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 An error is defined as the difference between the actual or true value and the measured value.
 An error should not be confused with a mistake, the mistake can be avoided, while the error can not be avoided but they can be reduced (minimized).
Types of Errors:
 1) Constant error, 2) Persistent or systematic errors 3) Accidental or random errors 4) Gross errors
Constant Error:
 When the results of a series of observations are in error by the same amount, the error is said to be a constant error.
 Systematic error due to faulty apparatus cause constant error.
Systematic Error:
 The error caused due to imperfect measurement technique, defective or imperfect apparatus or some personal reasons is called systematic error.

Causes of systematic error:
 Imperfection in the apparatus: For e.g. metre scale has an actual length of 99 cm but being used as 100 cm.
 Defective apparatus: a) The zero of a scale is not matching with the pointer. b) Micrometer screw gauge may have an error if the zero marking on the circular scale is not matching with the zero marking of the main scale when the jaws are closed.
 Personal errors: Peculiar habits of the person during measurement can cause errors. most such errors are in the same direction. If a proper formula is not used for calculation then it will lead to an introduction of error.
Random Errors:
 an error in measurement caused by factors which vary from one measurement to another is called random error.

Causes of random errors:
 Fluctuating conditions: for e.g. variation in temperature or in the environment may introduce an error in the measurement. In heat or electrical experiments, there is a possibility of such errors.
 Small disturbances: Small disturbances like vibrations may introduce error in the measurement.
 Error of judgement: There may be variation in the estimation of a measurement.
 Failure to define a quality: If the quality of the object whose measurements are to be made there is a possibility of the introduction of an error.
 Errors can be minimized by taking a number of readings and then finding the average of the readings taken.
Terminology w.r.t. Types of Errors:
Most probable value:
 When the sufficiently large number of readings are taken, then the mean of these readings is called as most probable value.
Absolute error:
 The magnitude of the difference between the most probable value (mean) and the individual measurement is called the absolute error of the measurement.
Final absolute error:
 The arithmetic mean of all the absolute errors is called as final absolute error
 Final absolute error = Sum of all absolute error / Total number of absolute errors
Relative error:
 The ratio of the absolute error in the measurement of a quantity with the most probable value is called as relative error
 Relative Error = Final absolute error / Most probable value
Percentage relative error:
 If relative error is multiplied by 100, the value obtained is called as percentage relative error.
 Percentage relative error = Relative error × 100
Examples of Calculation of Percentage Relative Error:
Example – 1:
 The length of metal plate was measured using vernier calipers of least count 0.01 cm. The readings obtained were 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error, the relative error and percentage error in the measurement of length.
 Solution:
Readings are 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm
The mean length = (3.11 cm + 3.13 cm + 3.14 cm + 3.14 cm) / 4 = 12.53 cm/4 =3.13 cm
Mean absolute error =[3.11 3.13 + 3.13 3.13 +3.14 3.13 +3.14 3.13] / 4
Mean absolute error =[0.02 + 0.00 + 0.01 + 0.01] / 4 = 0.04 / 4 = 0.01 cm
Relative Error = Final absolute error / Mean value = 0.01 / 3.13 = 0.00319
Percentage relative error = Relative error × 100 = 0.00319 × 100 = 0.319 %
Example 2:
 The weight of a body is measured using physical balance and readings obtained were 5.04 g, 5.06 g, 4.97 g, 5.00 g, 4,93 g. Find the percentage error in the measurement.
 Solution:
5.04 g, 5.06 g, 4.97 g, 5.00 g, and 4,93 g
The mean length = (5.04 g + 5.06 g + 4.97 g + 5.00 g + 4,93 g) / 5 = 25 g/5 =5.00 g
Mean absolute error =[5.04 – 5.00 + 5.06 5.00 +4.97 5.00 +5.00 5.00 + 4.93 – 5.00] / 5
Mean absolute error =[0.04 + 0.06 + 0.03 + 0.00 + 0.07 ] / 5 = 0.2 / 5 = 0.04 g
Relative Error = Final absolute error / Mean value = 0.04 / 5 = 0.008
Percentage relative error = Relative error × 100 = 0.008 × 100 = 0.8 %
Example – 3:
 An object was weighed by a physical balance and following readings are obtained: 5.04 g, 5.06 g, 4.97 g, 5 g, 4.93 g. Find a) mean value b) absolute error and c) percentage error
 Solution:
Readings are 5.04 g, 5.06 g, 4.97 g, 5 g, 4.93 g
The mean weight= ( 5.04 g + 5.06 g + 4.97 g + 5 g + 4.93 g) / 5 = 25 g/5 = 5 g
Mean absolute error =[5.04 – 5 + 5.06 – 5 +4.97 – 5 +5 – 5 + 4.93 – 5] / 5
Mean absolute error =[0.04 + 0.06 + 0.03 + 0 + 0.07] / 5 = 0.2 / 5= 0.04 g
Relative Error = Final absolute error / Mean value = 0.04 / 5 = 0.008
Percentage relative error = Relative error × 100 = 0.008 × 100 = 0.8 %
Ans: a) mean value = 5 g, b) absolute error = 0.04 g, c) percentage error = 0.8 %
Example – 4:
 The values of radius of glass rod measured by three students are 0.301 cm, 0.323 cm and 0.325 cm. Find the mean radius.
 Solution:
Readings are 0.301 cm, 0.323 cm and 0.325 cm.
The mean weight= ( 0.301 cm + 0.323 cm + 0.325 cm.) / 3 = 0.949 cm/3 = 0.316 cm
Mean absolute error =[0.301 – 0.316 + 0.323 – 0.316 +0.325 – 0.316 ] /3
Mean absolute error =[0.015 + 0.007 + 0.009] / 3 = 0.031/ 3= 0.010 cm
Relative Error = Final absolute error / Mean value = 0.010 / 0.316 = 0.0316
Percentage relative error = Relative error × 100 = 0.0316 × 100 = 3.16 %
Ans: percentage error = 3.16 %
Example – 5:
 The measurement of diameter of a wire is 0.74 mm. If the least count of the instrument is 0.01 cm, calculate the percentage error in the measurement.
 Solution:
Measurement of diameter of wire = 0.74 mm
Least count of the instrument = 0.01 cm = 0.1 mm
Hence possible error in measurement = 0.1 mm
Percentage error = (Possible error / Measurement) × 100
∴ Percentage error = (0.1/ 0.74) × 100 = 13.51 %
Ans: Percentage error = 13.51 %
Propagation of Error:
Propagation of Error in Addition:
 Suppose a result x is obtained by addition of two quantities say a and b i.e. x = a + b
 Let Δ a and Δ b are absolute errors in the measurement of a and b and Δ x be the corresponding absolute error in x.
∴ x ± Δ x = ( a ± Δ a) + ( b ± Δ b)
∴ x ± Δ x = ( a + b ) ± ( Δ a + Δ b)
∴ x ± Δ x = x ± ( Δ a + Δ b)
∴ ± Δ x = ± ( Δ a + Δ b)
∴ Δ x = Δ a + Δ b
Thus maximum absolute error in x = maximum absolute error in a + maximum absolute error in b
 Thus, when a result involves the sum of two observed quantities, the absolute error in the result is equal to the sum of the absolute error in the observed quantities.
Propagation of Error in Subtraction:
 Suppose a result x is obtained by subtraction of two quantities say a and b i.e. x = a – b
 Let Δ a and Δ b are absolute errors in the measurement of a and b and Δ x be the corresponding absolute error in x.
∴ x ± Δ x = ( a ± Δ a) – ( b ± Δ b)
∴ x ± Δ x = ( a – b ) ± Δ a – + Δ b
∴ x ± Δ x = x ± ( Δ a + Δ b)
∴ ± Δ x = ± ( Δ a + Δ b)
∴ Δ x = Δ a + Δ b
Thus the maximum absolute error in x = maximum absolute error in a + maximum absolute error in b.
 Thus, when a result involves the difference of two observed quantities, the absolute error in the result is equal to the sum of the absolute error in the observed quantities.
Propagation of Error in Product:
 Suppose a result x is obtained by the product of two quantities say a and b i.e. x = a × b ……….. (1)
 Let Δ a and Δ b are absolute errors in the measurement of a and b and Δ x be the corresponding absolute error in x.
∴ x ± Δ x = ( a ± Δ a) x ( b ± Δ b)
∴ x ± Δ x = ab ± a Δ b ± b Δ a ± Δ aΔ b
∴ x ± Δ x = x ± a Δ b ± b Δ a ± Δ aΔ b
∴ ± Δ x = ± a Δ b ± b Δ a ± Δ aΔ b …… (2)
Dividing equation (2) by (1) we have
 The quantities Δa/a, Δb/b and Δx/x are called relative errors in the values of a, b and x respectively. The product of relative errors in a and b i.e. Δa × Δb is very small hence is neglected.
 Hence maximum relative error in x = maximum relative error in a + maximum relative error in b
 Thus maximum % error in x = maximum % error in a + maximum % error in b
 Thus, when a result involves the product of two observed quantities, the relative error in the result is equal to the sum of the relative error in the observed quantities.
Propagation of Error in Quotient:
 Suppose a result x is obtained by the quotient of two quantities say a and b. i.e. x = a / b ……….. (1)
 Let Δ a and Δ b are absolute errors in the measurement of a and b and Δ x be the corresponding absolute error in x.
The values of higher power of Δ b/b are very small and hence can be neglected.
Now the quantity (Δ aΔ b / ab)is very small. hence can be neglected.
 The quantities Δa/a, Δb/b and Δx/x are called relative errors in the values of a, b and x respectively.
 Hence maximum relative error in x = maximum relative error in a + maximum relative error in b
 Thus maximum % error in x = maximum % error in a + maximum % error in b
 Thus, when a result involves the quotient of two observed quantities, the relative error in the result is equal to the sum of the relative error in the observed quantities.
Propagation of Error in Product of Powers of Observed Quantities:
 Let us consider the simple case . Suppose a result x is obtained by following relation x = a^{n} ……….. (1)
 Let Δ a be an absolute error in the measurement of a and Δ x be the corresponding absolute error in x.
 The values of higher power of Δa/a are very small and hence can be neglected.
 The quantities Δa/a and Δx/x are called relative errors in the values of a and x respectively.
 Hence the maximum relative error in x = n x maximum relative error in a. i.e. maximum relative error in x is n times the relative error in a.
 Consider a general relation
 The quantities Δa/a, Δb/b, Δc/c, and Δx/x are called relative errors in the values of a, b, c and x respectively.
 Thus maximum % error in x is
Examples Explaining Propagation of Error:
Example – 1:
 The lengths of two rods are recorded as 25.2 ± 0.1 cm and 16.8 ± 0.1 cm. Find the sum of the lengths of the two rods with the limit of errors.
 Solution:
We know that in addition the errors get added up
The Sum of Lengths = (25.2 ± 0.1) + (16.8 ± 0.1) = (25.2 + 16.8) ± (0.1 + 0.1) = 42.0 ± 0.2 cm
Example – 2:
 The initial temperature of liquid is recorded as 25.4 ± 0.1 °C and on heating its final temperature is recorded as 52.7 ± 0.1 °C. Find the increase in temperature.
 Solution:
We know that in subtraction the errors get added up
The increase in temperature = (52.7 ± 0.1) + (25.4 ± 0.1) = (52.7 – 25.4) ± (0.1 + 0.1) = 27.3 ± 0.2 °C.
Example – 3:
 During the study, the flow of a liquid through a narrow tube by experiment following readings were recorded. The values of p, r, V and l are 76 cm of Hg, 0.28 cm, 1.2 cm^{3 }s^{1} and 18.2 cm respectively. If these quantities are measured to the accuracies of 0.5 cm of Hg, 0.01 cm, o.1 cm^{3 }s^{1} and 0.1 cm respectively, find the percentage error in the calculation of η if formula used is
 Solution:
Example – 4:
 The percentage errors of measurements in a, b, c and d are 1%, 3%, 4% and 2% respectively. These quantities are used to calculate value of P. Find the percentage error in the calculation of P, If the formula used is
 Solution:
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