Video Lecture
Theory For Notes Making
Accuracy
It is the extent to which a reported measurement approaches the true value of the quantity measured. Problems with accuracy are due to errors.
Precision
Precision is how consistent results are when measurements are repeated. Precise values differ from each other because of random error, which is a form of observational error. It is the degree of exactness or refinement of a measurement. Precision describes the limitation of the measuring instrument.
For example
if actual value of a quantity is 60 units and we take 3 readings of it using an instrument and we get 50.40, 50.45,50.36 then as the readings are very close to each other it is said to be precise but since the readings are not close to actual value (i.e.60) so readings are not accurate.
Hence precision is independent of accuracy. You can be very precise but inaccurate. You can also be accurate but imprecise.
In an experiment accuracy and precision both matter and it’s best when measurements are both precise and accurate.
Least Count
Smallest value of a physical quantity which can be measured accurately by an instrument.
In any scale it is given by the distance between two nearest graduations or markings.
For example in a meter scale the distance between two nearest markings is 1mm, hence the least count of a meter scale is 1mm
Important Point
The maximum possible error present in any reading is equal to the least count of the instrument used.
Significant Figures
Each measurement involves errors due to various reasons. Hence we consider significant figures in the measured value to find its accuracy. Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement or vice versa.
In every experiment the measured results has a number that includes all reliably known digits and first uncertain digit .The combination of reliable digits and first uncertain digit are called the significant figures.
Example:
If a mass is measured as 5.26 gm then 5 and 2 are reliable while 6 is uncertain. Thus the measured value has three significant figures.
Rules of counting significant figures
(1) All non zero digits are significant.
Example:
1367 has four significant digits.
(2) All zeros present between 2 non zero digits are significant, irrespective of the position of the decimal point.
Example:
305009 has 6 significant figures.
(3) If there is no decimal point, all zeros to the right of the right-most non zero digit are considered to be significant only if they come from a measurement. (Note that a measured value from experiment must be written with its proper units)
Example:
86000 has only 2 significant digits. The three zeros which are underline are non-significant. Now consider a experimentally find length 91000 m, it has 5 significant digits. (Note that here 86000 is only a number whereas 91000 m is a measured value from some experiment because it is written with its unit i.e. meter)
(4) All zeros to the right of a decimal point but to the left of non-zero digits are considered to be non significant, provided there should be no non zero digit to the left of the decimal point.
Example:
0.00305 has 3 significant figures. The two zeros which are underline are non-significant
(5) All zeros are significant if they are placed to the right of a decimal point and to the right of a non zero digit.
Example:
0.04080 has 4 significant figures ( The underlined zero is non-significant according to 4th rule)
whereas 10.000 has 5 significant figures
(6)The number of significant figures does not alter in different units.
Example:
If we want to write 260 kg in different units, we can write it 2.60 x 105 gm, but in both the cases there are 3 significant figures.
(7) In case a number is written with 10n (like 32.65 x 108) then ignore 10n while finding the significant figure and use normal rules on the remaining number.
Illustration
State the number of significant figures in the following –
(a) 0.0038020
(b) 153400
(c) 58000 m
(d) 6.435 x 102 m
(e) 3.2 x 10-6 farad
(f) 8.400110
Solution (a) 5 (Rule No.- 4) (b) 4 (Rule No.-3)
(c) 5 (Rule No.-3) (d) 4 (Rule No.-7)
(e) 2 (Rule No.-7) (f) 7 (Rule No.-5)
Rules for Rounding off a quantity during mathematical operations
(1) If all the digits to be dropped or discarded are such that the first discarded digit is less than 5, the remaining digits are left unchanged.
Example:
8.321398 can be written in 4 significant figures as 8.321 ( because the first discarded digit (3) is less than 5
(2) If the digit to be dropped or discarded is 5 followed by digits other than zero, then the preceding digit should be raised by 1.
Example:
9.25007, on being rounded off to first decimal, became 9.3
(3) If the digits to be discarded is 5 or 5 followed by zero the preceding digit remains unchanged if it is even and the preceding digit is raised by 1 if it is odd.
Example:
3.6500 will become 3.6 and 4.7500 will become 4.8 in 2 significant figures.
Arithmetic operations with significant figures:
(1) Addition and subtraction In addition and subtraction, the number of decimal places in the result is the smallest number of decimal places of terms in the operation.
Example:
Let us consider the sum of following measurements.
2.47 sec., 9.6 sec. and 12.275 sec.
2.47 + 9.6 + 12.275 = 24.345
So the sum will be 24.3 sec as 9.6 sec has only 1 digit after the decimal point while the others are having more than one digit. So we apply rounding off rule to keep one digit after decimal. Hence we drop or discard 45 and apply rule no.(1) for rounding off
Multiplication and Division:
In the result of multiplication or division, the number of significant figures is same as the smallest number of significant figures among the numbers.
Example :
Lets Multiply 4.25 and 2.1.
So 4.25 x 2.1 = 8.925 ( Now apply rule no.1 to round off keeping only 2 significant figures in the final answer )
So the result is 8.9 as there are only 2 significant digits in 2.1
The same procedure is followed for division.
Illustration
Multiply 107.88 by 0.610 and express the result upto the correct number of significant figure.
(A) 65.8068 (B) 65.807
(C) 65.81 (D) 65.8
Solution
Number of significant figures in multiplication is three corresponding to the minimum number so 107.88´0.610 = 65.8068 = 65.8
\ Answer is (D)
Illustration
Compute the following length regard to significant figures.
(i) 876 + 0.4382
(ii) 8.0 – 0.42
(iii) 4.6\times 0.128
(iv) \frac{{0.9995\times 1.53}}{{1.592}}
Solution: Here,
(i) 876+0.4382=876.4382=876
(ii) 8.0-0.42\,\,\,\,=7.58\,\,\,=7.6 (round of to one place of decimal, as in 8.0)
(iii) 4.6\times 0.128=0.5888 = 0.59
(iv) \frac{{0.9995\times 1.53}}{{1.592}}=0.96057\,\,\,=0.961
Concept Of Errors In Measurement
In an experiment the difference in true value and measured value of a physical quantity is called error in measurement of that physical quantity.
There are mainly three categories of experimental error as given below.
(i) Systematic errors
(ii) Random errors
(iii) Gross errors
Important terms used in the calculation of errors
The calculation of errors is mainly based on the following three terms which are given below.
1. Absolute error
2. Mean absolute error
3. Fractional error
Absolute error
Let in an experiment n number of observations are made to measure a particular physical quantity.
Let the observations are a1, a2 , a3 , ………an.The arithmetic mean of the measured value is
{{a}_{m}}=\frac{{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+…….{{a}_{n}}}}{n}=\frac{1}{n}\sum\limits_{{i=1}}^{{i=n}}{{{{a}_{i}}}}
Now the absolute error in each observation is defined as the difference between the arithmetic mean and the measured value.
Hence by definition, absolute error in each observation can be written as,
\Delta {{a}_{1}}={{a}_{m}}-{{a}_{1}},\,
\Delta {{a}_{2}}={{a}_{m}}-{{a}_{2}}
\Delta {{a}_{3}}={{a}_{m}}-{{a}_{3}}
.
.
.
\Delta {{a}_{n}}={{a}_{m}}-{{a}_{n}}
The absolute errors may be positive or negative in different readings or observations.
Mean absolute error
It is the arithmetic mean of the magnitudes of absolute errors in all the measurements of the quantity
\Delta {{a}_{m}}=\frac{{|\Delta {{a}_{1}}|+|\Delta {{a}_{2}}|\,+\,……+|\Delta {{a}_{2}}|}}{n} =\frac{1}{n}\times \sum\limits_{{i=1}}^{{i=n}}{{|\Delta {{a}_{i}}|}}
Fractional error and Percentage error
It is defined as the ratio of mean absolute error to the mean value of the quantities measured.
Fractional error =\frac{{\Delta {{a}_{m}}}}{{{{a}_{m}}}}
Percentage error =\frac{{\Delta {{a}_{m}}}}{{{{a}_{m}}}}\times 100\%
Propagation Of Errors In Mathematical Operations
We know in an experiment differenct quantities are measured . Due to limitations of instruments or experimenatal conditions or experimenter himself the measurements always have some error.When ever these measured quantities with error are used in mathematical operations like addition,subtraction,multiplication or division etc the results will also have some error. So here we will see how the errors affect the result during mathematical operations.
Effect of errors in addition and subtraction
Addition of two quantities
Let we have a formula Z=A+B,
where A and B are two physical quantities and Z be the resultant of their addition.
But during the experiment the quantities A and B are measured with some error.
If DA and DB be the errors in A and B respectively, hence the measured values of A is A\pm \Delta Aand that of B is B\pm \Delta B. Due to errors in A and B,let DZ be the error in Z.
Therefore, Z\pm \Delta Z\,\,=(A\pm \Delta A)+(B\pm \Delta B)
A+B\pm \Delta Z\,\,=A\pm \Delta A+B\pm \Delta B ( because Z=A+B)
Hence \Delta Z=\,\pm \,\Delta A\pm \Delta B
Therefore the maximum absolute error in Z is given by
\displaystyle \Delta Z=\Delta A+\Delta B
Subtraction of two quantities
Let Z=A-B
Then, Z\pm \Delta Z=(A\pm \Delta A)-(B\pm \Delta B),
A-B\pm \Delta Z\,\,=A\pm \Delta A-B\pm \Delta B ( because Z=A-B)
Hence \Delta Z=\,\pm \,\Delta A\pm \Delta B
Therefore the maximum absolute error in Z is given by
\displaystyle \Delta Z=\Delta A+\Delta B
which is same as that of summation.
Note that the maximum possible error in the result in case of addition and subtraction is same and is equal to the addition of absolute errors of the two quantities which are added or subtracted . Hence we can say that errors are always additive in nature.
Effect of errors in multiplication and division
Multiplication
Suppose Z = A.B and the measured values of A and B are A ± DA and B ± DB. Then
Z ± DZ = (A ± DA) (B ± DB)
or Z ± DZ = AB ± B DA ± A DB ± DA DB
Dividing L.H.S. and R.H.S. by Z and putting Z= AB in the R.H.S. we have
\displaystyle 1\pm \frac{{\Delta Z}}{Z}=1\pm \frac{{\Delta A}}{A}\pm \frac{{\Delta B}}{B}\pm \left( {\frac{{\Delta A}}{A}} \right)\left( {\frac{{\Delta B}}{B}} \right).
Since DA and DB are small we shall ignore their product i.e. the last term, so we get
\displaystyle \frac{{\Delta Z}}{Z}=\frac{{\Delta A}}{A}\pm \frac{{\Delta B}}{B}
Hence the maximum fractional error in Z
\displaystyle \frac{{\Delta Z}}{Z}=\left( {\frac{{\Delta A}}{\text{A}}} \right)+\left( {\frac{{\Delta B}}{\text{B}}} \right)
Similarly, we can easily verify that this is true for division also.
So, when two or more quantities multiplied or divided, the fractional error in the result is the sum of the fractional errors of the quantities multiplied or divided
Let Z=\frac{A}{B}
Then, (Z\pm \Delta Z)\,\,=\frac{{A\pm \Delta A}}{{B\pm \Delta B}}
On solving we can easily prove that the maximum value of fractional error is
\displaystyle \frac{{\Delta Z}}{Z}=\left( {\frac{{\Delta A}}{\text{A}}} \right)+\left( {\frac{{\Delta B}}{\text{B}}} \right)
which is same as that in case of multiplication
Operations involving some quantities with power
Let Z={{A}^{2}}
Then, (Z\pm \Delta Z)\,\,={{(A\pm \Delta A)}^{2}}
Z\pm \Delta Z={{A}^{2}}\pm {{(\Delta A)}^{2}}\pm 2A.\Delta A
neglecting (DA)2 and dividing whole equation by Z
\frac{{Z\pm \Delta Z}}{Z}=\frac{{{{A}^{2}}\pm 2A.\Delta A}}{{{{A}^{2}}}}
so 1\pm \frac{{\Delta Z}}{Z}=1\pm \frac{{2A.\Delta A}}{{{{A}^{2}}}}
therefore \displaystyle \frac{{\Delta Z}}{Z}=\pm 2\left( {\frac{{\Delta A}}{A}} \right)
the digit 2 on RHS is actually the power of A in the given equation Z={{A}^{2}}
Hence note that the power of A is multiplied to the fractional error or percentage error of A.
Therefore if Z={{A}^{n}} then the maximum fractional error of Z is given by
\displaystyle \frac{{\Delta Z}}{Z}=\pm n\left( {\frac{{\Delta A}}{A}} \right)
Error in the result having product or division of quantities:
Let Z=\frac{{{{A}^{p}}{{B}^{q}}}}{{{{C}^{r}}}}
taking log on both sides we get
ln(Z)=p.ln(A) + q.ln(B) –r.ln(c)
on differentiation we get
\frac{{dZ}}{Z}=p.\frac{{dA}}{A}+q.\frac{{dB}}{B}+r.\frac{{dC}}{C}
For small change dZ\approx \Delta Z.
Þ \displaystyle \frac{{\Delta Z}}{Z}=p\frac{{\Delta A}}{A}+q\frac{{\Delta B}}{B}+r\frac{{\Delta C}}{C}
Illustration
Find the fractional error in Z, if \displaystyle Z=\sqrt{{\frac{{XY}}{M}}}
Solution
\displaystyle \frac{{\Delta \text{Z}}}{\text{Z}}=\frac{1}{2}\frac{{\Delta X}}{\text{X}}+\frac{1}{2}\frac{{\Delta Y}}{\text{Y}}+\frac{1}{2}\frac{{\Delta M}}{\text{M}}
Illustration
Find maximum possible percentage error in \displaystyle X=\frac{{{{a}^{\ell }}{{b}^{m}}}}{{{{y}^{p}}{{z}^{k}}}}
Solution
\displaystyle \frac{{\Delta X}}{\text{X}}\times 100=\left( {\ell \frac{{\Delta a}}{\text{a}}+m\frac{{\Delta b}}{\text{b}}+p\frac{{\Delta y}}{\text{y}}+k\frac{{\Delta z}}{\text{z}}} \right)\times 100
Illustration
In the relation x = 3yz2, x, y and z represents various physical quantities, if the percentage error in measurement of y and z is 3% and 1% respectively, then final maximum possible percentage error in x.
Solution
\displaystyle \frac{{\Delta x}}{\text{x}}\times 100=\left( {\frac{{\Delta y}}{\text{y}}+2\frac{{\Delta z}}{\text{z}}} \right) ´ 100
= 3% + 2 x 1% = 5%
Illustration
Calculate percentage error in the determination of g=4{{\pi }^{2}}l/{{t}^{2}} where l and t are measured with 2% and 3% errors respectively.
Solution
Here, g=4{{\pi }^{2}}l/{{t}^{2}}
\frac{{\Delta g}}{g}\times 100=\,\pm \left( {\frac{{\Delta l}}{l}\times 100+2\frac{{\Delta t}}{t}\times 100} \right) =\,\,\,\pm \left( {2+2\times 3} \right)\,=\,\pm 8\%
Illustration
In an experiment two capacitor measured are (1.3\pm 0.1)\mu \mathbf{F} and (2.4\pm 0.2)\,\mu \mathbf{F}. Calculate the capacitance in parallel with percentage error.
Solution
Here, {{C}_{1}}=(1.3\pm 0.1)\,\mu \mathbf{F} and {{C}_{2}}=(2.4\pm 0.2)\,\mu \mathbf{F}
In parallel, {{C}_{p}}={{C}_{1}}+{{C}_{2}}=1.3+2.4=3.7\,\mu \mathbf{F}
\Delta {{C}_{p}}=\,\,\pm \,(\Delta {{C}_{1}}+\Delta {{C}_{2}})
=\,\pm \,\,(0.1+0.2)\,\,=\,\pm \,0.3
% age error =\,\,\pm \frac{{0.3}}{{3.7}}\times 100\,\,\,=\pm \,8.1\%
Hence, {{C}_{p}}=(3.7\,\pm \,0.3)\,\mu \mathbf{F}
=3.7\,\mu F\,\,\,\pm \,\,\,8.1\,\%
Objective Assignment
1.
A wire is of mass (0.3 ± 0.003)gm. The radius is (0.5 ± 0.005)mm and length is (6.0 ± 0.06)cm then % error in density is
(a) 3 (b) 4
(c) 6 (d) -2
Ans: (b)
2.
Error in the measurement of radius of sphere is 2%. Then error in the calculation of volume will be
(a) 1% (b) 5%
(c) 3% (d) 6%
Ans: (d)
3.
The period of oscillation of a simple pendulum is given by T=2\pi \sqrt{{\frac{l}{g}}} where l is about 100 cm and is known to have 1mm The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is
(a) 0.1% (b) 1%
(c) 0.2% (d) 0.8
Ans: (c)
4.
What is the number of significant figures in 0.310×103
(a) 2 (b) 3
(c) 4 (d) 6
Ans: (b)
5.
A body travels uniformly a distance of (13.8 ±2) m in a time (4.0 ± 0.3) s. The velocity of the body within error limits is
(a) (3.45 ± 0.2) ms-1 (b) (3.45 ± 0.3) ms-1
(c) (3.45 ± 0.4) ms-1 (d) (3.45 ± 0.5) ms–1
Ans: (b)
6.
The decimal equivalent of 1/20 upto three significant figures is
(a) 0.0500 (b) 0.05000
(c) 0.0050 (d) 5.0 × 10-2
Ans: (a)
7.
Accuracy of measurement is determined by
(a) Absolute error (b) Percentage error
(c) Both (d) None of these
Ans: (b)
8.
If L= 2.331cm and B = 2.1cm, then L+ B =?
(a) 4.431 cm (b) 4.43 cm
(c) 4.4 cm (d) 4 cm
Ans: (c)
9.
The number of significant figures in all the given numbers 25.12, 2009, 4.156 and 1.217 x 10-4 is
(a) 1 (b) 2
(c) 3 (d) 4
Ans: (d)
10.
A physical quantity is given by X={{M}^{a}}{{L}^{b}}{{T}^{c}}. The percentage error in measurement of M,L and T are \alpha ,\beta and \gamma respectively. Then maximum percentage error in the quantity X is
(a) a\alpha +b\beta +c\gamma
(b) a\alpha +b\beta -c\gamma
(c) \frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }
(d) None of these
Ans : (a)
Subjective Assignment
1.
If error in measuring diameter of a circle is 4%, find the error in calculating area of circle.
Ans : 8 %
2.
To study the flow of a liquid through a narrow tube the following formula is used
\displaystyle \eta =\frac{{\pi p{{r}^{4}}}}{{8v\ell }}where the letters have their usual meanings. The values of p, r, v and l are measured to be 76 cm of Hg, 0.28 cm, 1.2 cm3s–1 and 18.2 cm respectively. If these quantities are measured to the accuracy of 0.5 cm of Hg, 0.01 cm, 0.1 cm3 s–1 and 0.1 cm respectively, find the percentage error in the value of h.
Ans : 23.84%
3.
If all measurements in an experiment are taken upto same number of significant figures then mention two possible reasons for maximum error
Ans : The maximum error will be due to
(i) measurement, which is least accurate.
(ii) measurement of the quantitiy which has maximum power in the expression
4.
Find out the result in proper significant figures, 291 x 0.03842 / 0.0080.
Ans : 1.4 ´ 103
5.
The radius of a sphere is (5.3 ± 0.1) cm. Find the percentage error in its volume.
Ans : 5.7%
6.
The time period of small oscillations of a spring mass system is given as T = \displaystyle 2\pi \sqrt{{\frac{m}{k}}}. What will be the accuracy in the determination of k if mass m is given as 10 kg with accuracy of 100 gm and time period is 0.5 sec measured for time of 100 oscillations with a watch of accuracy of 1 sec.
Ans : ±5%
7.
If RI = 20 ± 2 W and R2 = 80 ± 5 W are connected in parallel. Then find the absolute error in the equivalent resistance
Ans : 3.72 W
8.
A physical quantity X is depending upon physical quantities A and B by the relation X = A2 – A3/B. In an experiment the value of A is measured as (2.0 ± 0.2) and of B is measured as (1.0 ± 0.1). Then represent the value of X.
Ans : -4.0 ± 2.4
9.
Calculate percentage error in the determination of g from the formula of time period of oscillations of a pendulum \displaystyle T=2\pi \sqrt{{\frac{l}{g}}}, where l and T are measured with maximum percentage error of 2 % and 3 % respectively.
Ans : 8%
10.
The percentage errors in the easurement of mass and speed are 2% and 3% respectively. How much will be the maximum error in the estimation of kinetic energy obtained by measuring mass and speed?
Ans : 8%
11.
The diameter of a given wire is measured by a screw gauze. The three measurements of the diameter give the reading in cm as 0.036, 0.035 and 0.037. What is the percentage error of the measurement?
Ans : 2.8%
12.
A physical quantity is described by the relation \displaystyle P\,=\,{{a}^{{1/2}}}\,{{b}^{2}}\,{{c}^{2}}\,{{d}^{{-4}}}. If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then , find the relative error in P.
Ans : 32%