`●` Rules for presenting data

`●` Scientific Notation

`●` Significant Figures and Determination of Number of Significant Figures

`●` Addition, Subtraction, Multiplication and Division of Significant Figures

`●` Rules for Rounding off the Numbers

`●` Dimensional Analysis

`●` Scientific Notation

`●` Significant Figures and Determination of Number of Significant Figures

`●` Addition, Subtraction, Multiplication and Division of Significant Figures

`●` Rules for Rounding off the Numbers

`●` Dimensional Analysis

To present the data, obtained experimentally or theoretically, some rules are followed for the sake of convenience

Since atoms and molecules have extremely low mass and are large in numbers. So, for dealing with such a large and small number scientific notation is used. It is also called exponential notation in which any number is represented as : eg : Andromeda galaxy (closest to milky way galaxy ) contains at least `2, 00 , 000, 000 , 000` stars. which is a large that causes problem so we write it in scientific notation `2.0xx10^(11)`.

`N xx 10^n`

where `n =` an exponent having positive or negative values.

`N =` any number between `1.000-9.999` is called digit term.

e.g. (i) `232.508` can be written as `2.32508xx10^2`.

(ii) `0.00016` can be written as `1.6xx10^(-4)`.

`N xx 10^n`

where `n =` an exponent having positive or negative values.

`N =` any number between `1.000-9.999` is called digit term.

e.g. (i) `232.508` can be written as `2.32508xx10^2`.

(ii) `0.00016` can be written as `1.6xx10^(-4)`.

Q 2684078857

Exponential notation in which any number can be represented in the form `N xx 1 0^n` , here `n` is an exponent having positive or negative values and N is a number called ....A... which varies between 1.000 and 9.999. Here, A refers to

(A)

non-digit term

(B)

digit term

(C)

numeral

(D)

base term

Exponential notation in which any number can be represented in the form `N xx 10^n`'. where `n` is an exponent having positive or negative values and `N` is a number called digit term which varies between `1.000` and `9.999`.

Correct Answer is `=>` (B) digit term

Q 2604078858

Few figures are expressed in scientific notation. Mark the incorrect one

(A)

`234000 = 2.34xx10^5`

(B)

`8008 = 8xx10^3`

(C)

`0.0048 = 4.8xx10^(-3)`

(D)

`500.0 = 5.00xx10^2`

`8008 = 8.008xx10^3`

Correct Answer is `=>` (B) `8008 = 8xx10^3`

Q 1933580442

Express the following In the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6,0012

Class 11 Exercise 1 Q.No. 18

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6,0012

Class 11 Exercise 1 Q.No. 18

(i) `4.8 xx 10^(-3)`

(ii) `2.34 xx 10^5`

(iii) `8.008 xx 10^3`

(iv) `5.000 xx 10^2`

(v) `6.0012 xx 10^10`

Both these operation follow the some rules i.e.

e.g. (i) `(5.6xx10^5)xx (6.9xx10^8) = (5.6xx6.9)(10^(5+8)) = (5.6xx6.9)xx10^(13) = 38.64xx10^(13)`

(ii) `(9.8xx10^(-2)) xx (2.5xx10^(-6)) = (9.8xx2.5)xx10^(-2+(-6)) = (9.8xx2.5) (10^(-2-6))` `= 24.50xx10^(-8)`

(iii) `(2.7xx10^(-3))/(5.5xx10^4) = (2.7 div 5.5 )(10^(-3-4)) = 0.4909 xx10^(-7)`

e.g. (i) `(5.6xx10^5)xx (6.9xx10^8) = (5.6xx6.9)(10^(5+8)) = (5.6xx6.9)xx10^(13) = 38.64xx10^(13)`

(ii) `(9.8xx10^(-2)) xx (2.5xx10^(-6)) = (9.8xx2.5)xx10^(-2+(-6)) = (9.8xx2.5) (10^(-2-6))` `= 24.50xx10^(-8)`

(iii) `(2.7xx10^(-3))/(5.5xx10^4) = (2.7 div 5.5 )(10^(-3-4)) = 0.4909 xx10^(-7)`

For both the operations, the numbers are written in the form which have some exponent. And after that coefficient are added or subtracted.

e.g. For Addition of `6.65xx10^4` and `8.95xx10^3`.

Step I : Make the exponent same. We can do it by changing the exponent of `6.65xx10^4` to `6.65xx10^3` or by changing the exponent of `8.95xx10^3` to `0.895xx10^2`.

Step II : Addition `(66.5xx10^3+8.95xx10^3) = 75.45xx10^3 = 7.545xx10^4`

or `(6.65xx10^4+0.895xx10^4) = 7.545xx10^4`

Similarly, subtraction of two numbers can be done in the same manner.

`2.5xx10^(-2) - 4.8xx10^(-3)`

`= (2.5 xx 10^(-2))-(0.48xx10^(-2))`

`= (2.5 - 0.48) xx 10^(-2) = 2.02xx 10^(-2)`

e.g. For Addition of `6.65xx10^4` and `8.95xx10^3`.

Step I : Make the exponent same. We can do it by changing the exponent of `6.65xx10^4` to `6.65xx10^3` or by changing the exponent of `8.95xx10^3` to `0.895xx10^2`.

Step II : Addition `(66.5xx10^3+8.95xx10^3) = 75.45xx10^3 = 7.545xx10^4`

or `(6.65xx10^4+0.895xx10^4) = 7.545xx10^4`

Similarly, subtraction of two numbers can be done in the same manner.

`2.5xx10^(-2) - 4.8xx10^(-3)`

`= (2.5 xx 10^(-2))-(0.48xx10^(-2))`

`= (2.5 - 0.48) xx 10^(-2) = 2.02xx 10^(-2)`

Precision : It is the closeness of various measurements for the same quantity.

Accuracy : It is the agreement of a particular value to the true value of the result

for example : value of g ( acceleration due to gravity `= 9.8065 m//s^2`) .

Now if the observed value are - `9.8714 , 9.8713 , 9.8712 , 9.8715` they are precise and if observed value are `9.81 , 9.79 , 9.82 , 9.77 => ` accurate.

`text(Importance of accuracy and percision)` :

(1) Accuracy comes in importance when you need to hit a target . Either you hit the target or don't hit it. It does no good to be precise if you miss all the shots .

(2). Precision gains importance in calculations when you use a measured value in calculation , you can be as precise as your least measurement from here arises the idea of significant figures ( which includes all the digits you know for certain plus the last digit , which contains come uncertainity).

For e.g., See Table 1.4.

Note : (i) The uncertainty in any value is indicated by mentioning the number of signification figures.

(ii) Significant figures are meaningful digits which have certainty.

(iii) Uncertainty is indicated by writing the certain digits and the last uncertain digit e.g. if there is a result written as `11.2` `mL`.

Formal Definition : The total number of digits in a number including the last digit whose value is uncertain is called

the number of significant figures.

Here, `11` is certain and `2` is uncertain. And uncertainty would be in the last digit. If uncertainty is not given then an uncertainty of `pm1` in the last digit is understood.

Accuracy : It is the agreement of a particular value to the true value of the result

for example : value of g ( acceleration due to gravity `= 9.8065 m//s^2`) .

Now if the observed value are - `9.8714 , 9.8713 , 9.8712 , 9.8715` they are precise and if observed value are `9.81 , 9.79 , 9.82 , 9.77 => ` accurate.

`text(Importance of accuracy and percision)` :

(1) Accuracy comes in importance when you need to hit a target . Either you hit the target or don't hit it. It does no good to be precise if you miss all the shots .

(2). Precision gains importance in calculations when you use a measured value in calculation , you can be as precise as your least measurement from here arises the idea of significant figures ( which includes all the digits you know for certain plus the last digit , which contains come uncertainity).

For e.g., See Table 1.4.

Note : (i) The uncertainty in any value is indicated by mentioning the number of signification figures.

(ii) Significant figures are meaningful digits which have certainty.

(iii) Uncertainty is indicated by writing the certain digits and the last uncertain digit e.g. if there is a result written as `11.2` `mL`.

Formal Definition : The total number of digits in a number including the last digit whose value is uncertain is called

the number of significant figures.

Here, `11` is certain and `2` is uncertain. And uncertainty would be in the last digit. If uncertainty is not given then an uncertainty of `pm1` in the last digit is understood.

Q 2624178951

If we write a result as `11.2 mL`. Match the items of Column I with Column II and choose the correct option from the codes given below

(A)

`A →3 , B →1 , C →2`

(B)

`A →1 , B →2 , C →3`

(C)

`A →3 , B →2 , C →1`

(D)

`A →2 , B →1 , C →3`

Correct Answer is `=>` (A) `A →3 , B →1 , C →2`

Q 2674178956

Least count of an analytical balance is `0.0002 g`. Which of the following is the correct recorded value?

(A)

`9.3171 g`

(B)

`9.31 7g`

(C)

`9.3175 g`

(D)

`9.3170g`

Result should be reported up to four decimal places such that fourth decimal digit should be even number (including zero i.e. 0, 2, 4, 6, 8)

Correct Answer is `=>` (D) `9.3170g`

Q 2644180053

The least count of an instrument is `0.01 cm`. Taking all precautions, the most possible error in the measurement can be

(A)

`0.005 cm`

(B)

`0.01 cm`

(C)

`0.0001 cm`

(D)

`0.1 cm`

As we know the least count of the instrument is equal to the most possible error of the instrument

Hence least count of the instrument will be `0.01 cm`

Correct Answer is `=>` (B) `0.01 cm`

Q 1923178941

What do you mean by significant figures?

Class 11 Exercise 1 Q.No. 16

Class 11 Exercise 1 Q.No. 16

The total number of digits in a number including

the last digit whose value is uncertain is called

the number of signiftcant figures.

Defination : (1). A leading zero is any 0 digit that comes before the first nonzero digit

(2). trailing zeros are a sequence of 0's after which no other digit follows.

(i) All non-zero digits are significant.

(ii) Zeros preceding to first non-zero digit are not significant. These zeros only indicates the position of decimal point only.

(iii) Zeros between two non-zero digit are significant.

(iv) Zero at the end or right of a number are significant if they are on the right side of the decimal point.

e.g. `0.200g` has `3` significant figures.

If there is no decimal point then the terminal zeros are not significant.

e.g. `tt((text{Number} , text{no. of significant figures} ) , (100 , 1) , (100. , 3) , (100.0 , 4))`

(v) Objects which can be counted have infinite significant figure as these are exact numbers. And can be represented as writing infinite number of zero after placing a decimal.

e.g. `2 = 2.000000` or `20 = 20.00000`

Q 2614280150

`285 cm` has .. . A ... significant figure(s) and `0.25 mL` has .. . B ... significant figure(s). Here, A and B refer to

(A)

A → three , B → two

(B)

A → two , B → three

(C)

A → four , B → two

(D)

A → three , B → three

285 cm has three significant figures and 0.25 mL has two significant figures

Correct Answer is `=>` (A) A → three , B → two

Q 2644280153

0.03 has .. . A ... significant figure(s) and 0.0052 has .. . B ... significant figure(s). Here, A and B refer to

(A)

A → one , B → two

(B)

A → two , B → one

(C)

A → two , B → four

(D)

A → one , B → three

0.03 has one significant figure and 0.0052 has two significant figures.

Correct Answer is `=>` (A) A → one , B → two

Q 2664280155

2.005 has .. . A ... significant figures. Here, A refers to

(A)

Four

(B)

Three

(C)

Two

(D)

One

2.005 has four significant figures.

Correct Answer is `=>` (A) Four

Q 2614380250

`1.00xx10^2` has ..............A............significant figures. Here, A refers to

(A)

Two

(B)

One

(C)

Three

(D)

zero

`1xx10^2` has one significant figure

`1.00xx10^2` has three significant figures

Correct Answer is `=>` (C) Three

Q 2684380257

How many significant figures are present in `0.010100 xx 10^3 ?`

(A)

`7`

(B)

`5`

(C)

`3`

(D)

`10`

`0.010100 xx 10^3` contains 5 significant figures

Correct Answer is `=>` (B) `5`

Q 1983680547

How many significant figures are present in

the following?

(i) 0.0025

(ii) 208

(iii) 5005

(iv) 1261000

(v)500.0

(vi) 2.00

Class 11 Exercise 1 Q.No. 19

the following?

(i) 0.0025

(ii) 208

(iii) 5005

(iv) 1261000

(v)500.0

(vi) 2.00

Class 11 Exercise 1 Q.No. 19

(i) 2

(ii) 3

(iii) 4

(iv) 3

(v) 4

(vi) 5

Q 1923080841

Round up the following upto three significant

figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

Class 11 Exercise 1 Q.No. 20

figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

Class 11 Exercise 1 Q.No. 20

(i) 34.2

(ii) 10.4

(iii) 0.0460

(iv) 2.80

The result obtained can not have more digits to the right of the decimal point than either of the original numbers.

e .g.

`12.11`

`18.0`

`1.012`

` underline overline(31.122)`

(Since 18.0 has only one digit after decimal point . So, the answer should be reported as `31.1`.)

e .g.

`12.11`

`18.0`

`1.012`

` underline overline(31.122)`

(Since 18.0 has only one digit after decimal point . So, the answer should be reported as `31.1`.)

Q 2644480353

What will be the answer in appropriate significant figures as a result of addition of `3.0223` and `5.041?`

(A)

`80.633`

(B)

`8.0533`

(C)

`8.063`

(D)

`806.33`

Since, `5 041` has only 3 digits after the decimal point, the result should be reported to 3 digits after decimal point.

Correct Answer is `=>` (C) `8.063`

Q 2614480359

Which of the following option is not correct?

(A)

`2.300 + 0.02017 + 0.02015 = 2.34032`

(B)

`126000` has 3 significant figures

(C)

`15.05 mus = 1.515xx10^(-50 s`

(D)

`0.0048 = 48 xx 1 0^(-3)`

`0.0048 = 4.8 xx 10^(-3)`

Correct Answer is `=>` (D) `0.0048 = 48 xx 1 0^(-3)`

The result obtained should have the significant figures equal to the number which has minimum number of significant figures.

e.g. `2.5xx1.25 = 3.125`

So, answer should be reported as `3.1`.

e.g. `2.5xx1.25 = 3.125`

So, answer should be reported as `3.1`.

Q 2674580456

`18.72 g` of a substance `'X'` occupies `1.81 cm^3`. What will be its density measured in correct significant figures?

(A)

`10.3 g//cm^3`

(B)

`10.34 g//cm^3`

(C)

`10.4 g//cm^3`

(D)

`10.3425 g//cm^3`

`text(Density) = text(Mass)/text(Volume) = (18.729)/(1.81 cm^3) = 10.34 g// cm^3`

Since. volume contains 3 significant figures hence answer will be `10.3 g cm^(-3)`

Correct Answer is `=>` (A) `10.3 g//cm^3`

(i) If the right most digit to be removed `> 5`, then preceding number is increased by one.

e.g. `1.386`

If we remove `6`, then it is rounded off to `1.39`.

(ii) If the right digit to be removed `< 5`, then the preceding number is not changed.

e.g. `4.334`

If we remove `4`, then it is rounded off to `4.33`.

(iii) If the right most digit to be removed `= 5`

Case I : If preceding number is even, then it is not changed.

Case II : If preceding number is odd, then it is increased by one.

e.g. `1.386`

If we remove `6`, then it is rounded off to `1.39`.

(ii) If the right digit to be removed `< 5`, then the preceding number is not changed.

e.g. `4.334`

If we remove `4`, then it is rounded off to `4.33`.

(iii) If the right most digit to be removed `= 5`

Case I : If preceding number is even, then it is not changed.

Case II : If preceding number is odd, then it is increased by one.

Q 2634580452

Which set of figures will be obtained after rounding off the following to three significant figures? 34.216, 0.04597, 10.4107

(A)

`34.3 , 0. 0461, 10.4`

(B)

`34.2, 0.0460, 10.4`

(C)

`34.20, 0.460, 10.40`

(D)

`34.21, 4.597, 1.04`

`34.216 = 34.2 , 0.04597 = 0.0460 , 10.4107 = 10.4`

Correct Answer is `=>` (B) `34.2, 0.0460, 10.4`

For converting units from one system to other a method was accomplished which is called `text(factor label method)` or `text(unit factor method)` or `text(dimensional analysis)`.

Note : Units can be handled just like other numerical part. It can be cancelled, divided, multiplied, squared etc.

Note : Units can be handled just like other numerical part. It can be cancelled, divided, multiplied, squared etc.

Q 2664267155

A piece of metal is `3` inch (represented by in) long. What is its length in `cm`?

We know that `1 ` in` = 2.54` cm

From this equivalence, we can write

`(1 i n ) /(2.54 cm) = 1 = (2.54 cm)/(1 i n )`

Thus `(1 i n)/(2.54 cm) ` equals 1 and `(2.54 cm)/(1 i n)` also equals 1. Both of these are called unit factors. If some number is multiplied by these unit factors (i.e. 1), it will not be affected otherwise.

Say, the 3 in given above is multiplied by the unit factor. So,

`3 i n = 3 i n xx (2.54 cm)/(1 i n) = 3xx2.54 cm = 7.62 cm`

Q 2604267158

How many seconds are there in 2 days

Here we know 1 day = 24 hours

`text(1 day)/text(24 h) = 1 = text(24 h)/ text(1 day)`

then `1h = 60 text(min)`

or `(1 h)/text(60 min) = 1 = text(60 min)/(1 h)`

so, for converting 2 days to seconds

i . e 2 days ........................ = ................. seconds

The unit factors can be multiplied in series in one step only as follows

`2text(day) xx (24 h)/text(1 day) xx text(60 min)/(1 h) xx (60 s)/text(1 min)`

` = 2xx24xx60xx60s`

` = 172800 s`

Q 2684267157

A jug contains `2L` of milk. Calculate the volume of the milk in `m^3`.

Since `1L = 1000 cm^3` and `1m = 100 cm ` which gives

`(1 m)/(100 cm) = 1 = (100 cm)/(1 m)`

To get `m^3` from the above unit factors the first unit factor is taken and it is cubed .

`((1 m)/(100 cm))^3 => (1 m^3)/(10^6 cm^3) = (1)^3 = 1`

Now `2L = 2xx1000 cm^3`

the above is multiplied by the unit factor

`2xx1000cm^3 xx (1 m^3)/(10^6 cm^3) = (2 m^3)/(10^3) = 2xx10^(-3) m^3`

Q 1923180941

If the speed of light is `3.0 xx 10^8 m s^(-1)`, calculate

the distance covered by light in `2.00` ns.

Class 11 Exercise 1 Q.No. 21

the distance covered by light in `2.00` ns.

Class 11 Exercise 1 Q.No. 21

Distance covered= Speed x Time

`= 3.0 xx 10^8ms^(-1) xx 2.00 ns`

`= 3.0 xx 10^8 ms^(-1) xx 2.00 ns xx (10^(-9)s)/(1 ns)`

`=6.00 xx 1-^(-1) m=0.600m`

Q 2634778652

Consider the following unit volumes of energy `A : 1 L`-atm, `B : 1` `erg`, `C : 1J`, `D : 1 ` kcal, Increasing order of these values is

(A)

`A = B = C = D`

(B)

`A < B < C < D`

(C)

`B < C < A < D`

(D)

`D < A < C < B`

`R = 0.0821 L` atm `mol^(-1) K^(-1) = 8.314xx10^7ergs mol^(-1) K^(-1)`

`= 8.314 J mol^(-1) K^(- 1) = 0.002 kcal mol^(-1 ) K^(-1)` ` = x` (assume)

`therefore 1 L ` atm= `x/(0.0821 ) = A`

`1 erg = x/(8.314xx10^7 ) = B`

`1J = x/(8.314) = C`

`1 kcal = x/(0.002) = D`

Thus, `B < C < A < D`

Correct Answer is `=>` (C) `B < C < A < D`