances, the light year ( $9.46 \times 1 0^{15}$ m) or parsec ( $3.08 \times 1 0^{16}$ m) is used.

Q8. What is 1 electron volt in joules?

One electron volt is the energy gained by an electron accelerated through a potential difference of $1$ volt. $1 eV = 1.6 \times 1 0^{- 19}$ J. This unit is used in atomic and nuclear physics because the joule is too large for such small energies.

Q9. What is the pitch of a screw gauge?

The pitch of a screw gauge is the linear distance moved along the pitch scale per one complete rotation of the thimble. For a standard screw gauge, pitch is typically $1$ mm or $0.5$ mm. The pitch and the number of circular divisions together fix the least count.

Previous year questions on Measurements

38 questions from past papers, each with a step-by-step solution.

Show all 38 questions

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