Vernier Calipers Made Easy for Students:
Supplementary Notes for IIT JEE Physics Book
Jitender Singh
Shraddhesh Chaturvedi
August 20, 2017
Contents
1 How much we know about the Vernier calipers?
Measurement is a fundamental part of all scientific experiments, including
Physics. At one extreme, our vast universe extends this measuring exercise to
lightyears, the distances so vast that we cannot see them from our eyes. On
the other extreme of the smallest distances, new discoveries are pushing it
down to femtometer (10^{15} m) or even less. These distances are so small
that we cannot see them from our eyes. At every 1  2 order of magnitude
change in distance, our instruments to measure the distances accurately
can differ. When the distances, we want to measure, are in the range of
10^{2} mm to 1 mm, we use Vernier Calipers and Screw Gauge for precise
measurement. In this article, we are going to focus on these measuring instruments
only.
In class 11^{th} Physics lab, we were trained to answer the following questions:
 How to find the Least Count (LC) or Vernier Constant?
 How to read Main Scale Reading (MSR) and Vernier Scale Reading
(VSR)?
 How to find the zero error?
 How to use the above data to get the final measurement?
Answers to these how to questions kept us content for 22 years. But this was not
adequate to solve the IIT JEE 2016 problem. We needed to figure out some more
interesting why questions like:
 Why least count is given by
 LC =   
 Why the measured value is given by
 Observed Value = MSR + LC × VSR   
 Why the zero error is subtracted from the observed value i.e.,
 True Value = Observed Value  Zero Error   
Let us start our journey with a Vernier calipers without zero error. When two
jaws are closed, 0^{th} mark on the Vernier scale is aligned with the 0^{th} mark on the
main scale as shown in the figure 1. Also note that 10^{th} mark on Vernier scale
coincides with the 9^{th} mark on main scale.
One main scale division (MSD) is the distance between two successive
marks on the main scale. It is given in the figure 1 that 1 MSD is equal
to 1mm. One Vernier scale division (VSD) is the distance between two
successive marks on the Vernier scale. It is given that 10VSD = 9MSD. Thus,
1VSD = (9∕10)MSD = (9∕10)1mm = 0.9mm i.e., distance between two successive
marks on the Vernier scale is 0.9mm.
Now, let us use this Vernier calipers to measure the diameter D of a marble ball
(marble balls usually have D = 1∕2in). The measurement is shown in the figure 2.
Here, x_{m0} is the distance between the jaw attached to main scale (left jaw) and the
0^{th} mark on the main scale and x_{v0} is the distance between the jaw attached to
Vernier scale (right jaw) and the 0^{th} mark on the Vernier scale. Observe that the
diameter is given by
where x is the distance between the 0^{th} mark on the main scale and the right
jaw.
Now, note that the 7^{th} mark on Vernier scale coincides with 1.9cm on the main
scale. This point is called the point of coincidence. The distance between the 0^{th}
mark on main scale and the point of coincidence is x_{m} and the distance
between the 0^{th} mark on Vernier scale and the point of coincidence is x_{v}.
Thus,
 x + x_{v0} + x_{v} = x_{m}  (2) 
Substitute x from equation 2 into equation 1 to get
 D = (x_{m}  x_{v})  (x_{m0}  x_{v0})  (3) 
The quantity (x_{m0} x_{v0}) is called the zero error of the Vernier calipers. Negative
of the zero error is called zero correction. Note that x_{m0} = x_{v0} in a Vernier calipers
without zero error (which is true in this case). Also, x_{m} = 19MSD = 19mm and
x_{v} = 7VSD = 7(9∕10) = 6.3mm. Substitute these values in equation 3 to get
D = 12.7mm = 1.27cm (half inch marble).
There is another easier way to get the measured value. The main scale reading
(MSR) is the first reading on the main scale immediately to the left of the zero of
Vernier scale (MSR = 12mm in this example). The Vernier scale reading (VSR) is
the mark on Vernier scale which exactly coincides with a mark on the main
scale (VSR = 7 in this example). Note that there are VSR divisions on the
main scale between MSR mark (i.e., mark on the main scale immediately
to the left of the zero of the Vernier scale) and the point of coincidence.
Thus,
 x_{v} = VSR × VSD,  (4)

 x_{m} = MSR + VSR × MSD.  (5) 
Corresponding values of the parameters for the zero errors are
 x_{v0} = VSR_{0} × VSD,  (6)

 x_{m0} = MSR_{0} + VSR_{0} × MSD.  (7) 
Substitute in equation 3 to get
D =    

  

=    

=    

=  MSR + VSR × LC  ZeroError.  (8) 
where LC = MSD  VSD is called the least count or Vernier constant. It is the
smallest length that can be measured accurately with a Vernier calipers. For the
given Vernier calipers
 LC = MSD  VSD = 1mm  9∕10mm = 0.1mm,  

 D = MSR + VSR × LC = 12mm + 7 × (0.1) = 12.7mm = 1.27cm.   
Note that Vernier calipers can be used to measure (1) outer dimensions like
diameter of a sphere or edge of a cube (2) inner dimensions like inner diameter of a
hollow cylinder and (3) depth of a hollow cylinder.
2 Worked Out Examples
Example 1: The jaws of the Vernier calipers shown in figure 3 are in contact with
each other. Find the zero error of this Vernier calipers.
Solution: The least count of given Vernier calipers is
 LC = MSD  VSD = 1  (9∕10) = 0.1mm.   
The main scale reading is MSR_{0} = 0mm and the Vernier scale reading is
VSR_{0} = 3. Thus,
 ZeroError = MSR_{0} + VSR_{0} × LC = 0 + 3 × 0.1 = 0.3mm.   
Example 2: The Vernier calipers of example 1 is used to measure the edge of a
cube. The readings are shown in the figure 4. Find the edge length of the
cube.
Solution: The readings are MSR = 25mm and VSR = 7. Thus,
a  = MSR + VSR × LS  ZeroError  

 = 25 + 7(0.1)  0.3 = 25.4mm = 2.54cm.   
Example 3: The jaws of the Vernier calipers, shown in figure 5, are in contact with
each other. Find the zero error of this Vernier calipers.
Solution: This is an interesting problem. What is MSR_{0}? It is the first reading on
the main scale immediately to the left of the zero of the Vernier scale. But there are
no marks on the main scale before zero of the Vernier scale. We claim that
MSR_{0} = 1mm (observe it carefully, why MSR_{0} is not equal to 2mm?). The
Vernier scale reading is VSR_{0} = 4 and the least count is LC = 0.1mm. Substitute
these values to get,
 ZeroError = MSR_{0} + VSR_{0} × LC = 1 + 4 × 0.1 = 0.6mm.   
Example 4: The Vernier calipers of example 3 is used to measure the edge of a
cube. The readings are shown in the figure 6. Find the edge length of the
cube.
Solution: The readings are MSR = 24mm and VSR = 8. Thus,
a  = MSR + VSR × LS  ZeroError  

 = 24 + 8(0.1)  (0.6) = 25.4mm = 2.54cm.   
Example 5: What is the LC of the Vernier calipers shown in the figure 7?
Solution: One main scale division is 1MSD = 1mm. Since 5VSD = 4MSD, we
get 1 VSD = (4∕5)MSD = 0.8mm. Thus, the least count of this calipers
is
LC  = MSD  VSD = 1.0  0.8 = 0.2mm.   
3 IIT JEE Solved Problems
Question. N divisions on the main scale of a Vernier calipers coincide with (N + 1)
divisions on its Vernier scale. If each division on the main scale is of a units,
determine the least count of instrument. (2003)
Solution. Given, a main scale division (MSD) of the Vernier calipers is
Since (N + 1) Vernier scale divisions (VSD) are equal to N main scale divisions,
we get
 1VSD = MSD = .  (2) 
The least count is given by
 LC = 1MSD  1VSD = .   
Question. The edge of a cube is measured using a Vernier calipers (9 divisions of
the main scale are equal to 10 divisions of Vernier scale and 1 main scale
division is 1mm). The main scale division reading is 10 and first division of
Vernier scale was found to be coinciding with the main scale. The mass of
the cube is 2.736g. Calculate the density in g∕cm^{3} up to correct significant
figures. (2005)
Solution. From the given data, one main scale division (MSD) is
Since 10 Vernier scale divisions (VSD) are equal to 9MSD, we get
The least count (LC) is given by
 LC = 1MSD  1VSD = 1.0  0.9 = 0.1mm.  (3) 
Given, the main scale reading (MSR) is 10 and the Vernier scale reading (VSR) is
1. The measured value of the edge is given by
 a = MSR × MSD + VSR × LC = 10 × 1 + 1 × 0.1 = 10.1mm.   
The measurement of a has three significant digits. The volume of the cube is
V = a^{3} = 1.03cm^{3} and the density is
 m∕V = 2.736∕1.03 = 2.6563 = 2.66g∕cm^{3},   
(after rounding off to three significant digits).
Question. A Vernier calipers has 1mm marks on the main scale. It has 20 equal
divisions on the Vernier scale which match with 16 main scale divisions. For this
Vernier calipers, the least count is (2010)
 0.02mm
 0.05mm
 0.1mm
 0.2mm
Solution. For the given Vernier calipers, one main scale division (MSD)
is
Since 20 Vernier scale divisions (VSD) are equal to 16MSD, we get
 1VSD = 16∕20 = 0.8mm.  (2) 
The least count (LC) is given by
 LC = 1MSD  1VSD = 1.0  0.8 = 0.2mm.   
Question. The diameter of a cylinder is measured using a Vernier calipers with no
zero error. It is found that the zero of the Vernier scale lies between 5.10cm and
5.15cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45cm.
The 24^{th} division of the Vernier scale exactly coincides with one of the main scale
divisions. The diameter of the cylinder is (2013)
 5.112cm
 5.124cm
 5.136cm
 5.148cm
Solution. From the given data, one main scale division (MSD) and one Vernier scale
division (VSD) are
 1MSD = 5.15cm  5.10cm = 0.05cm,  (1)

 1VSD = cm = 0.049cm.  (2) 
The least count (LC) of the given Vernier calipers is
 LC = 1MSD  1VSD = 0.001cm.  (3) 
For the given measurement, main scale reading (MSR) is 5.10cm and the
Vernier scale reading (VSR) is 24. Hence, the diameter D of the cylinder
is
 D = MSR + VSR × LC = 5.10 + 24 × 0.001 = 5.124cm.   
Question. Consider a Vernier calipers in which each 1cm on the main scale is
divided into 8 equal divisions and a screw gauge with 100 divisions on its
circular scale. In the Vernier calipers, 5 divisions of the Vernier scale coincide
with 4 divisions on the main scale and in the screw gauge, one complete
rotation of the circular scale moves it by two divisions on the linear scale.
Then, (2015)
 If the pitch of the screw gauge is twice the least count of the Vernier
calipers, the least count of the screw gauge is 0.01mm.
 If the pitch of the screw gauge is twice the least count of the Vernier
calipers, the least count of the screw gauge is 0.005mm.
 If the least count of the linear scale of the screw gauge is twice the least
count of the Vernier calipers, the least count of the screw gauge is 0.01mm.
 If the least count of the linear scale of the screw gauge is twice the
least count of the Vernier calipers, the least count of the screw gauge is
0.005mm.
Solution. In given Vernier calipers, each 1cm is equally divided into 8 main scale
divisions (MSD). Thus, 1MSD = 1∕8 = 0.125cm. Further, 4 main scale divisions
coincide with 5 Vernier scale divisions (VSD) i.e., 4MSD = 5VSD. Thus,
1VSD = 4∕5MSD = 0.8 × 0.125 = 0.1cm. The least count of the Vernier calipers is
given by
 LC = 1MSD  1VSD = 0.125  0.1 = 0.025cm.  (1) 
In screw gauge, let l be the distance between two adjacent divisions on the linear
scale. The pitch p of the screw gauge is the distance travelled on the linear scale
when it makes one complete rotation. Since circular scale moves by two divisions on
the linear scale when it makes one complete rotation, we get p = 2l. The least count
of the screw gauge is defined as ratio of the pitch to the number of divisions on the
circular scale (n) i.e.,
 lc = p∕n = 2l∕100 = l∕50.  (2) 
If p = 2LC = 2(0.025) = 0.05cm, then l = p∕2 = 0.025cm. Substitute l in
equation 2 to get the least count of the screw gauge
 lc = 0.025∕50 = 5 × 10^{4}cm = 0.005mm.  (3) 
If l = 2LC = 2(0.025) = 0.05cm then equation 2 gives
 lc = 0.05∕50 = 1 × 10^{3}cm = 0.01mm.  (4) 
Question. There are two Vernier calipers both of which have 1cm divided into 10
equal divisions on the main scale. The Vernier scale of one of the calipers (C_{1}) has
10 equal divisions that correspond to 9 main scale divisions. The Vernier
scale of the other caliper (C_{2}) 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 C_{1} and C_{2}, respectively
are (2016)
 2.85 and 2.82
 2.87 and 2.83
 2.87 and 2.86
 2.87 and 2.87
Solution. In both calipers C_{1} and C_{2}, 1cm is divided into 10 equal
divisions on the main scale. Thus, 1 division on the main scale is equal to
x_{m1} = x_{m2} = 1cm∕10 = 0.1cm. In calipers C_{1}, 10 equal divisions on the Vernier scale
are equal to 9 main scale divisions. Thus, 1 division on the Vernier scale of C_{1} is
equal to x_{v1} = 9x_{m1}∕10 = 0.09cm. In calipers C_{2}, 10 equal divisions on the Vernier
scale are equal to 11 main scale divisions. Thus, 1 division on the Vernier scale of C_{2}
is equal to x_{v2} = 11x_{m2}∕10 = 0.11cm.
Let main scale reading be MSR and v^{th} division of the Vernier scale coincides with m^{th}
division of the main scale (m is counted beyond MSR). The value measured by this
calipers is
 X = MSR + x = MSR + mx_{m}  vx_{v}.  (1) 
In calipers C_{1}, MSR_{1} = 2.8cm, m_{1} = 7 and v_{1} = 7 and in calipers C_{2},
MSR_{2} = 2.8cm, m_{2} = 8 and v_{2} = 7. Substitute these values in equation 1 to
get
 X_{1} = MSR_{1} + m_{1}x_{m1}  v_{1}x_{v1} = 2.8 + 7(0.1)  7(0.09) = 2.87cm.  

 X_{2} = MSR_{2} + m_{2}x_{m2}  v_{2}x_{v2} = 2.8 + 8(0.1)  7(0.11) = 2.83cm.   
The Vernier calipers are generally of type C_{1} having m = v and least count
LC = x_{m}  x_{v}. For these calipers, equation 1 gives X = MSR + v × LC.
Question. During Searle’s experiment, zero of the Vernier scale lies between
3.20 × 10^{2}m and 3.25 × 10^{2}m of the main scale. The 20^{th} division of the Vernier
scale exactly coincides with one of the main scale divisions. When an additional load
of 2kg is applied to the wire, the zero of the Vernier scale still lies between
3.20 × 10^{2}m and 3.25 × 10^{2}m of the main scale but now 45^{th} division of Vernier
scale coincides with one of the main scale divisions. The length of the thin metallic
wire is 2m and its crosssectional area is 8 × 10^{7}m^{2}. The least count of the Vernier
scale is 1.0 × 10^{5}m. The maximum percentage error in the Young’s modulus of the
wire is ……. (2014)
Solution. The difference between the two measurements by Vernier scale gives
elongation of the wire caused by the additional load of 2kg. In the first measurement,
main scale reading is MSR = 3.20 × 10^{2}m and Vernier scale reading is VSR = 20.
The least count of Vernier scale is LC = 1 × 10^{5}m. Thus, the first measurement by
Vernier scale is
L_{1}  = MSR + VSR × LC  

 = 3.20 × 10^{2} + 20(1 × 10^{5}) = 3.220 × 10^{2}m.  (1) 
In the second measurement, MSR = 3.20 × 10^{2}m and VSR = 45. Thus, the
second measurement by Vernier scale is
 L_{2} = 3.20 × 10^{2} + 45(1 × 10^{5}) = 3.245 × 10^{2}m.  (2) 
The elongation of the wire due to force F = 2g is
 l = L_{2}  L_{1} = 0.025 × 10^{2}m.  (3) 
The maximum error in the measurement of l is Δl = LC = 1 × 10^{5}m. Young’s
modulus is given by Y = . The maximum percentage error in the measurement of
Y is
 × 100 = × 100 = × 100 = 4%.   
The readers are encouraged to compute the values of Y and ΔY . Hint:
Y = 2 × 10^{11}N∕m^{2} and ΔY = 0.08 × 10^{11}N∕m^{2}.
4 Exercise Problems
Problem 1. The jaws of the Vernier calipers shown in figure 8 are in contact with
each other. Find the zero error of this Vernier calipers. Ans.
1.9mm
Problem 2. The jaws of the Vernier calipers shown in figure 9 are in contact with
each other. Find the zero error of this Vernier calipers. Ans.
1.2mm
Problem 3. The zero error of the Vernier calipers shown in figure 10 is
0.9mm. What is the diameter of the sphere being measured in figure 10.
Ans.
3.14cm
Problem 4. The zero error of the Vernier calipers shown in figure 11 is
0.5mm. What is the diameter of the sphere being measured in figure 11.
Ans.
3.14cm
Problem 5. Choose the wrong statement for zero error and zero correction
 If the zero of the Vernier scale does not coincide with the zero of the main
scale then the instrument is said to be having a zero error.
 Zero error is positive when the zero of Vernier scale lies to the left of the
zero of the main scale.
 Zero correction has a magnitude equal to zero error but sign is opposite
to that of the zero error.
 None of the above is wrong.
Ans. (B)
Problem 6. What is Vernier constant?
 It is the value of one main scale division divided by the total number of
divisions on the main scale.
 It is the value of one Vernier scale division divided by the total number of
divisions on the Vernier scale.
 It is the difference between value of one main scale division and one Vernier
scale division.
 It is also the least count of the Vernier scale.
Ans. (C), (D)
Problem 7. The smallest division on main scale of a Vernier calipers is 1mm and
10 Vernier scale division coincide with 9 main scale divisions. While measuring the
length of a line, the zero mark of the Vernier scale lies between 10.2cm and 10.3cm
and the third division of Vernier scale coincide with a main scale division.
(a) Determine the least count of the calipers, and (b) Length of the line.
Ans. 0.01cm, 10.23cm
Problem 8. The main scale of a Vernier calipers 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 coincides with a main scale division.
Find (a) least count of the Vernier calipers and (b) radius of the cylinder.
Ans. 0.005cm, 1.76cm
Problem 9. Least count of a Vernier calipers is 0.01cm. When the two jaws of the
instrument touch each other the 5^{th} division of the Vernier scale coincide with a main
scale division and the zero of the Vernier scale lies to the left of the zero of the main
scale. Furthermore while measuring the diameter of a sphere, the zero mark of the
Vernier scale lies between 2.4cm and 2.5cm and the 6^{th} Vernier division
coincides with a main scale division. Calculate the diameter of the sphere.
Ans. 2.51cm
Problem 10. 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 halfadegree (= 0.5°)
then the least count of the instrument is AIEEE
2009
 one minute
 half minute
 one degree
 half degree
Ans.
(A)
Problem 11. A student measured the length of a rod and wrote it as 3.50 cm.
Which instrument did he use to measure it?
 A meter scale
 A Vernier caliper where the 10 divisions in Vernier scale matches with
9 division in main scale and main scale has 10 divisions in 1 cm
 A screw gauge having 100 divisions in the circular scale and pitch as
1 mm
 A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
Ans.
(B)
Problem 12. 19 divisions on the main scale of a Vernier calipers coincide with
20 divisions on the Vernier scale. If each division on the main scale is of 1cm,
determine the least count of instrument.
Ans. 0.05cm
Problem 13. The angle of a prism is measured by a spectrometer. The main scale
reading is 58.5 degree and Vernier scale reading is 9 divisions. Given that 1 division
on main scale corresponds to 0.5 degree and 30 divisions on the Vernier scale match
with 29 divisions on the main scale. The angle of the prism from the above data
is (AIEEE
2012)
 58.59 degree
 58.77 degree
 58.65 degree
 59 degree
Ans.
(C)
Problem 14. 1cm on the main scale of a Vernier calipers is divided into 10 equal
parts. If 10 divisions of Vernier coincides with 8 small divisions of main scale, then
the least count of the calipers is
 0.01cm
 0.05cm
 0.005cm
 0.02cm
Ans. (D)
Problem 15. The jaws of a Vernier calipers touch the inner wall of calorimeter
without any undue pressure. The position of zero of Vernier scale on the main scale
reads 3.48 cm. The 6th division of Vernier scale division coincides with a main scale
division. Vernier constant of calipers is 0.01 cm. Find actual internal diameter of
calorimeter, when it is observed that the Vernier scale has a zero error of
0.03cm.
 3.37 cm
 3.57 cm
 3.42 cm
 3.54 cm
Ans.
(B)
Problem 16. In a travelling microscope 1 cm on main scale is divided into
20 equal divisions and there are 50 divisions on the Vernier scale. What is the least
count of microscope? Ans.
0.001 cm
Problem 17. The Vernier constant of a travelling microscope is 0.001cm. If
49 main scale divisions coincide with 50 Vernier scale divisions, then the value of
1 main scale division is
 0.1mm
 0.4mm
 0.5mm
 1mm
Ans. (C)
Problem 18. The thin metallic strip of Vernier calipers moves downward from top
to bottom in such a way that it just touches the surface of beaker. Main
scale reading of caliper is 6.4 cm whereas its Vernier constant is 0.1 mm.
The 4th division of Vernier scale coincides with a main scale division. The
actual depth of beaker is (when zero of Vernier coincides with zero of main
scale)
 6.64 cm
 6.42 cm
 6.44 cm
 6.13 cm
Ans. (C)
Problem 19. In an instrument, there are 25 divisions on the Vernier scale which
coincides with 24 divisions of the main scale. 1 cm on main scale is divided into
20 equal parts. The least count of the instrument is
 0.002 cm
 0.05 cm
 0.001 cm
 0.02 cm
Ans.
(A)
Problem 20. 1cm of main scale of a Vernier calipers is divided into 10 divisions. If
least count of the calipers is 0.005cm, then the Vernier scale must have
 10 divisions
 20 divisions
 25 divisions
 50 divisions
Ans. (B)
Problem 21. In a Vernier calipers, 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
 1.804 cm
 1.840 cm
 1.800 cm
 None of these
Ans. (D)
Problem 22. Diameter of a steel ball is measured using a Vernier calipers which
has divisions of 0.1 cm 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 (1) MSR = 0.5cm, VSD = 8 (2) MSR = 0.5cm, VSD = 4 (3)
MSR = 0.5cm, VSD = 6. If the zero error is 0.03cm, then mean corrected diameter
is
 0.53 cm
 0.56 cm
 0.59 cm
 0.52 cm
Ans. (C)
Problem 23. Each division on the main scale is 1mm. Which of the following
Vernier scales gives Vernier constant equal to 0.01mm?
 9mm divided into 10 divisions
 90mm divided into 100 divisions
 99mm divided into 100 divisions
 9mm divided into 100 divisions
Ans. (C)
Problem 24. In a Vernier calipers 1cm of the main scale is divided into 20 equal
parts. 19 divisions of the main scale coincide with 20 divisions on the Vernier scale.
Find the least count of the instrument. Ans. 0.025cm
Problem 25. In a vernier calipers, 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 calipers is (AMU PMT
2009)
 x


Ans. (C)
5 More…
You can make a low cost Vernier calipers by drawing main scale and Vernier scale on
strips of paper etc. The lines on main scale may be separated by 1cm. To get lines on
the Vernier scale, you can divide 9 divisions on main scale into 10 equal divisions (it
is a test for your skills in geometry).
Take a scales of 30 cm length and another scale of 15 cm length. The bigger scale
is main scale of your Vernier calipers and smaller scale is Vernier scale. Draw 30
parallel lines on main scale at 1 cm distance each. Draw 10 parallel lines on
Vernier scale at 0.9 cm each. Your Vernier caliper is ready (see figure 12).
Take another scale (or any other rectangular piece) as support. The object,
whose length is to measured, is placed between the support and Vernier scale.
A Vernier thruster is a rocket engine used on a spacecraft for fine adjustments to
the velocity of a spacecraft. The name is derived from Vernier calipers (named after
Pierre Vernier) which have a primary scale for gross measurements, and a secondary
scale for fine measurements.
References