Vernier caliper

Measurement by Vernier caliper

A quick guide on how to read a vernier caliper. A vernier caliper outputs measurement readings in centimetres (cm) and it is precise up to 2 decimal places (E.g. 1.23 cm).

Note: The measurement-reading technique described in this post will be similar for vernier calipers which output measurement readings in inches

Measurement Reading Technique For Vernier Caliper

In order to read the measurement readings from vernier caliper properly, you need to remember two things before we start. For example, if a vernier caliper output a measurement reading of 2.13 cm, this means that:

• The main scale contributes the main number(s) and one decimal place to the reading (E.g. 2.1 cm, whereby 2 is the main number and 0.1 is the one decimal place number)
• The vernier scale contributes the second decimal place to the reading (E.g. 0.03 cm)

Let’s examine the image of the vernier caliper readings above. We will just use a two steps method to get the measurement reading from this:

• To obtain the main scale reading: Look at the image above, 2.1 cm is to the immediate left of the zero on the vernier scale. Hence, the main scale reading is 2.1 cm
• To obtain the vernier scale reading: Look at the image above and look closely for an alignment of the scale lines of the main scale and vernier scale. In the image above, the aligned line correspond to 3. Hence, the vernier scale reading is 0.03 cm.

In order to obtain the final measurement reading, we will add the main scale reading and vernier scale reading together. This will give 2.1 cm 0.03 cm = 2.13 cm.

In a nutshell

Use the following formula:

$$\text{Obtained reading}=\text{Main scale reading}\text{Vernier scale reading}$$

Let’s go through another example to ensure that you understand the above steps:

Main scale reading: 10.0 cm (Immediate left of zero)

Vernier scale reading: 0.02 cm (Alignment of scale lines)

Measurement reading: 10.02 cm

Compensating For Zero Error

In a nutshell

Use the following formula:

$\text{Correct reading}=\text{Obtained reading}–\text{Zero error}$

where $\text{zero error}$ can be either negative (the “0” of vernier scale is left of the “0” of the main scale) or positive (the “0” of vernier scale is right of the “0” of the main scale)

Explanation

Now, we shall try with zero error. If you are not familiar on how to handle zero error for vernier calipers, I suggest that you read up on Measurement of Length.

The reading on the top is the measurement obtained and the reading at the bottom is the zero error. Find the actual measurement. (Meaning: get rid of the zero error in the measurement or take into account the zero error)

Measurement with zero error: 3.34 cm

Zero error: – 0.04 cm (negative because the vernier scale is to the left)

Measurement without zero error: $3.34–(–0.04)=3.38$ cm

If you do not understand the subtraction of the negative zero error from the measurement, please read on. Since the zero error is -0.04 cm, this means that all measurements taken by the vernier calipers will be SMALLER by 0.04 cm. Hence, you will have to ADD 0.04 cm to ALL measurements in order to get the TRUE measurement. The subtraction is done in the above case is to have an elegant way of obtaining a resultant addition: $3.340.04=3.38$ and to make it COMPATIBLE with positive zero error. This means that once you have determined the nature of the zero error (positive or negative), you can just subtract the zero error and be sure that your final answer is correct.

Consider a zero error of 0.04 cm. With my method, $3.34–\left(0.04\right)=3.30$ cm.

Normal method: Since the zero error is 0.04 cm, this means that all measurements taken by the vernier calipers will be larger by 0.04 cm. Hence, you will have to SUBTRACT 0.04 cm from ALL measurements in order to get the true measurement. The final calculation will be $3.34–0.04=3.30$ cm, which is the same as my method.

Numerical problems based on Vernier callipers

Question 1.

Diameter of a steel ball is measured using a vernier callipers which has divisions of 0.1cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as :

If the zero error is – 0.03 cm, then mean corrected diameter is:

(1) 0.53 cm

(2) 0.56 cm

(3) 0.59 cm

(4) 0.52 cm

Solution —

MSD = 0.1 cm

VSD = 9/10 MSD = 0.9 × 0.1= 0.09 cm

Least count = MSD – VSD = 0.1 – 0.09 = 0.01 cm

Total reading = Main Scale Reading Vernier coincidence × Least Count

1st reading = 0.5 8 × 0.01 = 0.58 cm

2nd reading = 0.5 4 × 0.01 = 0.54 cm

3rd reading = 0.5 6 × 0.01 = 0.56 cm

Average reading = (0.58 0.54 0.56)/3 = 0.56 cm

Corrected reading = reading – zero error  = 0.56 – (-0.03) = 0.56 0.03 = 0.59 cm

Question 2.

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is:

(1) 5.112 cm

(2) 5.124 cm

(3) 5.136 cm

(4) 5.148 cm

Solution —

Smallest main scale division = 0.05 cm (5.15 – 5.10)

Main scale reading = 5.10 cm (comes just before the zero of the Vernier scale)

Vernier coincidence = 24

Least count =0.05 – 2.45/50 = 0.001 cm

Diameter

= Main Scale Reading Vernier coincidence × Least Count

= 5.10 24 × 0.001 = 5.124 cm

Question 3.

There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2, respectively, are:

(1) 2.87 and 2.86

(2) 2.85 and 2.82

(2) 2.87 and 2.87

(3) 2.87 and 2.83

Solution —

MSD = 1/10 cm = 0.1 cm

For first vernier caliper,

10 VSD = 9 MSD

⇒ VSD = 9/10 MSD = 0.9 × 0.1 = 0.09 cm

Reading

= main scale reading upto coinciding main scale division − n × VSD

= 3.5 – 7 × 0.09 = 3.5 – 0.63 = 2.87 cm

For second vernier caliper,

10 VSD = 11 MSD

⇒ VSD = 11/10 MSD = 1.1 × 0.1 = 0.11 cm

Reading = 3.6 – 7 × 0.11 = 3.6 – 0.77 = 2.83 cm

Why we not used the formula here that we were using in previous problems,

Here is the formula we were using,

Least count = MSD – VSD

Reading = Main Scale Reading Vernier coincidence × Least Count

We derived these formula for Vernier callipers having n divisions of the Vernier scale equal to n-1 divisions of the main scale.

If we apply these formula on first vernier caliper then we will get the same result, but if we apply these formula on the second vernier caliper then we will get negative least count (0.1 − 0.11 = −0.01cm), which is not possible, that is why we cannot apply these formula in this case. 

Question 4.

The main scale of a vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of vernier scale coinsides with a main scale division. Find:

(i). Least count

(ii). Radius of cylinder

Solution —

(i).

MSD = 0.1 cm

20 VSD = 19 MSD

⇒ VSD = 19/20 MSD = 19/20 × 0.1 cm = 0.095cm

Least Count = MSD – VSD = 0.1 – 0.095 cm = 0.005 cm

(ii).

Main scale reading = 35 mm = 3.5 cm

Diameter

= Main scale reading vernier coincidence × Least count

= 3.5 4 × 0.005

= 3.5 0.02

= 3.52 cm

Radius = Diameter/2 = 3.52/2 = 1.76 cm

Question 5.

In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (=0.5°) then the least count of the instrument is:

(A) one minute

(B) half minute

(C) one degree

(D) half degree

Solution —

MSD = 0.5°

30 VSD = 29 MSD

Least Count

= Length of 1 main scale division / No. of divisions of Vernier scale

= 0.5 ° / 30

= 0.5 × 60 min / 30

= 1 min

Question 6.

A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier callipers, the least count is:

(A) 0.02 mm

(B) 0.05 mm

(C) 0.1 mm

(D) 0.2 mm

Solution —

MSD = 1 mm

20 VSD = 16 MSD

⇒ VSD = 16/20 MSD = 0.8 MSD = 0.8 mm

Least Count = MSD – VSD = 1 – 0.8 = 0.2 mm

Question 7.

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

(A) A metre scale

(B) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm

(C) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm

(D) A screw guage having 50 divisions in the circular scale and pitch as 1 mm

Solution —

As the measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm

For vernier callipper with MSD = 1 mm and 9 MSD = 10 VSD,

Least count = 1 MSD – 1 VSD = 0.1 mm

Question 8.

In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n – 1) divisions of the main scale. The least count (in cm) of the callipers is:

Solution —

MSD = x cm

n VSD = (n-1) MSD

⇒ VSD = (n-1)/n × MSD = (n-1)/n × x cm

Least Count

= MSD – VSD

= x – (n-1)/n × x

= x/n

Question 9.

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degree

Vernier scale reading : 09 divisions

Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data

(A) 58.59 degree

(B) 58.77 degree

(C) 58.65 degree

(D) 59 degree

Solution —

MSD = 0.5°

LC = MSD/n = 0.5 / 30 = 1/60

Total Reading

= Main scale reading vernier coincidence × LC

= 58.5 9 × 1/60

= 58.5 0.15

= 58.65°

Question 10.

The jaws of a vernier callipers touch the inner wall of calorimeter without any undue pressure. The position of zero of vernier scale on the main scale reads 3.48. The 6th of vernier scale division is coinciding with any main scale division. Vernier constant of callipers is 0.01 cm. Find actual internal diameter of calorimeter, when it is observed that the vernier scale has a zero error of – 0.03 cm.

(A) 3.37 cm

(B) 3.57 cm

(C) 3.42 cm

(D) 3.54 cm

Solution —

Main scale reading = 3.48 cm

LC = 0.01 cm

Vernier coincidence = 6

Reading

= Main scale reading Vernier coincidence × LC

= 3.48 6 × 0.01

= 3.48 0.06

= 3.54 cm

Corrected reading

= reading – zero error

= 3.54 – (-0.03)

= 3.54 0.03

= 3.57 cm

Question 11.

The thin metallic strip of vernier callipers moves downward from top to bottom in such a way that it just touches the surface of beaker. Main scale reading of calliper is 6.4 cm whereas its vernier constant is 0.1 mm. The 4th of vernier scale division is coinciding with main scale division. The actual depth of beaker is (when zero of vernier coincides with zero of main scale)

(A) 6.64 cm

(B) 6.42 cm

(C) 6.44 cm

(D) 6.13 cm

Solution —

Reading = 6.4 4 × 0.01 cm = 6.44 cm

Question 12.

In an instrument, there are 25 divisions on the vernier scale which coincides with 24th division of the main scale. 1 cm on main scale is divided into 20 equal parts. The least count of the instrument is:

(A) 0.002 cm

(B) 0.05 cm

(C) 0.001 cm

(D) 0.02 cm

Solution —

MSD = 1/20 = 0.05 cm

25 VSD = 24 MSD

⇒ VSD = 24/25 MSD = 24/25 × 0.05 = 0.048 cm

Least Count = MSD – VSD = 0.05 – 0.048 = 0.002 cm

Question 13.

In a vernier calliper, there are 10 divisions on the vernier scale and 1 cm on main scale is divided in 10 parts. While measuring a length, the zero of the vernier scale lies just ahead of 1.8 cm mark and 4th division of vernier scale coincides with a main scale division. The value of length is:

(A) 1.804 cm

(B) 1.840 cm

(C) 1.800 cm

(D) None of these

Solution —

Length of 1 main scale division = 1/10 = 0.1 cm

No. of vernier scale divisions = 10

Main scale reading = 1.8 cm

Coinciding vernier scale division = 4th

Least Count

= Length of smallest main scale division / No. of vernier scale divisions

= 0.1/10

= 0.01 cm

(vernier calliper has a least count 0.01 cm. So, measurement is accurate only upto

three significant figures.)

length

= Main scale reading Coinciding vernier scale division × LC

= 1.8 4 × 0.01

= 1.8 0.04

= 1.84 cm