To Measure The Diameter Of Small Spherical Cylindrical Body Using Vernier Callipers

There are precision instruments - Vernier calipers - that are used for measuring small linear dimensions very accurately. If one is to get the diameter of small spherical or cylindrical body using Vernier calipers, one should grasp the meaning of both main scale and Vernier scale. Calipers include fixed main scale and moving Vernier scale, which makes it possible to carry out precise measurements.

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This Story also Contains

1. Important Terminologies
2. Aim of the Experiment
3. Theory
4. Procedure
5. Observations:
6. Precaution
7. Solved Examples Based on To Measure The Diameter Of Small Spherical Cylindrical Body Using Vernier Callipers
8. Summary

The Vernier caliper is a device which greatly aids in achieving accuracy. In various practical and theoretical problems within the framework of Class 12, NEET, and JEE Main exams, students preparing for such exams must have a thorough understanding as well as perfect mastery over its application because it's among their most important tools. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fourteen questions have been asked on this concept.

A Vernier caliper is an instrument that measures internal or external dimensions and distances. It allows you to take more precise measurements than you could with regular rulers.

Important Terminologies

LEAST COUNT AND ZERO ERROR

The magnitude of the smallest measurement that can be measured by an instrument accurately is called its least count (L.C.).

The difference between one main scale division (M.S.D.) and one vernier scale division is called the least count.

i.e. L.C. = One M.S.D. – One V.S.D.

Zero Error

If there is no object between the jaws (i.e., jaws are in contact), the vernier should give zero reading. But due to some extra material on the jaws, even if there is no object between the jaws, it gives some excess reading. This excess reading is called zero error.

Zero correction: Zero correction is an invert of zero error.

Zero correction = – (Zero error)

Actual reading = observed reading – excess reading (zero error)
= observed reading + zero correction

Aim of the Experiment

To measure the diameter of a small spherical cylindrical body using Vernier Callipers.

Apparatus

Vernier callipers, a spherical body (pendulum bob) or a cylinder and a magnifying lens.

Theory

If with the body between the jaws, the zero of the vernier scale lies ahead of the Nth division of the main scale, then the main scale reading (M.S.R.) = N

If with division of the vernier scale coincides with any division of the main scale, then the vernier scale reading (V.S.R.)

= n × (L.C.) ( Here, L.C. is the least count of vernier callipers)

= n × (V.C.) ( Here, V.C. is vernier constant of vernier callipers)

And total reading, T.R. = M.S.R. + V.S.R. = N + n × (V.C.)

Procedure

1. Determine the vernier constant (V.C.) i.e. least count (L.C.) of the vernier callipers and record it stepwise.
2. Bring the movable jaw BD in close contact with the fixed jaw AC and find zero error. Do it three times and record them. If there is no zero error, record zero error nil.
3. Open the jaws, place the sphere or cylinder between the two jaws A and B and adjust the jaw DB, such that it gently grips the body without any undue pressure on it. Tight the screw S attached to the vernier scale V.

4. Note the position of the zero mark of the vernier scale on the main scale. Record the main scale reading just before the zero mark of the vernier scale. This reading ( 1 ST) is called the main scale reading (M.S.R.).
(Note the number ( n ) of the vernier scale division which coincides with some division of the main scale.
6. Repeat steps 4 and 5 after rotating the body by $90^{\circ}$ for measuring the diameter in a perpendicular direction.

7. Repeat steps 3,4,5 and 6 for three different positions. Record the observations in each set in a tabular form.
8. Find the total reading and apply zero correction.
9. Take the mean of different values of diameter and show that in the result with the proper unit.

Observations:

1. Vernier constant (least count) of the Vernier Callipers:

1 M.S.D. = 1 mm

10 vernier scale divisions = 9 main scale divisions
$
\begin{aligned}
& \text { i.e. } \quad 10 \text { V.S.D. }=9 \text { M.S.D. } \\
& \therefore \quad 1 \mathrm{~V} . \mathrm{S} . \mathrm{D} .=\frac{9}{10} \mathrm{M} . \mathrm{S} . \mathrm{D} . \\
& \text { Vernier Constant (V.C) }=1 \text { M.S.D. }-1 \text { V.S.D. }=1 \text { M.S.D. }-\frac{9}{10} \text { M.S.D. } \\
& =\left(1-\frac{9}{10}\right) \text { M.S.D. }=\frac{1}{10} \times 1 \text { M.S.D. } \\
& =\frac{1}{10} \times 1 \mathrm{~mm}=0.1 \mathrm{~mm}=0.01 \mathrm{~cm}
\end{aligned}
$

2. Zero error: (i) ........ cm (ii) ............cm (iii) ..........cm

Mean Zero Error (e) = ............ cm

Mean Zero Correction (c) = - (Mean Zero Error)

= .......... cm

Precaution

• Make the motion of the vernier scale on the main scale very smooth (by oiling).

• Find the vernier constant and zero error carefully and record them properly.

• Grip the body between the jaws formerly but gently.

• Take at least observation at three different places at right angles at each place.

• No parallax should be there.

Solved Examples Based on To Measure The Diameter Of Small Spherical Cylindrical Body Using Vernier Callipers

Example 1: A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading : 0 mm

Circular scale reading: 52 divisions

Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale.

The diameter (in cm) of wire from the above data is :

1) 0.052

2) 0.52

3) 0.026

4) 0.005

Solution:

Vernier constant (least count) of the Vernier Callipers:

1 M.S.D. = 1 mm

10 vernier scale divisions = 9 main scale divisions

i.e. 10 V.S.D. = 9 M.S.D.
$
\begin{aligned}
& \therefore \quad 1 \text { V.S.D. }=\frac{9}{10} \text { M.S.D. } \\
& \text { Vernier Constant (L.C.) }=1 \text { M.S.D. }-1 \text { V.S.D. }=1 \text { M.S.D. }-\frac{9}{10} \text { M.S.D. } \\
& =\left(1-\frac{9}{10}\right) \text { M.S.D. }=\frac{1}{10} \times 1 \text { M.S.D. } \\
& =\frac{1}{10} \times 1 \mathrm{~mm}=0.1 \mathrm{~mm}=0.01 \mathrm{~cm} \\
& \text { Least count of screw gauge }=\frac{1}{100} \mathrm{~mm}=0.01 \mathrm{~mm}
\end{aligned}
$

$\begin{aligned} & \text { Diameter }=\text { Division on circular scale } \times \text { least count }+ \text { main scale reading } \\ & \qquad=52 \times \frac{1}{100}+100=0.52 \mathrm{~mm} \\ & \text { diameter }=0.052 \mathrm{~cm}\end{aligned}$

Example 2: 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

1) one minute

2) half minute

3) one degree

4) half degree

Solution:

As we learnt in Vernier calliper

Least count $=1$ M.S.D. -1 V.S.D.
Given 29 divisions of main scale coincide with 30 divisions of Vernier scale.

Least count $=1$ M.S.D. -1 V.S.D.

$
\begin{aligned}
& =1 M . S . D .-\frac{29}{30} M . S . D .=\frac{1}{30} M \cdot S \cdot D \\
& =\frac{1}{30} \times 0.5 \text { degree }=1 \text { minute }
\end{aligned}
$
Hence, the answer is the option (1).

$
\begin{aligned}
& \text { so } 29 \mathrm{~V} . \mathrm{S} . \mathrm{D} .=30 \mathrm{M} . \mathrm{S} . \mathrm{D} \text {. } \\
& 1 \mathrm{~V} . \mathrm{S} . \mathrm{D} .=\frac{29}{30} \mathrm{M} . \mathrm{S} . \mathrm{D} \text {. }
\end{aligned}
$

Example 3: The least count of the main scale of vernier calipers is 1 mm. Its vernier scale is divided into 10 divisions and coincides with 9 divisions of the main scale. When jaws are touching each other, the 7^th division of the vernier scale coincides with a division of the main scale, and the zero of the vernier scale is lying right side of the zero of the main scale. When this vernier is used to measure the length of a cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4^th VSD coincides with a main scale division. The length of the cylinder is (in cm) : (VSD is vernier scale division)

1) 3.07

2) 3.21

3) 3.20

4) 2.99

Solution:

$\begin{aligned} & \text { Least count }=1 \mathrm{~mm} \text { or } 0.01 \mathrm{~cm} \\ & \text { Zero error }=0+0.01 \times 7=0.07 \mathrm{~cm} \\ & \text { Reading }=3.1+(0.01 \times 4)-0.07 \\ & =3.1+0.04-0.07 \\ & =3.1-0.03 \\ & =3.07 \mathrm{~cm}\end{aligned}$

Hence, the answer is (3.07).

Example 4: One main scale division of a vernier calipers is 'a' cm and the $\mathrm{n}^{\text {th }}$ division of the vernier scale coincides with $(n-1)^{t h}$ division of the main scale. The least count of the calipers in mm is :

1) $\frac{10 a}{n}$
2) $\left(\frac{n-1}{10 n}\right) a$
3) $\frac{10 n a}{(n-1)}$
4) $\frac{10 a}{(n-1)}$

Solution:

As the $\mathrm{n}^{\text {th }}$ division of the vernier scale coincides with $(n-1)^{\text {th }}$ division of the main scale

$
(\mathrm{n}-1) \mathrm{a}=\mathrm{n}\left(\mathrm{a}^{\prime}\right)
$
So $\quad \Rightarrow a^{\prime}=\frac{(n-1) a}{n}$
Where $\mathrm{a}^{\prime}$ is the One main scale division of vernier calipers.

$
\begin{aligned}
& \text { S0 } \\
& \therefore \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\
& =\left(\mathrm{a}-\mathrm{a}^{\prime}\right) \mathrm{cm} \\
& =\mathrm{a}-\frac{(\mathrm{n}-1) \mathrm{a}}{\mathrm{n}} \\
& =\frac{\mathrm{na}-\mathrm{na}+\mathrm{a}}{\mathrm{n}}=\frac{\mathrm{a}}{\mathrm{n}} \mathrm{cm} \\
& L C=\frac{10 a}{n} \mathrm{~mm}
\end{aligned}
$

Hence, the answer is the Option (1).

Example 5: The vernier scale used for measurement has a positive zero error of 2.0 mm. If while taking a measurement it was noted that '0' on the vernier scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of a measurement is _______ cm. (least count = 0.01 cm)

1) 8.56 cm

2) 8.54 cm

3) 8.58 cm

4) 8.36 cm

Solution:

Positive zero error $=0.2 \mathrm{~mm}$
Main scale reading $=8.5 \mathrm{~cm}$
Vernier scale reading $=6 \times 0.01=0.06 \mathrm{~cm}$
Final reading $=8.5+0.06-0.02=8.54 \mathrm{~cm}$
Hence, the answer is the Option (2).

Summary

To measure the object’s small cylindrical or spherical body using Vernier callipers, one should first adjust the calipers to zero and ensure the jaws are tightly closed so as to avoid any gap appearing between them. The object should then be inserted in between these jaws before closing them softly until their surfaces touch only. At this point, read the main scale and Vernier scale observing where they coincide. This gives you both the main and Vernier scales lengths as is appropriate for this purpose.