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The study of Physics Topics involves the exploration of matter, energy, and the forces that govern the universe.

## What is an Example of a Significant Figure? What are the Rules for Rounding off Digits?

Measurement of a physical quantity cannot be free from errors. Errors in measurements are usually of two types—

Systematic error: Generally two types of systemetic errors are known:

Instrumental error: This arises due to defective instru-ments.

Personal error: This arises from incorrect setting of the instruments, and incorrect recording of data.

Systematic errors can be minimised or eliminated by properly indentifying the sources of errors.

Random error: This type of error arises randomly due to known and unknown reasons that are entirely beyond our control. Random errors cannot be eliminated totally.

### Calculation of Errors

Even when a particular physical quantity is measured many times under identical conditions using the same method, the results may not be identical. This dispersion arises due to errors and cannot be eliminated totally.

Actual of true value: Let the values of measurement of a physical quantity, measured n times using the same instrument and the same method be x_{1}, x_{2}, x_{3}, ……, x_{n}. The average of the measurements is then considered to be the actual or true value of the physical quantity.

Hence, actual value, \(\bar{x}\) = \(\frac{x_1+x_2+x_3+\cdots+x_n}{n}\)

Error: It is not sufficient to write the actual value \(\bar{x}\) as the absolute value in a measurement. The extent of uncertainly associated with \(\bar{x}\) needs to be mentioned.

Hence, absolute value is x = \(\bar{x}\) ± ∈; where ∈ the uncertainty, can be calculated as

∈ = \(\frac{\left|x_1-\bar{x}\right|+\left|x_2-\bar{x}\right|+\cdots+\left|x_n-\bar{x}\right|}{n}\)

i.e., ∈ is the average of the differences between the measured value and the actual value, ∈ is also referred to as the mean absolute error. It is to be noted,

|x_{1} – \(\bar{x}\)|, |x_{2} – \(\bar{x}\)|,…….|x_{n} – \(\bar{x}\)| ≥ 0

Fractional error or relative error: It is the ratio of the mean absolute error to the absolute value, i.e., \(\frac{\epsilon}{x}\).

Percentage error: It is obtained by multiplying the fractional error by 100. Hence, percentage error = \(\frac{\epsilon}{x}\) × 100.

Propagation of errors: When the value of a physical quantity involves a number of measurements, the resultant error depends on

- the error associated with each individual measurement, and
- the mathematical operations (addition, subtraction, multiplication, division etc.) required to arrive at the final value.

Here, we shall discuss the second factor. It is said that the errors in measurement propagate with the said mathematical operations.

Let, Δa, Δb, Δc, be the mean absolute errors, respectively, in the measurements of a. b, c, … . Then in the determination of x, due to mathematical operations among a, b, c…., Δx = maximum absolute error, f = \(\frac{\Delta x}{x}\) = maximum fractional error and p = \(\frac{\Delta x}{x}\) × 100 = maximum percentage error.

i) Error due to addition: If x = a + b + c + …, then,

Δx = Δa + Δb + Δc + ……;

f = \(\frac{\Delta a+\Delta b+\Delta c+\cdots}{a+b+c+\cdots}\)

p = \(\frac{\Delta a+\Delta b+\Delta c+\cdots}{a+b+c+\cdots}\) × 100

ii) Error due to subtraction: If x = a – b, then,

Δx = Δa + Δb ; f = \(\frac{\Delta x+\Delta b}{|a-b|}\); p = \(\frac{\Delta x+\Delta b}{|a-b|}\) × 100

It is to be noted that if a and b have very close values, then due to the low value of |a – b|, the error due to subtraction tends to become very high.

iii) Error due to multiplication: If x = ab, then,

f = \(\frac{\Delta a}{a}\) + \(\frac{\Delta b}{b}\) ; p = (\(\frac{\Delta a}{a}\) + \(\frac{\Delta b}{b}\)) × 100

iv) Error due to division: If x = \(\frac{a}{b}\), then,

f = \(\frac{\Delta a}{a}\) + \(\frac{\Delta b}{b}\) ; p = (\(\frac{\Delta a}{a}\) + \(\frac{\Delta b}{b}\)) × 100

v) Error due to powers of a, b, c, …….: If

x = \(\frac{a^k c^m}{b^l}\) = a^{k}b^{-l}c^{m}, then

f = k\(\frac{\Delta a}{a}\) + l\(\frac{\Delta b}{b}\) + m\(\frac{\Delta c}{c}\)

p = (k\(\frac{\Delta a}{a}\) + l\(\frac{\Delta b}{b}\) + m\(\frac{\Delta c}{c}\)) × 100

It is to be noted that the error is high for high values of k, l, m. For example if m is high but k and l are comparatively low in the expression of x, then a considerable amount of error tends to arise from the measurement of c. So, c should be measured more precisely than a or b.

### Significant Figures

In a measurement, the digits in the measured value is said to be significant figures when all the digits except the last one are reliably accurate.

Example: Suppose you measure the length of a rod with the help of an ordinary scale (in such a scale the distance between two divisions is 0.1 cm). You observe that the length of the rod is more than 20.1 cm but less than 20.2 cm. in general, if you look carefully, you will find that the length is either closer to 20.1 cm or 20.2 cm. If ills closer to 20.1 cm, you can write it as 20.1 cm. But you know that the last digit 1 is inaccurate, i.e., uncertain. Thus, the number of significant figures of this measurement is 3.

Number of significant figures in any reading or measurement indicates how accurate the reading or measurement is.

The following points need to be taken into consideration while determining the number of significant figures in a reading or measurement.

i) All non-zero digits are significant.

ii) To count the number of significant figures or digits, we begin from the leftmost non-zero digit and count all the digits up to the rightmost digit; e.g., 27.7 has 3 significant figures.

iii) Zeros between two non-zero digits are significant; e.g., 207.007 has 6 significant figures.

iv) Zeros between the decimal point and the first non-zero digit to its right are not significant; e.g., 0.00207 has 3 significant figures.

v) Zeros on the right of the decimal point are significant if there is at least one non-zero digit to its left; e.g., 277.0 has 4 significant digits. Note that the number of significant figures of 277 and 277.0 are 3 and 4 respectively. The former denotes that only the last 7 is uncertain while the latter denotes that only 0 is uncertain.

vi) Zeros added to the right of a measured value, while changing the unit, are not significant; e.g., when 277.0 kg is written as 277000 g, the number of significant digits remains the same, i.e., 4 only. Note that we should write a mass as either 2.770 × 10^{2} kg or 2.770 × 10^{5}g to avoid the error in counting significant figures.

vii) During the multiplication of two numbers or during division of one number by another, the number of significant figures of the product or quotient respectively should be equal to that of the number with less number of significant figures.

For example, let an object of mass 10 g have a volume of 3 cm^{3}. Hence its density is supposed to be 10 ÷ 3 = 3.33… g ᐧ cm^{-3}. But since the number of significant figures of volume is 1, the quotient has to be written as 3g ᐧ cm^{-3} i.e., with one significant figure. Again if the volume of an object is 4.23 cm^{3} and density is 11 g ᐧ cm^{-3}, the mass will be 4.23 × 11 = 46.53 g. But since the number of significant figures of density is 2, the product has to be written as 46 g i.e., with two significant figures only.

viii) During addition of two numbers or subtraction of one number from another, the number of digits to the right of the decimal point in the sum or difference should be equal to that of the number with less number of digits to the right of the decimal point. In this case, it is immaterial how many significant figures the two numbers contain.

For example, if the length of two rods are 5 m and 1.25 m, then the actual sum of the lengths of the rods is 6.25 m. But since for 5 m (number of significant figures = 1), the number of digits to the right of the decimal point is 0, the sum is to be expressed as:

5(number of significant figures = 1) + 1.25 (number of significant figures = 3) = 6 (number of significant figures = 1)

Few more examples for determining the number of significant figures are given below.

1. 1738 N | Four significant figures |

2. 0.024 | two significant figures |

3. 8.0023 g ᐧ cm^{-3} |
five significant figures |

4. 2.520 × 10^{2} m |
four significant figures |

5. 0.0627 N ᐧ m^{-3} |
four significant figures |

6. 4200 | two significant figures |

7. 51.00 m | five significant figures |

8. 3.69 × 10^{24} kg |
three significant figures |

9. 0.008 m^{2} |
one significant figures |

10 0.2370 g ᐧ cm^{-3} |
four significant figures |

### Rules For Rounding Off Digits

The accuracy of a result obtained by mathematical calculation can never be greater than the accuracy of original physical measurements. Therefore, the non-significant figures should be dropped from the result.

The following rules are adopted while dropping figures in rounding off to the appropriate digit:

i) If the digit to be dropped is less than 5 then the preceding digit is kept unchanged. For example, if a number 3.454 is to be rounded off to three significant figures the digit to be dropped is 4 which is less than 5. Hence the preceding digit, namely 5, is not changed. Therefore, it should be written as 3.45.

ii) If the digit to be dropped is more than 5, then the preceding digit is increased by 1. For example 3.458 is rounded off as 3.46 to three significant figures.

iii) If the digit to be dropped happens to be 5 or 5 followed by zero(s), then the preceding digit to be retained is increased by 1 if it is odd; if it is even then it remains unchanged. For example. 3.475 or 3.4750 or 3.47500, when rounded off to the second decimal place, will be written as 3.48. For the numbers like 3.48, 3.4850 or 3.48500, all will be rounded off to the same decimal place and will be written as simply 3.48.

iv) If the digit to be dropped happens to be 5 followed by some non-zero digit at any place then the preceding digit up to which the rounding off is desired will be increased by 1 (no matter if it is odd or even).

For example, if 3.485010 or 3.485125 when rounded off up to second decimal place will be written as 3.49. Similarly 3.475010 when rounded off up to second decimal place will be written as 3.48.

### Numerical Examples

**Example 1.**

In an experiment of simple pendulum a student made several observations for the period of oscillation. His readings turned out to be 2.63s, 2.56s, 2.42s, ills and 2.80 s. Find

(i) mean time period of oscillations or most accurate value of time period,

(ii) absolute error in each reading,

(iii) mean absolute error,

(iv) fractional error and

(v) percentage error.

**Solution:**

i) The mean time period of oscillation,

T = \(\frac{2.63+2.56+2.42+2.71+2.80}{5}\)s

= \(\frac{13.12}{5}\)s = 2.624 s ≈ 2.62 s

(rounded off to 2nd decimal place)

ii) Taking 2.62 s as the true value, the absolute errors

(true value – measured value) in the live readings are:

(2.62 – 2.63) s = -0.01 s; (2.62 – 2.56) s = 0.06 s;

(2.62 – 2.42) s = 0.20 s; (2.62 – 2.71) s = -0.09 s and

(2.62 – 2.80) s = -0.18 s

iii) The (maximum) mean absolute error is,

(δT)_{max} = \(\frac{(0.01+0.06+0.20+0.09+0.18)}{5}\)s

= \(\frac{0.54}{5}\) = 0.108 s ≈ 0.11 s

iv) The (maximum) fractional error is,

(\(\frac{\delta T}{T}\))_{max} = \(\frac{0.11 \mathrm{~s}}{2.62 \mathrm{~s}}\) ≈ 0.04

v) The maximum percentage error is,

(\(\frac{\delta T}{T}\))_{max} × 100 = 0.04 × 100 = 4%

∴ The value of T should be written as (2.62 ± 0.11) s

**Example 2.**

The measured length and breadth of a rectangle are written as (5.7 ± 0.1) cm and (3.4 ± 0.2) cm respectively. Calculate the area of the rectangle with error limits.

**Solution:**

Given l = 5.7 cm and Δl = 0.1 cm; b = 3.4 cm and Δb = 0.2 cm. The area of the rectangle without error limit is,

A = l × b = (5.7 × 3.4) cm^{2} = 19.38 cm^{2} ≈ 19.4 cm^{2}

Next, the fractional error in A is,

\(\frac{\Delta A}{A}\) = \(\frac{\Delta l}{l}\) + \(\frac{\Delta b}{b}\) = \(\frac{0.1}{5.7}\) + \(\frac{0.2}{3.4}\) ≈ (o.02 + 0.06) = 0.08

∴ ΔA = 0.08 × A ≈ 0.08 × 19.4 cm^{2} ≈ 1.6 cm^{2}

Hence, the area of the rectangle with error limit is (19.4 ± 1.6) cm^{2}.

**Example 3.**

The potential difference across the ends of a wire has been measured to be (100 ± 5) volt and the current in the wire as (10 ± 0.2) ampere. What is the percentage error in the computed resistance of the wire?

**Solution:**

Given, V = (100 ± 5) vòlt and I = (10 ± 0.2) ampere

Now, R = \(\frac{V}{I}\)

The maximum percentage error in R is,

\(\left(\frac{\Delta R}{R}\right)\)_{max} × 100 = (\(\frac{\Delta V}{V}\) × 100) + (\(\frac{\Delta I}{I}\) × 100)

= (\(\frac{5}{100}\) × 100) + (\(\frac{0.2}{10}\) × 100)

= 5% + 2% = 7%

**Example 4.**

A student performing Searle’s experiment for finding the Young’s modulus Y of the material of a wire takes the following observations:

Length of the wire (L) = 2.890 m, diameter of the wire (D) = 0.082 cm, mass suspended from the wire (M) = 3.00 kg, extension in the length of wire (l) = 0.087 cm.

Calculate the maximum permissible error in the value of Y.

**Solution:**

The Young’s modulus of the material is given by,

Y = \(\frac{4 M g L}{\pi D^2 l}\)

Here, M = 3.00 kg ∴ ΔM = 0.01 kg

L = 2.890 m ∴ ΔL = 0.01 m

D = 0.082 cm ∴ ΔD = 0.01 m

l = 0.087 cm ∴ Δl = 0.01 m

The maximum permissible percentage error is Y is,

### Accuracy And Precision of Measuring Instruments

Accuracy: Accuracy of a measuring instrument is decided by the closeness of the measured value of any physical quantity to the actual value which is known beforehand. Suppose the mass of a 100 g body, when measured using a common balance, reads 95 g. The measurement is, therefore, not accurate.

Precision: An instrument is precise when it repeats almost the same value when a physical quantity is measured a number of times. Precision, therefore, denotes how close the measured values of a physical quantity are with respect to one another. Suppose the mass of a 100 g body, when measured five times using a common balance, reads 90g, 96g, 92g, 93g, 97g. The measurement is, therefore, not precise.

### Comparison between accuracy and precision:

1. By using an instrument only once, we can determine its accuracy But to know its precision we need to take a number of measurements.

2. Accuracy denotes how close the measured value is, relative to the actual value. Precision denotes how close the measured values are, relative to one another,

3. Accuracy of an instrument depends upon the method of measurement but precision depends on random factors.

Example: Let us consider the examples in the following table, taking 100 g as standard:

Set of measurements | Inference |

90 g, 96 g, 92 g, 93 g, 97 g | 1. The instrument has neither accuracy nor precision. |

95 g, 94 g, 95 g, 95 g, 96 g | 2. The instrument has good precision but no accuracy. |

102 g, 98 g, 100 g, 100 g, 99 g | 3. The instrument has good accuracy but no precision. |

100 g, 101 g, 99 g, 100 g, 100 g | 4. Both precision and accuracy of the instrument are very good. |

The difference between accuracy and precision can be explained pictorially in the following simple example: Supposer B is a target board and T is the target point marked on it [Fig.]. A shooter hits the target using two different rifles. The dots on the figure are the bullet marks. Now an observation of the target boards clearly shows that:

i) the 1st rifle is more accurate since the bullet marks are all around the target point [Fig.(i)]. But the precision is low as the points are scattered over a large area. This means that the quality of the rifle is fairly low.

ii) the 2nd rifle is highly inaccurate since the bullet marks are far from the target point [Fig.(ii)]. But the precision is very high because the marks are very close to one another. This indicates that it is a good quality rifle, but its initial settings are somehow defective. On proper adjustments, all the bullet marks can he brought close to the target point. In that case, the accuracy and precision would both be high.

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