Video Lecture

Theory For Making Notes

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\,\%

 

Practice Questions (Basic Level)

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) ms1

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) 

Practice Questions (JEE Main Level)

1.

A physical quantity x is calculated from the relation x={{a}^{3}}{{b}^{2}}/\sqrt{{cd}} . Calculate percentage error in x, if a,b,c\,\And d measured respectively with an error of 1%, 3%, 4% and 2%.

(a) 12%                               

(b) 6%           

(c) 14%                                 

(d) 4%

Ans : (a)

2.

We have error in the measurement of length, radius, mass and current of a wire are 2%, 3%, 2% and 1% then error in its density will be

(a) 11%                             

(b) 8%                                  

(c) 10%                             

(d) 7%

Ans : (c)

3.

If a copper wire is stretched to make its radius decrease by 0.1%, then the percentage increase in resistance is approximately

(a) 0.1 %                         

(b) 0.2 %                            

(c) 0.4%                         

(d) 0.8%

Ans : (c)

 4.

The diameter of a wire is measured with a screw gauge having 50 divisions on circular scale and by one complete rotation of circular scale, main scale moves 0.5 mm. If reading of screw gauge is 0.250 cm. The minimum percentage error in the reading will be

(a) 0.4                                 

(b) 0.8           

(c) 4                                     

(d) 5

Ans : (a)

5.

The length, width and thickness of a block are (100.0 ± 0.1) cm, (10.00 ± 0.01) cm and (1.000 ± 0.001) cm respectively. The maximum possible error in its volume will be

(a) ±  0.111 cm3               

(b) ± 0.012 cm3                 

(c) + 0.03 cm3                 

(d) none of these

Ans : (d)

6.

The random error in the arithmetic mean of 100 observations is x, then random error in the arithmetic mean of 400 observations would be

(a)   4x                                 

(b)   \frac{1}{4}x

(c)   2x                                 

(d)   \frac{1}{2}x

Ans : (b)

7.

The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will be

(a)   1%                                  

(b)   2%

(c)   3%                                  

(d)   4%

Ans : (c)

8.

A physical quantity P is given by P= \frac{{{{A}^{3}}{{B}^{{\frac{1}{2}}}}}}{{{{C}^{{-4}}}{{D}^{{\frac{3}{2}}}}}}. The quantity which brings in the maximum percentage error in P is

(a)   A                                     

(b)   B

(c)   C                                     

(d)   D

Ans : (c)

9.

If the length of rod A is 3.25 ±01 cm and that of B is 4.19 ± 0.01 cm then the rod B is longer than rod A by 

(a)   0.94  ± 0.00 cm            

(b)   0.94  ± 0.01 cm

(c)   0.94  ± 0.02 cm            

(d)   0.94  ± 0.005 cm

Ans : (c)

10.

Read the statements and choose the right option

Statement A    

Out of three measurements l = 0.7 m, l = 0.70 m and l = 0.700 m, the last one is most accurate.

Statement B               

In every measurement, only the last significant digit is not accurately known

(a) Only statement A is correct

(b) Only statement B is correct

(c) Both the statements are wrong

(d) Both the statements are correct

Ans : (d)

Practice Questions (JEE Advance Level)

1.

In the context of accuracy of measurement and significant figures in expressing results of experiment, which of the following is/are correct

(1) Out of the two measurements 50.14 cm and 0.00025 ampere, the first one has greater accuracy

(2) If one travels 478 km by rail and 397 m by road, the total distance travelled is 478 km.

(a)   Only (1) is correct       

(b)   Only (2) is correct

(c)   Both are correct          

(d)   None of them is correct.

Ans : (a)

2.

The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is (5.00±05) Newton and weight in water is (4.00±0.05) Newton. Then the relative density along with the maximum permissible percentage error is

(a)     5.0 ±11%                       

(b)   5.0 ±1%

(c)     5.0 ±6%                         

(d)   1.25 ±5%

Ans : (a)

3.

Two resistances of 400 W and 800 W connected in series with a 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000W is used to measure the potential difference across 400 W. The error in the measurement of potential difference in volts approximately is

(a)  0.01                             

(b)  0.02                              

(c) 0.04                             

(d) 0.05

Ans : (c)

4.

If there is a positive error of 50% in the measurement of velocity of a body, then the error in the measurement of kinetic energy is

(a) 25 %                             

(b) 50 %                              

(c) 100 %                           

(d) 125 %

Ans : (d)

Comprehension ( Q.no. 5 – 7 )

The diameter of a wire as measured by a screw gauge was found to be 1.328, 1.330, 1.326, 1.334 and 1.336 cm.  Then answer the following questions based on above information.

5.

Express the result in terms of absolute error.

(a) 1.330 ± 0.004 cm                     

(b) 1.330 + 0.004 cm

(c) 1.330 – 0.004 cm                     

(d) 1.326 ± 0.005 cm

Ans : (a)

6.

What is the fractional error in the measurement of the diameter of the wire,

(a) 0.001                             

(b) 0.002

(c) 0.003                             

(d) 0.004

Ans : (a)

7.

What is the possible pitch length and no. of divisions on the circular scale of the screw gauge?

(a) 1 mm, 50 divisions             

(b) ½ mm, 50 division

(c) ½ mm, 100 division           

(d) 1 mm, 200 division

Ans : (b)

8.

 

The graph between object distance (u) and image distance (v) is as shown in the figure. What is the focal length of the convex lens

(a) 5.00 cm,  ±0.05 cm 

(b) 5.00 cm, ±0.10 cm

(c) 0.5 cm, ± 0.05 cm    

(d) 0.5 cm, ±0.10 cm

Ans : (a)