Measurement is a fundamental part of all scientific experiments, including Physics. At one extreme, our vast universe extends this measuring exercise to light-years, 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 femto-meter (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:
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 1 mm. One Vernier scale division (VSD) is the distance between two successive marks on the Vernier scale. It is given that 10 VSD = 9 MSD. Thus, 1 VSD = (9/10) MSD = 0.9 mm i.e., distance between two successive marks on the Vernier scale is 0.9 mm.
Now, let us use this Vernier calipers to measure the diameter D of a marble ball (marble balls usually have D = 1/2 in). 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 \begin{align} D=x_{m0}+x \end{align} 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 1.9 cm 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, \begin{align} x+x_{v0}+x_v=x_m \end{align} Substitute x from above equation into the first equation to get \begin{align} D=(x_m-x_v)-(x_{m0}-x_{v0}) \end{align}
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, $x_m=19\mathrm{MSD}=19$ mm and $x_v=7\mathrm{VSD}=7(9/10)=6.3$ mm. Substitute these values in above equation to get D = 12.7 mm =1.27 cm (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 = 12 mm 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, \begin{align} x_v&=\mathrm{VSR}\times\mathrm{VSD},\\ x_m&=\mathrm{MSR}+\mathrm{VSR}\times\mathrm{MSD} \end{align} Corresponding values of the parameters for the zero errors are \begin{align} x_{v0}&=\mathrm{VSR_0}\times\mathrm{VSD},\\ x_{m0}&=\mathrm{MSR_0}+\mathrm{VSR_0}\times\mathrm{MSD} \end{align} Substitute in the expression for D to get \begin{align} D=\mathrm{MSR}+\mathrm{VSR}\times\mathrm{LC}-\mathrm{Zero\,Error} \end{align} 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 \begin{align} \mathrm{LC}&=\mathrm{MSD}-\mathrm{VSD} \nonumber\\ &=1-9/10=0.1\,\mathrm{mm}, \\ D&=\mathrm{MSR}+\mathrm{VSR}\times\mathrm{LC}\nonumber\\ &=12+7(0.1)=1.27\,\mathrm{mm} \end{align}
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.
The jaws of the Vernier calipers shown in figure are in contact with each other. Find the zero error of this Vernier calipers.
Solution: The least count of given Vernier calipers is \begin{align} \mathrm{LC}&=\mathrm{MSD}-\mathrm{VSD} \\ &=1-9/10=0.1\,\mathrm{mm} \end{align}
The main scale reading is MSR_{0} = 0 mm and the Vernier scale reading is VSR_{0} = 3. Thus, \begin{align} \mathrm{Zero\, Error}&=\mathrm{MSR_0}+\mathrm{VSR_0}\times\mathrm{LC} \\ &=0+3\times 0.1=0.3\,\mathrm{mm} \end{align}
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 = 25 mm and VSR = 7. Thus, \begin{align} a&=\mathrm{MSR}+\mathrm{VSR}\times\mathrm{LC}- \mathrm{Zero\, Error} \\ &=25+7\times 0.1-0.3=25.4,\mathrm{mm}. \end{align}
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} = -1 mm (observe it carefully, why MSR_{0} is not equal to -2 mm?. The Vernier scale reading is VSR_{0} = 4 and the least count is LC = 0.1 mm. Substitute these values to get, \begin{align} \mathrm{Zero\, Error}&=\mathrm{MSR_0}+\mathrm{VSR_0}\times\mathrm{LC}\\ &=-1+4\times 0.1=-0.6\,\mathrm{mm}. \end{align}
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 = 24 mm and VSR = 8. Thus, \begin{align} a&=\mathrm{MSR+VSR\times LC-Zero\,Error} \\ &=24+8(0.1)-(-0.6)=25.4\,\mathrm{mm}. \end{align}
What is the LC of the Vernier calipers shown in the figure 7?
Solution: One main scale division is 1 MSD = 1 mm. Since 5 VSD = 4 MSD, we get 1 VSD = (4/5) MSD = 0.8. Thus, the least count of this calipers is LC = MSD - VSD = 1 - 0.8 = 0.2 mm.
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.
Solution: Given, a main scale division (MSD) of the Vernier calipers is $N=a$. Since (N + 1) Vernier scale divisions (VSD) are equal to N main scale divisions, we get \begin{align} 1\,\mathrm{VSD}=\frac{N}{N+1}\mathrm{MSD}=\frac{Na}{N+1} \end{align} The least count is given by \begin{align} \mathrm{LC}=1\mathrm{MSD}-1\mathrm{VSD}=a/(N+1). \end{align}
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 1 mm). 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 2.736 g. Calculate the density in g/cm^{3} up to correct significant figures.
Solution> From the given data, one main scale division (MSD) is 1 MSD = 1 mm. Since 10 Vernier scale divisions (VSD) are equal to 9 MSD, we get 1 VSD = 9/10 MSD = 0.9 mm. The least count (LC) is given by LC = 1 MSD - 1 VSD = 1.0 - 0.9 = 0.1 mm.
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 \begin{align} a&=\mathrm{MSR}+\mathrm{VSR}\times\mathrm{LC}\\ &=10+1\times 0.1=10.1\,\mathrm{mm}. \end{align}
The measurement of a has three significant digits. The volume of the cube is V = a^{3} = 1.03 cm^{3} and the density is \begin{align} \frac{m}{V}=\frac{2.736}{1.03}=2.66\, \mathrm{g/cm^3} \end{align} (after rounding off to three significant digits).
A Vernier calipers 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 calipers, the least count is
Solution: For the given Vernier calipers, one main scale division (MSD) is 1 MSD = 1 mm. Since 20 Vernier scale divisions (VSD) are equal to 16 MSD, we get 1 VSD = 16/20 MSD = 0.8 mm. The lease count is given by LC = 1 MSD - 1 VSD = 1 - 0.8 =0.2 mm.
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 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24^{th} division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
Solution: From the given data, one main scale division (MSD) and one Vernier scale division (VSD) are 1 MSD = 5.15 - 5.10 = 0.05 cm and 1 VSD = 2.45/50 = 0.049 mm. The least count of the given Vernier Calipers is LC = 1 MSD - 1 VSD = 0.001 cm.
For the given measurement, main scale reading (MSR) is 5.10 cm and the Vernier scale reading (VSR) is 24. Hence, the diameter D of the cylinder is \begin{align} D&=\mathrm{MSR}+\mathrm{VSR}\times\mathrm{LC} \\ &=5.10+24\times 0.001=5.124\,\mathrm{cm}. \end{align}
Consider a Vernier calipers in which each 1 cm 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,
Solution: In given Vernier calipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = 1/8 = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.1 cm. The least count of the Vernier calipers is given by LC = 1 MSD - 1 VSD = 0.125 - 0.1 = 0.025 cm.
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., \begin{align} \mathrm{lc}=p/n=2l/100=l/50. \end{align}
If $p=2\mathrm{LC}=2(0.025)=0.05$ cm, then $l=p/2=0.025$ cm. Substitute in above equation to get the least count of the screw gauge lc = 0.005 mm.
If $l=2\mathrm{LC}=2(0.025)=0.05$ cm, then above equation gives lc = 0.01 mm.
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 (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
Solution: In both calipers C_{1} and C_{2}, 1 cm is divided into 10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to $x_{m1}=x_{m2}=1/10=0.1$ cm. 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 $x_{v1}=9x_{m1}/10=0.09$ cm. 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 $x_{v2}=11x_{m2}/10=0.11$ cm.
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 \begin{align} X&=\mathrm{MSR}+x &=\mathrm{MSR}+mx_m-vx_v \end{align}
In calipers C_{1}, MSR_{1} = 2.8 cm, m_{1} = 7 and v_{1} = 7, and in calipers C_{2}, MSR_{2} = 2.8 cm, m_{2} = 8 and v_{2} = 7.Substitute these values in above equation to get \begin{align} X_1&=\mathrm{MSR}_1+m_1 x_{m1} -v_1 x_{v1}=2.87\,\mathrm{cm}\\ X_2&=\mathrm{MSR}_2+m_2 x_{m2} -v_2 x_{v2}=2.83. \end{align}
Answer: 1.9 mm
Answer: -1.2 mm
Answer: 3.14 cm
Answer: 3.14 cm
Answer: (B)
Answer: (C), (D)
Answer: 0.01 cm, 10.23 cm
Answer: 0.005 cm, 1.76 cm
Answer: 2.51 cm
Answer: (A)
Answer: (B)
Answer: 0.05 cm
Answer: (C)
Answer: (D)
Answer: (B)
Answer: 0.001 cm
Answer: (C)
Answer: (C)
Answer: (A)
Answer: (B)
Answer: (D)
Answer: (C)
Answer: (C)
Answer: 0.025 cm
Answer: (C)
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.