To Measure The Thickness Of The Given Sheet Using Screw Gauge

Taking the measurements of a sheet with the help of a screw gauge is a very accurate and simple technique due to the numerous. A micrometre screw gauge, or simply a screw gauge, is a measuring tool specifically designed for taking measurements of very small dimensions with very high accuracy levels. This instrument comprises a spindle which moves along the calibrated screw when it is turned. So as to take the thickness measurement of a sheet, one needs to position the sheet in between the anvil and spindle of the screw gauge.

One of the most basic experiments in Physics, which shows precision measurement methodologies, is measuring the width of a provided strip using a micrometre screw gauge. This is significant to learners who seek to join Class 12, NEET, and JEE Main exams since it aids them to know about micrometre screw gauges applied commonly in various types of engineering and scientific practices. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fifteen questions have been asked on this concept.

Aim

To measure the thickness of the given sheet using a screw gauge

Apparatus

Screw gauge, Sheet, half-metre scale and magnifying lens.

Theory

1. If we place the sheet between plane faces A and B, the edge of the cap lies ahead of the Mb division of the linear scale. Then, linear scale reading (L.S.R.) =N.
If the nth division of the circular scale lies over the reference line.
Then, circular scale reading (C.S.R) = n x (L.C.) (Here, L.C. is the least count of screw gauge)

Total reading (TR) = LS.R. + C.S.R. =N+n (L . C .)

Procedure

1. Note the number of divisions on the circular scale.

2. Give five complete rotations to the screw.

3. Note the linear distance moved by the screw.

4. Find the pitch and L.C. of the screw gauge.

5. Find the zero error and zero correction by moving the screw only in one direction in such a way that studs A and B just touch each other.

6. Now grip the given sheet in the gap A and B of the screw gauge.

7. Turn the screw head till the ratchet arrangement gives a click.

8. Note the readings of linear scale and circular scale and find the observed thickness using the relation, observation thickness = L.S.R. + C.S.R.

9. Add the zero correction to the observed thickness to find the corrected diameter.

10. Repeat steps 6 to 9 to find the thickness from four more different places.

Calculations

$\begin{aligned} & \text { C.S. } R=\text { circular scale reading } \\ & \begin{aligned} \text { Total reading } & =\text { M.S.R }+ \text { C.S.R } \\ & =M S R+L C \times C S R\end{aligned}\end{aligned}$

L.C = least count

C.S.R = Circular scale reading

Precautions

1. To avoid undue pressure; the screw should always be rotated by ratchet R and not by cap K.
2. The screw should move freely without friction.
3. The zero correction, with proper sign should be noted very carefully and added algebraically.
4. For the same set of observations, the screw should be moved in the same direction to avoid the back-lash error of the screw.

Solved Examples Based on Measure the Thickness of the Given Sheet Using Screw Gauge

Example 1: A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45^th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness (in mm) of the sheet if the main scale reading is 0.5 mm and the 25^th division coincides with the main scale line?

1) 0.80

2) 0.75

3) 0.70

4) 0.50

Solution:

To measure the thickness of the given sheet using a screw gauge

$\begin{aligned} & \text { Least count }=\frac{\text { pitch }}{\text { no.of division on circular scale }} \\ & \qquad=\frac{0.5 \mathrm{~mm}}{50} \\ & \text { L.C. }=0.01 \mathrm{~mm} \\ & - \text { ve zero error }=-5 \times L C=-0.05 \mathrm{~mm}\end{aligned}$

Measured value=main scale reading + screw gauge reading - zero error

$=0.5 \mathrm{~mm}+(25 \times 0.01-(-0.05)) \mathrm{mm}=0.80 \mathrm{~mm}$

Hence, the answer is (0.80).

Example 2: In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness (in cm) of the wire is :

1) 0.2150

2) 0.4300

3) 0.3150

4) 0.0430

Solution:

To measure the thickness of the given sheet using a screw gauge

number of divisions on the circular scale =100

The number of full rotations given to screw=5

Distance moved by the screw=0.25
$
\begin{aligned}
& \text { Pitch }=\frac{\text { Distance moved by the screw }}{\text { Number of full rotations given to screw }} \\
& \text { Pitch }=\frac{0.25}{5} \\
& \text { Pitch } \\
& \text { Least count }=\overline{\text { Number of divisions on circular scale }} \\
& \text { leastcount }=\frac{0.25}{5 \times 100} \mathrm{~cm}=5 \times 10^{-4} \mathrm{~cm} \\
& \text { Reading }=\text { main scale divisions } * \text { pitch }+(\text { least count*circular scale divisions })=4 \times 0.05 \mathrm{~cm}+30 \times 5 \times 10^{-4} \mathrm{~cm} \\
& =(0.2+0.0150) \mathrm{cm}
\end{aligned}
$

= 0.2150cm

Hence, the answer is 0.2150cm.

Example 3: If the screw on the screw gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale, the least count of the screw gauge is: (in cm)

1) 0.001

2) 0.01

3) 0,02

4) 0.001

Solution:

$\begin{aligned} & \text { Pitch }=\frac{3}{6}=0.5 \mathrm{~mm} \\ & \text { L.C }=\frac{0.5 \mathrm{~mm}}{50}=\frac{1}{100} \mathrm{~mm}=0.01 \mathrm{~mm}=0.001 \mathrm{~cm}\end{aligned}$

Hence, the answer is the option (1).

Example 4:The number of circular divisions on the shown screw gauge is 50. It moves 0.5 mm on the main scale for one complete rotation and the main scale has 1/2 mm marks. The diameter of the ball is : (please give your answer in mm)

1) 1.2

2) 2.2

3) 2.25

4) 1.25

Solution:

To measure the thickness of the given sheet using a screw gauge

$\begin{aligned} & \text { C.S.R }=\text { circular scale reading } \\ & \text { Total reading } \\ & =\text { M.S.R }+ \text { C.S.R } \\ & =M S R+L C \times C S R \\ & \text { L.C }=\text { least count } \\ & \text { C.S.R }=\text { Circular scale reading } \\ & \qquad 5 \times \frac{0.5}{50}=0.05 \mathrm{~mm} \\ & \text { Zero error }= \\ & \text { Actual measurement }=2 \times 0.5 \mathrm{~mm}+25 \times \frac{0.5}{50}-0.05 \mathrm{~mm} \\ & \quad=1 \mathrm{~mm}+0.25 \mathrm{~mm}-0.05 \mathrm{~mm}=1.20 \mathrm{~mm}\end{aligned}$.

Hence, the answer is the option (1).

Example 5: The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5mm and 48 respectively, the thickness of this sheet is: (in mm)

1) 5.725

2) 5.740

3) 5.950

4) 5.755

Solution:

$\begin{aligned} & L C=\frac{\text { Pitch }}{\text { Number of division }} \\ & \begin{aligned} & L C=0.5 \times 10^{-2} \mathrm{~mm} \\ &+ \text { ve error }=3 \times 0.5 \times 10^{-2} \mathrm{~mm}=0.015 \mathrm{~mm}\end{aligned} \\ & \begin{aligned} \text { Reading } & =\mathrm{MSR}+\mathrm{CSR}-(+ \text { ve error) } \\ & =5.5+0.24-0.015 \\ & =5.725 \mathrm{~mm}\end{aligned}\end{aligned}$

Hence, the answer is the option (1).

Summary

One way to know the depth of a sheet, using a micrometre screw gauge is by applying the following steps: Begin by determining the least count (e) of the instrument which is usually 0.01mm. Put the sheet between the anvil and spindle of the gauge. Turn the thimble round until the sheet is tightly held without any clearance. Determine the main scale reading on millimetres (mm) at which the thimble’s edge coincides with the main scale.