Significant Figures

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Rules for Identifying Significant Figures

Significant figures are the digits in a number that indicates reliability of a measurement. These are identified by following rules

  1. All non-zero digits are significant.
  2. Zeros between non-zero digits are significant.
  3. Trailing zeros (the right most zeros) are significant when there is a decimal point in the number.
The underlined digits in 0.010200 are significant.

Scientific Notation

The change of units cannot change the significant digits. Thus, 1.20 m = 12.0 cm = 120 mm all should have three significant digits. But number 120 has two significant digits. To avoid this ambiguity, measurements are recorded in scientific notation as $x\times{10}^{y}$, where $x$ is a number between 1 and 10 with correct number of significant digits. Thus, $\underline{1.20}\;\mathrm{m}=\underline{1.20}\times{10}^{1}\;\mathrm{cm}=\underline{1.20}\times{10}^{2}\;\mathrm{mm}$ is unambiguous (the exponent $y$ is not considered for counting the significant digits).

The order of magnitude of a number is the exponent $y$ in its scientific notation (after rounding off $x$ to 1 if $x\leq 5$ and to 10 if $5\leq x\leq 10$). Thus, order of magnitude of nuclear density $2.3\times{10}^{17}\;\mathrm{kg/m^3}$ is 17 and order of magnitude of the earth radius $6.3\times{10}^{6}\;\mathrm{m}$ is 7.

Significant Figures in Multiplication or Division

In multiplication or division, the final result should retain as many significant digits as are there in the original number with the least significant digits. For example if $m=\underline{2.736}\;\mathrm{g}$ and $V=\underline{1.03}\;\mathrm{cm^3}$ then density is $m/V = 2.736/1.03 = 2.6563 = \underline{2.66} \;\mathrm{g/cm^3}$.

Significant Figures in Addition or Subtracion

In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places. For example, $1.23\;\mathrm{m}+1.2\;\mathrm{m}$ should be recorded as 2.4 m and $1.23\;\mathrm{m}-1.2\;\mathrm{m}$ should be recorded as 0.0 m.

Convention for Rounding Off

In intermediate step of a multi-step calculation, retain one digit more than the significant digits required in final results and round off at the end. The convention for rounding off is: preceding digit is raised by 1 if the insignificant digit to be dropped is more than 5, and is left unchanged if the latter is less than 5. What if insignificant digit is 5? If preceding digit is even, then insignificant digit is dropped, if it is odd, the preceding digit is raised by 1. Thus, both 2.665 and 2.655 are rounded off to 2.66.

Solved Problems on Significant Figures

Problem from JEE Mains (AIEEE) 2012

A student measured the diameter of a wire using a screw gauge with the least count 0.001 cm. The measured value should be recorded as

  1. 5.3200 cm
  2. 5.3 cm
  3. 5.32 cm
  4. 5.320 cm

Solution: The value of uncertain digit in recorded measurement should be equal to the least count 0.001 cm. Thus, (D) is correct.

Problem from NCERT

The mass of a box measured by a grocer's balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. The total mass of the box and the difference in the masses of the gold pieces should be recorded as

  1. 2340.32 g, 0.02 g
  2. 2.34032 kg, 0.02 g
  3. 2.3 kg, 0 g
  4. 2.340 kg, 0.02 g

Solution: The mass measured by grocer's balance, $m_b=2.300 \;\mathrm{kg}$, has four significant digits and the uncertainty in measurement (least count of the balance) is $0.001\;\mathrm{kg}=1 \;\mathrm{g}$. This can be recorded as $(2300\pm 1)\mathrm{g}$ as change of units does not change number of significant digits. The masses of gold pieces, $m_{g1}=20.15\;\mathrm{g}$ and $m_{g2}=20.17\;\mathrm{g}$, both has four significant digits but the uncertainty in measurement is 0.01 g.

The total mass of the box is \begin{align} m_{t}&=m_b+m_{g1}+m_{g2}\\\nonumber &=2300+20.15+20.17\\\nonumber &=2340.32 \;\mathrm{g}.\nonumber \end{align} As per rules of significant digits for addition and subtraction, numbers of digits after decimal points in the final result should be equal to the minimum number of digits after decimal point in the values being added (which is zero in this case). Thus, total mass should be recorded as $m_t=(2340\pm1)\;\mathrm{g}$ or 2.340 kg. Similarly, the differences in masses of the gold pieces should be recorded as $m_{g2}-m_{g1}=20.17-20.15=0.02\;\mathrm{g}$.

Thus (D) is correct option.

Problem from IIT JEE 2003

The edge of a cube is measured by a scale of least count 1 mm. The measured value is $l=1.2 \;\mathrm{cm}$. The volume of the cube should be recorded as

  1. $(1.728\pm0.003)\;\mathrm{cm^3}$
  2. $(1.73\pm0.004)\;\mathrm{cm^3}$
  3. $(1.7\pm0.4)\;\mathrm{cm^3}$
  4. $(1.7\pm0.3)\;\mathrm{cm^3}$
  5. $(1.728\pm0.003)\;\mathrm{cm^3}$

Solution: The measured length is $a=(1.2\pm 0.1)\;\mathrm{cm}$, where $\Delta a=0.1\;\mathrm{cm}$ is the least count of the scale. The volume of the cube is $V=a^3=(1.2)^3=1.728\;\mathrm{cm^3}$. The error in volume is $\Delta V=(3\Delta a/a)V=0.432\;\mathrm{cm^3}$. Thus, volume should be reported as $(1.7\pm 0.4)\;\mathrm{cm^3}$ upto two significant digits.

Questions on Significant Digits

Question 1: The respective number of significant figures for numbers 23.023, 0.0003 and $2.1\times{10}^{-3}$ are

A. 5, 1, 2
B. 5, 1, 5
C. 5, 5, 2
D. 4, 4, 2

Related Topic

  1. Vernier Calipers Made Easy
  2. Screw Gauge

References and External Links

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  2. 100 Solved Problems on Units, Dimensions and Measurement, Jitender Singh and Shraddhesh Chaturvedi
  3. NCERT Class 11 Physics Part 1