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#### 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)

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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So nice understanding by your

Thanks

It helpsd me a lot

Nice

I understand in very easy way .

Thanks a lot.

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