How to Calculate Least Count , Zero Correction of Vernier Callipers & Screw Gauge ?

Measurement of Length :

The simplest method measuring the length of a straight line is by means of a meter scale. But there exists some limitation in the accuracy of the result:

(i) the dividing lines have a finite thickness.

(ii) naked eye cannot correctly estimate less than 0.5 mm

For greater accuracy , devices like

(a) Vernier callipers

(b) micrometer scales (screw gauge) are used .

Vernier Callipers :

It consists of a main scale graduated in cm/mm over which an auxiliary scale (or Vernier scale) can slide along the length. The division of the Vernier scale being either slightly longer and shorter than the divisions of the main scale.

How to Calculate Least count of Vernier Callipers ? 

The least count or Vernier constant (v.c.) is the minimum value of correct estimation of length without eye estimation.
If N division of vernier coincides with (N-1) division of main scale, then

$\large 1 VSD = \frac{N-1}{N} MSD$

( Where VSD = Vernier scale division & MSD = Main scale division )

Least Count or Vernier constant is

L.C = 1 MSD – 1 VSD

$ \displaystyle = (1-\frac{N-1}{N}) MSD $

$ \displaystyle = \frac{1}{N} MSD$

, which is equal to the value of the smallest division on the main scale divided by total number of divisions on the vernier scale.

What is the Zero error of Instrument ?

If the zero marking of main scale and vernier callipers do not coincide, necessary correction has to be made for this error which is known as zero error of the instrument.
Positive & Negative Zero Error : If the zero of the vernier scale is to the right of the zero of the main scale the zero error is said to be positive and the correction will be negative and vice versa.

Solved Example : Consider the following data :
10 main scale division = 1cm, 10 vernier division = 9 main scale divisions , zero of vernier scale is to the right of the zero marking of the main scale with 6 vernier divisions coinciding with main scale divisions and the actual reading for length measurement is 4.3 cm with 2 vernier divisions coinciding with main scale graduations. Estimate the length .

Solution: In this case, vernier constant = (1mm/10) = 0.1 mm

Zero error = 6 × 0.1 = + 0.6 mm

Correction = -0.6 mm

Actual length = (4.3 + 2 × 0.01) + correction

= 4.32 – 0.06 = 4.26 cm

Solved Example :  When the two jaws of a vernier caliper are in touch, zero of vernier scale lies to the right of zero of main scale and coinciding vernier division is 3. If vernier constant is 0.1 mm, What is the zero correction ?

Solution: As zero of Vernier scale is to the right of zero of main scale, therefore, zero correction is negative. Its magnitude = 3 × 0.1 mm = 0.3 mm = 0.03 cm

Solved Example : The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the least count of instrument .

Solution :  least count of instrument = 1 main scale division – 1 vernier scale division

$\large L.C = 0.5 mm – \frac{49}{50} (0.5 mm) $

= 0.01 mm

Screw Gauge (or Micrometer Screw)

In general vernier callipers can measure accurately upto 0.01 cm and for greater accuracy micrometer screw devices e.g. screw gauge, spherometer are used. These consist of accurately cut screw which can be moved in a closely fitting fixed nut by turning it axially. The instrument is provided with two scales:


(i) The main scale or pitch scale M graduated along the axis of the screw.
(ii) The cap-scale or head scale H round the edge of the screw head.

Constants of the Screw Gauge :

(a) Pitch : The translational motion of the screw is directly proportional to the total rotation of the head. The pitch of the instrument is the distance between two consecutive threads of the screw which is equal to the distance moved by the screw due to one complete rotation of the cap.
Thus for 10 rotation of cap ≡ 5 mm, pitch = 0.5 mm

(b) Least count: In this case also, the minimum (or least) measurement (or count) of length is equal to one division on the head scale which is equal to pitch divided by the total cap divisions.
Thus in the aforesaid Illustration, if the total cap division is 100, then least count = 0.5 mm/100 = 0.005 mm

$\large Least \; Count = \frac{Pitch}{Total \; no. \; of \; divisions \; on \; circular \; scale}$

Zero Error: In a perfect instrument the zero of the main scale coincides with the line of graduation along the screw axis with no zero-error, otherwise the instrument is said to have zero-error which is equal to the cap reading with the gap closed.

Positive & negative Zero Error : This error is positive when zero line or reference line of the cap lies above the line of graduation and vice-versa. The corresponding corrections will be just opposite

Solved Example : The circular scale of a screw gauge has 200 divisions. When it is given 4 complete rotations, it moves through 2 mm. What is the least count of screw gauge ?

Solution: $\large Pitch = \frac{2}{4} mm = 0.5 mm $

$\large Least \; Count = \frac{Pitch}{200} = \frac{0.5}{200}$

= 0.25 × 10-2 mm

0.25 × 10-3 cm

Solved Example: While measuring diameter of a wire using a screw gauge, the main scale reading is 7 mm and zero of circular scale is 35 division above the reference line. If screw gauge has a zero error of -0.003 cm, What is the correct diameter of wire ? (given least count =0.001 cm)

Solution: Observed diameter = 0.7 cm + 35 (0.001) cm

= 0.735 cm

Corrected diameter

=(0.735 + 0.003) cm = 0.738 cm

Solved Example: When a screw gauge is completely closed, zero of circular scale is 4 divisions below the reference line of graduation. If least count of screw gauge is 0.001 cm, What is the zero correction ?

Solution: As zero of circular scale is below the reference line of graduation, zero correction is negative.

Its magnitude is 4 × 0.001 cm = 0.004 cm

How to Calculate Speed of Sound using Resonance Column ?

A tuning fork of known frequency (f) is held at the mouth of a long tube, which is dipped into water as shown in the figure.

The length (l1) of the air column in the tube is adjusted until it resonates with the tuning fork. The air temperature and humidity are noted.

The length of the tube is adjusted again until a second resonance length (l2) is found (provided the tube is long).

Then, l2 – l1 = λ/2, provided l1 , l2 are resonance lengths for adjacent resonances.

∴ λ = 2( l2 – l1), is the wavelength of sound.

Since the frequency f , is known; the velocity of sound in air at the temperature (θ) and humidity (h) is given by v = f λ = 2(l2 – l1)f

It is also possible to use a single measurement of the resonant length directly, but, then it has to be corrected for the “end effect”:

λ(fundamental) = 4( l1 + 0.3 d ), where d = diameter

Errors: The major systematic errors introduced are due to end effects in (end correction) and also due to excessive humidity.

Random errors are given by

$\Large \frac{\delta c}{c} = \frac{\delta (l_2 – l_1)}{l_2 – l_1} = \frac{\delta l_2 + \delta l_1}{l_2 – l_1} $

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