Physics > Basic Mathematics and Measurements > 11.0 Length Measuring Instruments
Basic Mathematics and Measurements
1.0 Introduction
2.0 Trigonometry
2.1 Values of trigonometric angles
2.2 Trigonometric identities
2.3 Trigonometric functions in different quadrants
3.0 Basic logarithmic functions
4.0 Differentiation
4.1 Derivatives of some simple functions
4.2 Rules of differentiation
4.3 Application of differentiation
4.4 Solved examples of differentiation
5.0 Integration
6.0 Graphs
6.1 Straight line
6.2 Circle
6.3 Ellipse
6.4 Parabola
6.5 Rectangular hyperbola
6.6 Exponential function
6.7 Logarithmic functions
7.0 Significant Figures
7.1 Rules to determine the significant figures
7.2 Rules for arthimetic operation with significant figures
8.0 Rounding off
9.0 Errors
9.1 Systematic error
9.2 Random errors
9.3 Least count error
9.4 Absolute error
9.5 Mean absolute error
9.6 Relative error or fractional error
9.7 Percentage error
10.0 Combination of errors
10.1 Addition of errors
10.2 Subtraction of errors
10.3 Multiplication of errors
10.4 Division of errors
10.5 Power
11.0 Length Measuring Instruments
11.1 Vernier Callipers
11.2 Zero error of vernier calliper
11.3 Vernier calliper solved examples
11.4 Screw Gauge
11.5 Zero error of screw gauge
11.6 Screw gauge solved examples
12.0 Questions
11.3 Vernier calliper solved examples
2.2 Trigonometric identities
2.3 Trigonometric functions in different quadrants
4.2 Rules of differentiation
4.3 Application of differentiation
4.4 Solved examples of differentiation
6.2 Circle
6.3 Ellipse
6.4 Parabola
6.5 Rectangular hyperbola
6.6 Exponential function
6.7 Logarithmic functions
7.2 Rules for arthimetic operation with significant figures
9.2 Random errors
9.3 Least count error
9.4 Absolute error
9.5 Mean absolute error
9.6 Relative error or fractional error
9.7 Percentage error
10.2 Subtraction of errors
10.3 Multiplication of errors
10.4 Division of errors
10.5 Power
11.2 Zero error of vernier calliper
11.3 Vernier calliper solved examples
11.4 Screw Gauge
11.5 Zero error of screw gauge
11.6 Screw gauge solved examples
Question: 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 $4^{th}$ division of vernier scale coincides with a main scale division.
Find:
(i). Least count
(ii). Radius of cylinder
Solution: Main scale of a vernier calliper is calibrated in $mm$ means its least count is $1\ mm$.
(i).
As stated in the solution,
$$19M.S.D. = 20V.S.D.$$$$1V.S.D. = \frac{{19}}{{20}}M.S.D.$$
As we know least count $(L.C.)$ is given by,
$$L.C. = 1M.S.D. - 1V.S.D.$$$$L.C. = 1M.S.D. - \frac{{19}}{{20}}M.S.D.$$$$L.C. = \frac{{M.S.D.}}{{20}}$$or$$L.C. = \frac{1}{{20}}mm$$$$L.C. = 0.05\,mm$$
$$19M.S.D. = 20V.S.D.$$$$1V.S.D. = \frac{{19}}{{20}}M.S.D.$$
As we know least count $(L.C.)$ is given by,
$$L.C. = 1M.S.D. - 1V.S.D.$$$$L.C. = 1M.S.D. - \frac{{19}}{{20}}M.S.D.$$$$L.C. = \frac{{M.S.D.}}{{20}}$$or$$L.C. = \frac{1}{{20}}mm$$$$L.C. = 0.05\,mm$$
(ii).
Vernier calliper readings are as follows,
$M.S.D.=35$ or $M.S.R.=35\ mm$
$n=4$
So, the diameter is,
$$D = M.S.R. + n \times L.C.$$$$D = \left( {35 + 4 \times 0.05} \right)mm$$$$D = 35.20\,mm$$
Radius can be calculated as,
$$R = \frac{D}{2} = \frac{{35.2}}{2}mm$$$$R = 17.6\,mm$$
$$D = M.S.R. + n \times L.C.$$$$D = \left( {35 + 4 \times 0.05} \right)mm$$$$D = 35.20\,mm$$
Radius can be calculated as,
$$R = \frac{D}{2} = \frac{{35.2}}{2}mm$$$$R = 17.6\,mm$$
