units and measurements class 11 assignment

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Units and Measurements Class 11 Notes PDF (Handwritten & Short Notes)

The Units and Measurements is one of the important chapters of the Class 11th Physics. Students can easily cover the chapter with the help of the Units and Measurements Class 11 notes. In the notes, all the concepts and topics are explained in a creative and precise manner. 

The Units and Measurements Class 11 notes are considered to be important study material throughout the preparation. As it helps students to understand the chapter Units and Measurements in many different ways. One of the important ways is by practising questions given in the chapter Units and Measurements notes. By practising many questions students can increase their overall score in the examination by performing too well in the examination. 

Units and Measurements Notes PDF

The Class 11 Physics notes is a written explanation for every passage or concepts included in the chapter Units and Measurements. These Units and Measurements notes are available in the Portable Document Format. The PDF of the Class 11 Physics Notes can be easily downloaded without any difficulty. With the help of the downloaded revision notes, Class 11 students can pay attention to the chapter Units and Measurements to be able to perform well in the board examinations as well as in the competition exam. 

How to Download the Units and Measurements Class 11 Notes?

To refer to the Units and Measurements Class 11 notes, students can look through the given steps. Those important steps are: 

• Open the Selfstudys website. 
• Click the navigation bar, a drop down menu will appear. 
• Select NCERT Books & Solutions from the given list. 
• Click NCERT Notes from the given list.

• A new page will appear, select Class 11th from the list of classes.

• Select Physics from the list of subjects.

• Again a new page will appear, select the chapter Units and Measurements to be able to download the revision Notes. 

Features of the Units and Measurements Class 11 Notes

Features are considered to be an important or noticeable part of the Units and Measurements Class 11 notes. Those important features are: 

• Explanations are Provided: In the Units and Measurements notes, explanations are provided to each and every topic as well as definitions. According to the explanations, students can understand the chapter Units and Measurements in a fine way. 
• Different Types of Questions are Included: Inside the Class 11 Units and Measurements notes, different types of questions are included. Those questions are: example questions, objective type questions, very short type answers, short type questions, long type questions, numerical problems, etc. 
• Simple Language: These Class 11 Physics notes are created by subject matter experts in a very easy and simple language. It is formed in an easy language so that Class 11 students can understand each topic and question of the chapter Units and Measurements. 
• Colourful Content are Given: Content included in the Class 11 Units and Measurements notes are in a colourful manner. This colourful content can attract many Class 11 students to cover the chapter Units and Measurements in a proper way. 
• Hints and Solutions are Provided: To tackle the questions given in Units and Measurements Class 11 Notes, students can refer to the Hints and Solutions given in the PDF file of Class 11 Revision Notes of Units and Measurements. With the help of hints and solutions, students can not only tackle the questions but can develop a good understanding in the overall chapter.

Advantages of the Units and Measurements Class 11 Notes

Completing the chapter with the help of Units and Measurements Class 11 notes can provide good results to students. This is one of the important advantage of Class 11 Physics notes, other than this are: 

• Helps to Memorise Topics: With the help of Class 11 Physics notes, students can easily memorise topics discussed in the chapter Units and Measurements. 
• Summary is Given: After covering all topics of the chapter Units and Measurements, a brief summary is given. Through the summary, students can get an idea about the important topics and definitions. 
• A Quick Revision Can Be Done: With the help of Class 11 Physics notes, students can have a quick revision of the chapter Units and Measurements. Through the last minute revision, students can identify the weak points for the chapter Units and Measurements. 
• All Topics are Covered: In the Units and Measurements notes, all the topics and concepts are covered in an elaborate manner. Understanding the topics and concepts in a detailed way can help students to increase their comprehensive skills. 
• Increases the Accuracy Level: Routine exercise of questions from the chapter Units and Measurements can help students to increase their accuracy level. Accuracy level mainly helps Class 11 students to increase their quality in giving answers for the chapter Units and Measurements.

When Is The Right Time To Go Through Units and Measurements Class 11 Notes?

After having a brief knowledge about the chapter Units and Measurements, students can look through the Units and Measurements Class 11 notes. With the help of notes, students can recall all topics and concepts of Units and Measurements in a better way. 

Although, generally, it is advised to all the students to use the revision notes on a weekly basis to recall the previous learnings. Along with this, using the Units and Measurements Class 11 notes at the time of exam preparation helps a lot so, using the Class 11 Physics Notes during exam preparation can be the right time.

Tips for Students to Cover the Chapter Units and Measurements Using Notes

It is a must that Class 11 students should follow some strategic tips to understand Units and Measurements. By understanding topics, students can score well in the questions related to the Units and Measurements. Below, we have mentioned some strategic tips, students can use -

• Look Through the Chapter: First tip is to look through all the topics included in the chapter Units and Measurements. Students can get a brief idea about all the topics and concepts of the Class 11th Physics chapter using Notes or Syllabus. 
•   Complete the Concepts: After getting some idea about the chapter and its topics, students can complete all the topics and concepts with the help of Units and Measurements Class 11 notes. However, notes are only ideal to recall the previou learning so, with the help of Units and Measurements Class 11 notes students can only revise the complete concepts of the Class 11 Physics chapter.
• Practising Questions: To understand topics of the chapter Units and Measurements, students need to practise questions in a fine way. Different kinds of questions are included in the Units and Measurements notes. Those questions given to practise are: objective questions, short type questions, very short type questions, short type questions, etc. 
• Note Down the Mistakes: While practising Units and Measurements questions from the given notes, students are advised to note down all the mistakes. By noting down the mistakes, students can easily improvise their preparation level for the chapter Units and Measurements. 
• Correction of Mistakes: Correcting the errors are equally important as solving questions from the chapter Units and Measurements. It can aid students to solve all their doubts regarding the chapter Units and Measurements. 
• Take Frequent Breaks: It is very necessary for Class 11 students to take frequent breaks while preparing for Units and Measurements. Frequent breaks can help students to focus and grab more information from the revision notes of Units and Measurements Class 11. 
• Proper Revision: Try to do proper revision of all the concepts and topics you have studied so far. Irregular revision can put you in devastating situations while recalling all the topics you have studied so far in Units and Measurements Class 11 notes. 
• Sleep Well and Eat Healthy: Taking care of yourself is very important during study and when you are trying to revise the topics like Units and Measurements Class 11 using notes then you must pay attention to the quality sleep and healthy diet so that your brain can efficiently manage all the topics you have studied so far.

Why is it Important for Class 11 Students to Refer to the Units and Measurements Class 11 Notes?

It is necessary for Class 11 students to refer to the Units and Measurements Class 11 notes as it can aid to progress in the chapter. It is also important to refer as it provides many important contents from concepts to important formulas and derivation. Necessary information included in the Class 11 Physics notes are: definitions, concepts, topics, questions, etc.

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• Physics Important Questions
• Class 11 Physics
• Chapter 1: Units Measurements

Chapter 1 - Units and Measurements

CBSE Class 11 Physics Important Questions Chapter 1 Units and Measurement are provided here for students. By practising these questions, students get thorough with the important topics of this chapter. Also, they boost their exam preparation for the annual exam. Below are some of the important  numerical on units and measurements class 11.

CBSE Class 11 Physics Important Questions Chapter 1 Units and Measurement

Hope you learned the  numerical on units and measurements class 11.  Also Access  CBSE Class 11 Physics Sample Papers.

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Units and Measurements Class 11 Notes - CBSE Physics Chapter 1

CBSE Class 11 Physics Chapter 1 Notes - Units and Measurements - Free PDF Download

The basis of any problem or concept in physics relies on the actual sense of measurement which is specified in Notes of Physics Class 11 Chapter 1. The idea of measurement is a comparison of the physical property of an object with a standard. Unit and Measurement Class 11 Notes PDF gives a comprehensive knowledge about all sorts of measurements, units, dimensions and errors encountered during measurement. To measure a physical quantity, one must know how many times a standard measure of that physical quantity is present in the object that is being measured. Vedantu’s Unit and Measurement Class 11 Notes provides a detailed description of standard international units used globally for an accurate representation of any quantity.

Units and Measurements & Basic Mathematics Class 11 Notes Physics - Basic Subjective Questions

Section–a (1 mark questions).

1. What is an error? 

Ans. An error is a mistake of some kind causing an error in your results, so the result is not accurate.

2. Find the number of the significant figures in $11\cdot 118\times 10^{-6}V$ .

Ans. The number of significant figures is 5 as 10 −6 does not affect this number.

3. A physical quantity P is given by $P=\dfrac{A^{5}B^{\dfrac{1}{2}}}{C^{-4}D^{\dfrac{3}{2}}}$ . Find the quantity which brings in the maximum percentage error in P.

Ans. Quantity C has maximum power. So it brings maximum error in P.

4. What is the difference between accuracy and precision?

Ans. Accuracy is an indication of how close a measurement is to the accepted value.

An accurate experiment has a low systematic error.

Precision is an indication of the agreement among a number of measurements.  

A precise experiment has a low random error.

5. Find the value of a for which $m=\dfrac{1}{\sqrt{3}}$ is a root of the equation $am^{2}+(\sqrt{3}-\sqrt{2})m-1=0$ .

Ans. Put $m=\dfrac{1}{\sqrt{3}}\Rightarrow \dfrac{a}{3}+(\sqrt{3}-\sqrt{2})\dfrac{1}{\sqrt{3}}-1=0$

$\Rightarrow a_+(3-\sqrt{6})-3=0$

So, a = $\sqrt{6}$

Section – B (2 Marks Questions)

6. A force F is given by $F=at+bt^{2}$ , where t is time. What are the dimensions of a and b?

Ans. From the principle of dimensional homogeneity $\left [ F \right ]=\left [ at \right ]\therefore \left [ a \right ]=\left [ \dfrac{F}{t} \right ]=\left [ \dfrac{MLT^{-2}}{T} \right ]=\left [ MLT^{-3} \right ]$

Similarly, $\left [ F \right ]=\left [ bt^{2} \right ]\therefore \left [ b \right ]=\left [ \dfrac{F}{t^{2}} \right ]=\left [ \dfrac{MLT^{-2}}{T^{2}} \right ]=\left [ MLT^{-4} \right ]$ .

7. Let l, r, c and v represent inductance, resistance, capacitance and voltage, respectively. Find the dimension of 1/rcv in SI units.  

Ans. Dimension of inductance = [M 1 L 2 T −2 A −2 ] = [l]

Dimension of capacitance = [M −1 L −2 T 4 A 2 ] = [c]

Dimension of resistance = [M 1 L 2 T −3 A −2 ] = [r]

Dimension of voltage = [M 1 L 2 T −3 A −1 ] = [v]

Dimension of l/rcv = 

M 1 L 2 T −2 A −2 ] / [M −1 L 2 T 4 A 2 ][M 1 L 2 T −3 A −2 ]

[M 1 L 2 T −3 A −1 ] = [ML 2 T −2 A −2 ]/[ML 2 T −2 A −1 ]

8. P represents radiation pressure, c represents speed of light and S represents radiation energy striking unit area per sec. Find the non zero integers x, y, z such that $P^{x}S^{y}c^{z}$ is dimensionless.

Ans. Try out the given alternatives.

When x = 1, y = −1, z = 1

$P^{x}S^{y}c^{z}=P^{-1}S^{-1}c^{1}=\dfrac{Pc}{S}$

$=\dfrac{\left [ ML^{-1}T^{-2} \right ]\left [ LT^{-1} \right ]}{\left [ \dfrac{ML^{2}T^{-2}}{L^{2}T} \right ]}=\left [ M^{0}L^{0}T^{0} \right ]$

9. The length and breadth of a metal sheet are 3.124 m and 3.002 m respectively. Find the area of this sheet upto correct significant figure is

Ans. $A=3\cdot 124m\times 3\cdot 002m$

$A=\dfrac{9\cdot 738248}{248m^{2}}$

$=9\cdot 378m^{2}$

10. A certain body weighs 22.42 gm and has a measured volume of 4.7 cc. The possible errors in the measurement of mass and volume are 0.01 gm and 0.1 cc. Then find the maximum error in the density.

Ans. $D=\dfrac{M}{V}\therefore %\dfrac{\Delta D}{D}=%\dfrac{\Delta M}{M}+%\dfrac{\Delta V}{V}=\left ( \dfrac{0\cdot 01}{22\cdot 42}-\dfrac{0\cdot 1}{4\cdot 7} \right )\times 100=2%$

PDF Summary - Class 11 Physics Units and Measurements & Basic Mathematics Notes (Chapter 1)

A unit can be defined as an internationally accepted standard for measuring quantities.

Measurement has been included of a numeric quantity along with a specific unit.

The units in the case of base quantities (such as length, mass etc.) are defined as Fundamental unitfs.

Derived units are the units that are the combination of fundamental units. 

Fundamental and Derived units constitute together as a System of Units.

An internationally accepted system of units can be defined as Système Internationale d’ Unites (This is how the International System of Units is represented in French) or SI. In 1971, it was produced and recommended by General Conference on Weights and Measures.

The table shown below is the list of 7 base units mentioned by SI.

There are two units along with it. They are, radian or rad (unit for plane angle) and steradian or sr (unit for solid angle). Both of these are dimensionless.

2.3.1. Parallax Method-Measurement of large distances:

Parallax can be defined as a displacement or difference in the apparent position of a body viewed along two various lines of sight and is calculated by the angle or semi-angle of inclination between those two lines. The basis is called the distance between the two viewpoints.

Calculating the distance of a planet using parallax method:

In the similar way,

$\text{ }\!\!\alpha\!\!\text{ =}\frac{\text{d}}{\text{D}}$

Where $\text{ }\!\!\alpha\!\!\text{ }$ be the planet’s angular size (angle subtended by d at earth) and d will be the diameter of the planet. If two diametrically opposite points of the planet are viewed, then $\text{ }\!\!\alpha\!\!\text{ }$ be the angle between the direction of the telescope.

2.3.2. Measuring very small distances:

For measuring the distances as low as size of a molecule, electron microscopes will be used. These will include electrons beams controlled by electric and magnetic fields.

Electron microscopes will be having a resolution of 0.6 Å or Angstroms.

Electron microscopes are used for resolving atoms and molecules when we use a tunnelling microscopy, it will be possible for estimating size of molecule. Calculating size of molecule of Oleic acid. Oleic acid will be a soapy liquid with large molecular size of the order of ${{10}^{-9}}\text{ m}$. The following steps are used in determining the size of molecule:

Dissolving $\text{1 c}{{\text{m}}^{\text{3}}}$ of oleic acid in alcohol for producing a solution of $\text{20 c}{{\text{m}}^{\text{3}}}$.Consider $\text{1 c}{{\text{m}}^{\text{3}}}$ of above solution and using alcohol dilute it to a concentration of $\text{20 c}{{\text{m}}^{\text{3}}}$. So, the concentration of oleic acid in the solution will become $\left( \frac{1}{20\times 20} \right)\text{ c}{{\text{m}}^{\text{3}}}$ of oleic acid/$\text{c}{{\text{m}}^{\text{3}}}$ of solution.

Then add lycopodium powder on the surface of water in a trough and keep one drop of above solution. In a circular molecular thick film, the oleic acid in the solution will spread over water.

Calculate the diameter of the above circular film by the use of below calculations.

When n –Number of drops of solution in water, t – Thickness of the film, V –Volume of each drop, A – Area of the film.

Total volume of n drops of solution $\text{=nVc}{{\text{m}}^{\text{3}}}$ 

The amount of Oleic acid in this solution$\text{=nV}\left( 120\times 20 \right)\text{c}{{\text{m}}^{\text{3}}}$

Thickness of the film =$\text{t=}\frac{\text{Volume of the film}}{\text{Area of the film}}$

$\text{t=}\frac{\text{nV}}{\text{20 }\!\!\times\!\!\text{ 20A}}\text{cm}$

Special Length units:

Measurement of Mass:

Mass can be defined usually in terms of kg but a unified atomic mass unit (u) will be used for atoms and molecules.

\[\text{1 u=}\frac{\text{1}}{\text{12}}\]of the mass of an atom of isotope of carbon 12 which will be included of the mass of electrons (\[\text{1}\text{.66 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-27}}}\text{ kg}\]).

Mass of planets is measured by the use of gravitational methods and mass of atomic particles are measured by the usage of the mass spectrograph (radius of trajectory will be proportional to the mass of charged particle which is in motion in uniform electric and magnetic field), apart from using balances for normal weights.

Range of Mass:

Measurement of Time:

Time has been calculated by the use of a clock. Atomic standard of time being now used, measured by the Cesium or Atomic clock, as a standard.

A second will be equivalent to 9,192,631,770 vibrations of radiation from the transition between two hyperfine levels of an atom of cesium-133 in a Cesium clock.

The caesium clock will be working on the vibration of the Cesium atom which is identical to vibrations of quartz crystal in a quartz wristwatch and balance wheel in a normal wristwatch.

National standard time and frequency will be maintained by 4 atomic clocks. Indian standard time will be maintained by a Cesium clock at National Physical Laboratory (NPL), New Delhi.

Caesium clocks will be perfectly accurate and the uncertainty will be very low 1 part in 1013 which means no more than \[\text{3  }\!\!\mu\!\!\text{ s}\] will be lost or gained in a year.

Range of Time:

Accuracy and Precision of Instruments:

Any uncertainty which is formed from the calculation by a measuring instrument can be defined as an error. This can be categorised as systematic or random.

The resolution of the measured value to the true value can be defined as the accuracy of a measurement.

The resolution of numerous of measurements of an identical quantity under the same conditions can be called precision.

Till 1 (less precise) and 2 (more precise) decimal places in the same order, has been used when the true value of a specific length is 3.678 cm and two instruments with various resolutions. When first measures the length as 3.5 and the second as 3.38 then the first will be having more accuracy but precision will be less while the second will be having less accuracy and more precision.

Types of Errors- Systematic Errors:

Systematic errors are errors that can either be positive or negative. The following types are:

1. Instrumental errors: which arouse from calibration error or imperfect design in the instrument. Zero error in a weighing scale, Worn off-scale are a few examples of instrument errors.

2. Imperfections in experimental techniques when the technique is not correct (measurement of the temperature of the human body by keeping thermometer under armpit resulting in lower temperature than actual can be considered as an example) and because of the external conditions such as temperature, wind, humidity, and these kinds of errors happen.

3. Personal errors: Errors happening because of human carelessness, lack of proper setting, taking down incorrect reading are defined as personal errors.

The removal of these errors will be:

By taking the proper instrument and properly calibrating it.

By experimenting under correct atmospheric conditions and techniques.

Avoiding human bias as far as possible.

Random Errors:

Errors which is happening at random with respect to sign and size are defined as Random errors.

These kind of errors happens because of the unpredictable fluctuations in experimental conditions such as temperature, voltage supply, mechanical vibrations, personal errors etc.

Least Count Error:

The smallest value which is measurable by the use of a measuring instrument is called its least count. The least count error is the error related to the least count of the instrument.

Least count errors is minimizable by the usage of equipments of higher precision/resolution and improving experimental techniques (take several readings of a measurement and then calculate a mean).

Errors in a series of Measurements. 

Assume that the values got in several measurement are \[{{\text{a}}_{\text{1}}},{{\text{a}}_{2}},{{\text{a}}_{3}}....{{\text{a}}_{\text{n}}}\].

Arithmetic mean,

\[{{\text{a}}_{\text{mean}}}\text{=}\frac{\left( {{\text{a}}_{\text{1}}}\text{+}{{\text{a}}_{\text{2}}}\text{+}{{\text{a}}_{\text{3}}}\text{+}...\text{+}{{\text{a}}_{\text{n}}} \right)}{\text{n}}\]

\[{{\text{a}}_{\text{mean}}}\text{=}\sum\limits_{\text{i=1}}^{\text{n}}{\frac{{{\text{a}}_{\text{i}}}}{\text{n}}}\]

Absolute Error can be defined as the magnitude of the difference between the true value of the quantity and absolute error of the measurement can be defined as the individual measurement value. It is represented as \[\left| \text{ }\!\!\Delta\!\!\text{ a} \right|\] (or Mod of Delta a). The mod value will be positive always even if \[\text{ }\!\!\Delta\!\!\text{ a}\] is negative. The individual errors will be:

$\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{1}}}\text{=}{{\text{a}}_{\text{mean}}}\text{-}{{\text{a}}_{\text{1}}}$ 

$\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{2}}\text{=}{{\text{a}}_{\text{mean}}}\text{-}{{\text{a}}_{2}}$$....................$

$ .....................$ 

$\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{n}}}\text{=}{{\text{a}}_{\text{mean}}}\text{-}{{\text{a}}_{\text{n}}}$ 

Mean absolute error can be explained as the arithmetic mean of all absolute errors. It has been represented as \[\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}\].

\[\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}=\left| \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{1}} \right|+\left| \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{2}} \right|+\left| \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{3}} \right|+....+\left| \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{n}}} \right|\]

\[\text{ }\!\!\Delta\!\!\text{

}{{\text{a}}_{\text{mean}}}\text{=}\sum\limits_{\text{i=1}}^{\text{n}}{\frac{\left| \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{i}}} \right|}{\text{n}}}\]

In the case of every single measurement, the value of ‘a’ is always in the range \[{{\text{a}}_{\text{mean}}}\pm \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}\]

So, \[\text{a=}{{\text{a}}_{\text{mean}}}\pm \text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}\]

Or, \[{{\text{a}}_{\text{mean}}}-\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}\le \text{a}\le {{\text{a}}_{\text{mean}}}+\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}\]

Relative Error can be defined as the mean absolute error divided by the mean value of the quantity measured.

Relative Error \[=\frac{\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}}{{{\text{a}}_{\text{mean}}}}\]

Percentage Error can be defined as the relative error expressed in percentage. It is denoted by \[\text{ }\!\!\delta\!\!\text{ a}\].

\[\text{ }\!\!\delta\!\!\text{ a=}\frac{\text{ }\!\!\Delta\!\!\text{ }{{\text{a}}_{\text{mean}}}}{{{\text{a}}_{\text{mean}}}}\text{ }\!\!\times\!\!\text{ 100}\] (Not in the updated syllabus)

Combinations of Errors

When a quantity is dependent on two or more other quantities, the combination of errors in the two quantities will be helping for determining and predicting the errors in the resultant quantity. There are various procedures for this.

Consider two quantities A and B have values as A ±ΔA and B ± ΔB. Let Z be the result and ΔZ is the error because of the combination of A and B.

Significant Figures

Every measurement gives us an output in a number that is included of reliable digits and uncertain digits.

Reliable digits added with the first uncertain digit can be defined as significant digits or significant figures. This is representing the precision of measurement which is dependent on least count of instrument used for measurement.

The period of oscillation of a pendulum is 1.62 s can be taken as an example. Here 1 and 6 will be the reliable and 2 is uncertain. Hence, the measured value will be having three significant figures.

Rules for the determination of number of significant figures

All non-zero digits will be significant.

Irrespective of decimal place, all zeros between two non-zero digits will be significant irrespective of decimal place.

Zeroes before non-zero digits and after decimal are not considered as significant, for a value less than 1. Zero present before decimal place in case of these number will be insignificant always.

Trailing zeroes in case of a number without any decimal place will be insignificant.

Trailing zeroes in case of a number with decimal place will be significant.

Cautions for removing ambiguities in calculating number of significant figures

Variation of units will not change number of significant digits. As an example, 

$\text{4}\text{.700 m=470}\text{.0 cm}$ 

$\text{              =4700 mm}$ 

Here, first two quantities are having 4 but third quantity is having 2 significant figures.

Make use of scientific notation for reporting measurements. Numbers must be shown in powers of 10 such as \[\text{a }\!\!\times\!\!\text{ 10b}\]where b is defined as order of magnitude. Example,

$\text{4}\text{.700 m = 4}\text{.700  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{2}}}\text{ cm }$

$\text{               = 4}\text{.700  }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ mm }$

$\text{               = 4}\text{.700  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-3}}}\text{ km}$

Here, as the power of 10 is being irrelevant, number of significant figures will be 4.

Multiplying or dividing exact numbers will be giving infinite number of significant digits. Example,\[\text{radius=}\frac{\text{diameter}}{\text{2}}\]. In this case, 2 can be represented as 2, 2.0, 2.00, 2.000 and so on.

Rules for Arithmetic operation with Significant Figures

Rules for Rounding off the uncertain digits

Rounding off will be essential for reducing the number of insignificant figures to hold to the rules of arithmetic operation with significant figures.

Rules for the determination of uncertainty in the results of arithmetic calculations

For calculating the uncertainty, below process must be used.

Do summation of a lowest amount of uncertainty in the original numbers. Example uncertainty for 3.2 will be \[\pm 0.1\] and for 3.22 will be \[\pm 0.01\].

Find out these in percentage also.

The uncertainties get multiplied/divided/added/subtracted after the calculations.

In the uncertainty, round off the decimal place for obtaining the end uncertainty result.

For example, for a rectangle, 

Suppose length,  \[\text{l=16}\text{.2 cm}\] and breadth, \[\text{b=10}\text{.1 cm}\]

After that, take \[\text{l=16}\text{.2}\pm \text{0}\text{.1 cm}\]or \[\text{l=16}\text{.2 cm}\pm \text{0}\text{.6  }\!\!%\!\!\text{ }\] and

breadth \[\text{=10}\text{.1 }\!\!\pm\!\!\text{ 0}\text{.1 cm}\]or \[\text{10}\text{.1 cm }\!\!\pm\!\!\text{ 1  }\!\!%\!\!\text{ }\]

When we multiply,

\[\text{area=length }\!\!\times\!\!\text{ breadth=163}\text{.62 c}{{\text{m}}^{\text{2}}}\text{ }\!\!\pm\!\!\text{ 1}\text{.6  }\!\!%\!\!\text{ }\]

Or \[\text{163}\text{.62}\pm \text{2}\text{.6 c}{{\text{m}}^{\text{2}}}\]

Hence after rounding off, area\[\text{=164}\pm \text{3 c}{{\text{m}}^{\text{2}}}\].

Therefore \[\text{3 c}{{\text{m}}^{\text{2}}}\]will be the uncertainty or the error in estimation.

1. In the case of a set experimental data of ‘n’ significant figures, the result must be accurate to ‘n’ significant figures or less (only in case of subtraction).

For example \[\text{12}\text{.9-7}\text{.06=5}\text{.84 or 5}\text{.8}\](when we round off to least number of decimal places of original number).

2. The relative error of a value of number mentioned to significant figures will be dependent on n and on the number itself.

As an example, say accuracy for two numbers 1.02 and 9.89 be \[\pm \text{0}\text{.01}\]. But relative errors are:

For \[\text{1}\text{.02,}\left( \frac{\text{ }\!\!\pm\!\!\text{ 0}\text{.01}}{\text{1}\text{.02}} \right)\text{ }\!\!\times\!\!\text{ 100  }\!\!%\!\!\text{ =}\pm \text{1  }\!\!%\!\!\text{ }\]

For\[\text{9}\text{.89,}\left( \frac{\text{ }\!\!\pm\!\!\text{ 0}\text{.01}}{\text{9}\text{.89}} \right)\text{ }\!\!\times\!\!\text{ 100  }\!\!%\!\!\text{ =}\pm 0.\text{1  }\!\!%\!\!\text{ }\]

Therefore, the relative error will be dependent upon number itself.

3. The results in the intermediate step of a multi-step computation should be found to have one significant figure more in all the measurement than the number of digits in the least precise measurement.

For example:\[\frac{1}{9.58}=0.1044\]

Now, \[\frac{1}{0.104}=9.56\] and \[\frac{1}{0.1044}=9.58\]

Therefore, taking one extra digit will be providing more precise outputs and reduces rounding off errors.

Dimensions of a Physical Quantity

The powers (exponents) to which base quantities are raised to represent that quantity can be defined as dimensions of a physical quantity. They are figured as the square brackets around the quantity.

Dimensions of the 7 base quantities has been considered as – Length [L], time [T], Mass [M], thermodynamic temperature [K], luminous intensity [cd], electric current [A] and amount of substance [mol].

For example,

$\text{Volume=Length }\!\!\times\!\!\text{ Breadth }\!\!\times\!\!\text{ Height}$

$text{=}\left[ \text{L} \right]\text{ }\!\!\times\!\!\text{ }\left[ \text{L} \right]\text{ }\!\!\times\!\!\text{ }\left[ \text{L} \right]\text{=}{{\left[ \text{L} \right]}^{\text{3}}}$ 

$\text{Force=Mass }\!\!\times\!\!\text{ Acceleration}$

$\text{=}\frac{\left[ \text{M} \right]\left[ \text{L} \right]}{{{\left[ \text{T} \right]}^{\text{2}}}}\text{=}\left[ \text{M} \right]\left[ \text{L} \right]{{\left[ \text{T} \right]}^{\text{-2}}}$

The other dimensions for a quantity will be always 0. As an example, in the case of volume only length has 3 dimensions but the mass, time

etc will be having 0 dimensions. Zero dimension is shown by superscript 0 like \[\left[ {{\text{M}}^{0}} \right]\].

Dimensions will not affect the magnitude of a quantity Dimensional formula and Dimensional Equation

The expression which is representing how and which of the base quantities represent the dimensions of a physical quantity is defined as Dimensional Formula.

An equation we got after equating a physical quantity with its dimensional formula is a Dimensional Equation.

Dimensional Analysis

The physical quantities which is having similar dimensions only can be added and subtracted. This can be named as the principle of homogeneity of dimensions.

Dimensions are multipliable and can be cancelled as normal algebraic methods.

Quantities on both sides should always have identical dimensions, in mathematical equations.

Arguments of special functions such as trigonometric, logarithmic and ratio of similar physical quantities will be dimensionless.

Equations will be uncertain to the extent of dimensionless quantities.

As an example, say Distance = Speed x Time. In Dimension terms, 

\[\left[ \text{L} \right]\text{=}\left[ \text{L}{{\text{T}}^{\text{-1}}} \right]\text{ }\!\!\times\!\!\text{ }\left[ \text{T} \right]\]

As the dimensions can be cancelled like we do in algebra, dimension \[\left[ \text{T} \right]\] will get cancelled and the equation will be \[\left[ \text{L} \right]\text{=}\left[ \text{L} \right]\].

Applications of Dimensional Analysis

When we check the Dimensional Consistency of equations

A dimensionally correct equation should be having identical dimensions on both sides of the equation.

There is no need for a dimensionally correct equation to be a correct equation but a dimensionally incorrect equation will be always incorrect. Dimensional validity can be tested but not calculate the correct relationship between the physical quantities.

Example, \[\text{x=}{{\text{x}}_{\text{0}}}\text{+}{{\text{ }\!\!\nu\!\!\text{ }}_{\text{0}}}\text{t+}\left( \frac{\text{1}}{\text{2}} \right)\text{a}{{\text{t}}^{\text{2}}}\]

Or, Dimensionally, \[\left[ \text{L} \right]\text{=}\left[ \text{L} \right]\text{+}\left[ \text{L}{{\text{T}}^{\text{-1}}} \right]\left[ \text{T} \right]\text{+}\left[ \text{L}{{\text{T}}^{\text{-2}}} \right]\left[ {{\text{T}}^{\text{2}}} \right]\]

Where, \[\text{x}\] be the distance travelled in time t,

\[{{\text{x}}_{0}}\]– starting position,

\[{{\nu }_{0}}\]- initial velocity,

\[\text{a}\]– uniform acceleration.

Dimensions on both sides will be [L] because [T] get cancelled out. Therefore this will be dimensionally correct equation.

Deducing relation among physical quantities

For deducing a relation among physical quantities, we must know the dependence of one quantity over others (or independent variables) and assume it as a product type of dependence.

Dimensionless constants will not be obtainable by the use of this method.

We can take an example, 

\[\text{T=k}{{\text{l}}^{\text{x}}}{{\text{g}}^{\text{y}}}{{\text{m}}^{\text{z}}}\]

$\left[ {{\text{L}}^{\text{0}}}{{\text{M}}^{\text{0}}}{{\text{T}}^{\text{1}}} \right]\text{=}{{\left[ {{\text{L}}^{\text{1}}} \right]}^{\text{x}}}{{\left[ {{\text{L}}^{\text{1}}}{{\text{T}}^{\text{-2}}} \right]}^{\text{y}}}{{\left[ {{\text{M}}^{\text{1}}} \right]}^{\text{z}}}$ 

$\text{                =}\left[ {{\text{L}}^{\text{x+y}}}{{\text{T}}^{\text{-2y}}}{{\text{M}}^{\text{z}}} \right]$

This means that, \[\text{x+y=0,-2y=1 and z=0}\]. So \[\text{x=}\frac{1}{2}\text{,y=-}\frac{1}{2}\text{ and z=0}\].

Hence the original equation will be reduced to \[\text{T=k}\sqrt{\frac{\text{l}}{\text{g}}}\].

Units and Measurements Notes Physics Chapter 1 - Free PDF Download

Topics covered under physics chapter 1 class 11 notes .

As mentioned in Vedantu’s Class 11 Physics Ch 1 Notes, an arbitrarily chosen and internationally accepted standard i.e. unit is used as a reference to express the measurement of a particular physical quantity.

Physical quantity = Numerical value x Unit. For example, Length of a ladder  = 5.5 m

Here 5.5 is a numerical value and m (metre) is the unit of length.

The units that can be expressed independently are called fundamental or base units. For example- mass is expressed in kilogram, length in metre, and time in second respectively are fundamental units.

The units for which a combination of the fundamental units is called derived units. Units of area and density which are m 2 , kg/m 3 respectively are examples of derived units.

The International System of Units

The base units for length, mass and time in unit systems are: 

CGS System: centimetre, gram and second

FPS System: foot, pound and second

MKS System: metre, kilogram and second

SI System: This system is internationally accepted for measurement (Système Internationale d’ Unites). Length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity are expressed in metre(m), kilogram(kg), second(s), ampere(A), Kelvin(K), mole(mol) and candela(cd) respectively.

Apart from the above units, there are two supplementary base units:

(i) radian (rad) for angle

(Image to be added soon)

(ii) steradian (sr) for a solid angle.

Parallax Method of Measurement of Large Distances

This method is used to measure large distances like that of planets and stars from earth.

Some Units of Large Distance

1 light-year = 1 ly = 9.46 × 10 15 m (the distance that light travels with a velocity of 3 × 10 8 m s –1 in 1 year) 

1 parsec = 3.08 × 10 16 m (distance at which average radius of earth’s orbit subtends an angle of 1 arc second)

Estimation of Very Short Distances: The Size of a Molecule

In Chapter 1 Physics Class 11 Notes, one can learn how to measure the size of a molecule.

Some units of short length:

1 fermi = 1 f = 10 –15 m 

1 angstrom = 1 Å = 10 –10 m 

Errors In Measurement

The uncertainty while measuring a physical quantity is called an error.

(i) Systematic Errors:

Instrumental errors

A flaw in experimental technique or procedure

Personal errors

(ii) Random Errors: Irregular errors due to unpredictable fluctuations in experimental conditions

Least Count Error: associated with the resolution of the instrument.

Absolute Error: a mean = (a 1 +a 2 +a 3 +...+a n ) / n

∆a  = a mean – measured value

Mean Absolute Error: ∆a mean = (|∆a 1 |+|∆a 2 |+|∆a 3 |+...+ |∆a n |)/n 

\[\sum_{i=1}^{n}\] |∆a i |/ni

Relative Error :   ∆a mean / a mean

Percentage Error: δa = (∆a mean / a mean ) × 100 %

Combination of Errors

An error of a sum or a difference

If Z=A+ B then  ∆Z =∆A + ∆B

if Z=A- B then ∆Z =∆A + ∆B

An error of a product or a quotient If Z= A×B or Z=A/B 

error in ‘Z’ = ∆Z/Z= (∆A/A) + (∆B/B)

Error in a quantity that has been raised to power:

then ∆Z/Z = K (∆A/A)

Arithmetic Operations with Significant Figures

For addition, subtraction, multiplication and division the result has the same number of significant figures as that of the number with a minimum number of decimal places.

Rounding off the Uncertain Digits

If dropping digit < 5, then the previous digit is left unchanged. 

If dropping digit > 5, then the previous digit is raised by 1

If dropping digit = 5 followed by non-zero digits, then the previous digit is raised by 1

Dimensions of Physical Quantities

Dimension of :

Electric current=[A]

Temperature=[K]

Luminous intensity=[cd]

Amount of substance = [mol]. 

Dimensional equations represent physical entities in terms of their base quantities. 

For example, speed = (distance/time)

Dimensional equation of speed [v]= [M 0 L T –1 ]

Importance of Vedantu’s Units and Measurements Class 11 Notes PDF Download

Vedantu's Units and Measurements Class 11 Notes offer students a valuable resource to grasp the fundamental concepts of physics. By using Vedantu's notes, students can enhance their problem-solving skills and develop a strong foundation in physics, empowering them to excel in their academic pursuits. Here are some points that explains the Importance of Units and Measurements Class 11 Notes PDF Download:

Vedantu's Class 11 Physics Chapter 1 Notes provide a comprehensive understanding of fundamental concepts in physics.

These notes cover topics such as SI units, dimensional analysis, and measurements with precision.

They offer clear explanations and examples to aid in conceptual clarity and application of theories.

Vedantu's notes facilitate effective revision and preparation for exams by condensing complex concepts into concise summaries.

With Class 11 Units and Measurements Notes, students can enhance their problem-solving skills and achieve mastery in the subject, laying a strong foundation for future studies in physics.

What does Vedantu’s Free Study Material - Class 11 Units and Measurements Notes Provide?

A hands-on experience in time management and paper pattern

A vivid idea about the topic based marks weightage along with hierarchical study plan

Explanatory videos and free mock test to enhance problem-solving skills

Specially designed model papers and practice test by our eminent educators

Class 11 Units and Measurements Notes suffice for all the basic concepts and doubt clearance regarding measurements, units, dimensions and errors. The principles of physical quantities in Ch 1 Physics Class 11 Notes go a long way in building a strong foundation for the upcoming chapters.

Download CBSE Class 11 Physics Revision Notes 2024-25 PDF

Also, check CBSE Class 11 Physics revision notes for other chapters:

Units and Measurement & Basic Mathematics Chapter-Related Important Study Materials

It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Class 11 Physics Chapter 1 Units and Measurements Notes offered by Vedantu is an excellent resource for students who want to excel in their physics studies. The Class 11 Units and Measurements Notes provide a comprehensive and detailed explanation of the concepts of units and measurement, including the SI units, dimensions, and errors, making it easier for students to understand and improve their physics skills. The notes also include practice exercises and questions that help students test their understanding of the chapter and prepare for their exams. Vedantu also provides interactive live classes and doubt-solving sessions to help students clarify their doubts and improve their understanding of the chapter. Overall, the Unit and Measurement Class 11 Notes PDF offered by Vedantu are an essential resource for students who want to improve their physics skills and score well in their exams.

FAQs on Units and Measurements Class 11 Notes - CBSE Physics Chapter 1

1. What Concepts Are Covered in the Second Chapter of Class 11 Units and Measurements Notes?

Class 11 Chapter - 2 Physics will strengthen the basic concepts of the following topics.

What is a unit?

Fundamental and Derived units

System of units

Propagation of Errors

You will also understand the need for S.I. units and how they are decided in this chapter.

2. Is NCERT Enough for Class 11 Physics?

NCERT builds the base of all the concepts. However, for staying updated with the latest trends of the questions being asked in the previous years, you need to look for online resources. We have provided all the materials on our website.

3. Write the Name of Seven Fundamental Dimensions.

Seven fundamental dimensions are as follows.

Temperature

Amount of light

Amount of Matter

Electric Current

4. What is Parallax Method Class 11? Define the Parallax Angle.

The term ‘parallax’ is considered the change in position of an object (at large distances) seen from two different positions. It is measured by the angle or semi-angle of inclination between those two positions.

In astronomy, it is the only method to determine the distance of stars outside the solar system. The parallax angle is used by astronomers to determine the distance from Earth to the stars. It is the angle between the Earth at one year time and the Earth six months later, as measured from the stars. 

5. Should I study from Vedantu Class 11 Physics notes for the second Chapter?

Students can study from the Class 11 Physics Chapter 1 Revision Notes. Class 11 Physics Revision Notes can help students to understand the concepts of Chapter 1. Class 11 Revision Notes are prepared by experienced Physics teachers. They collect information from different sources that are reliable and informative. All textbook questions are covered here with answers. Students can easily study for Class 11 Physics from the notes available free of cost on Vedantu website and mobile app.

6. What are the important topics given in Units and Measurements Class 11 Notes PDF Download?

Class 11 Physics Chapter 1 is based on units and measurement. Students will study the different units and measurement systems. They will study different units used for measuring length, time, and distance. The important topics given in Class 11 Physics Chapter 1 include absolute errors, SI units, significant figures, and dimensional analysis. It is an important chapter for the exams. Students can understand the topics given in Class 11 Physics Chapter 1 from the notes given on Vedantu app and website.

7. What do you understand by dimensional analysis according to Chapter 1 of Class 11 Physics?

Dimensional analysis means measuring the size and shape of objects. It means giving the dimensions of an object using mathematical calculations. It helps to know the quantity of an object or the size of an object. The quantities having the same dimensions can be added or subtracted. We can also compare the quantities with the same dimensions. If two physical quantities have the same dimensions, they are equal to each other.

8. How can Vedantu help me in preparing for Chapter 1 of Class 11 Physics?

Vedantu is a learning website that offers free NCERT Solutions, important questions, and notes for all classes. Students of Class 11 can prepare Chapter 1 of Physics from the Vedantu website. They can study from the revision notes, important questions, and NCERT solutions given for Class 11 Physics. Students can easily understand the concepts of Class 11 Physics from the simple notes given on Vedantu.

9. How can I score high marks in Class 11 Physics Chapter 1 Units and Measurements?

Students can score high marks in Chapter 1 of Class 11 Physics by downloading the Revision Notes on their computers. They can download the Class 11 Physics Notes for Chapter 1 free of cost from Vedantu. Students can understand the concepts of Class 11 Physics Chapter 1 from the notes to score high marks. Simply visit the official Vedantu website and choose the subject and the chapter of your choice. You will notice a download PDF option. Clicking on it will save the solutions on your device and you can refer

10. What is the short note of unit and measurement?

In science, units and measurement act as the rulers and scales for our observations. Units are standard references for properties like length (meters) or time (seconds), while measurement assigns a specific value using that unit (e.g., 2 meters). Standardized units (like the International System of Units) ensure everyone speaks the same scientific language, fostering accurate and comparable results across experiments. From simple comparisons to complex calculations, units and measurement are the foundation for quantifying our world.

11. What are the units of measurement class 11 Physics?

In Class 11, you'll delve deeper into units of measurement, particularly focusing on the International System of Units (SI). Here's a quick breakdown:

The SI system is the globally accepted standard for units of measurement. It consists of seven base units:

Meter (m): Length

Kilogram (kg): Mass

Second (s): Time

Ampere (A): Electric current

Kelvin (K): Thermodynamic temperature

Mole (mol): Amount of substance

Candela (cd): Luminous intensity

Derived Units: These are units formed by combining base units. Examples include:

Meter per second (m/s): Velocity

Square meter (m²): Area

Kilogram per cubic meter (kg/m³): Density

Joule (J): Energy (derived from kg∙m²/s²)

Unit Prefixes: Prefixes like kilo (10^3), centi (10^-2), and nano (10^-9) are used to represent multiples or fractions of base units for convenience.

12. Which topics are important in Unit and Measurement Class 11 Notes?

In Units & Measurements (Class 11), master the SI system (meter, kilogram, second, etc.) and how it forms derived units (velocity, area). Learn prefixes (kilo-, centi-) for convenience.  Dimensional analysis and unit conversions might also be covered, depending on your curriculum.

13. What is the importance of measurement in Unit and Measurement Class 11 Notes?

Without measurement, physics would be all talk and no action. Physics (Class 11) is all about quantifying the world. Measurement is key! It lets you:

Describe things with numbers (length, time, force)

Build physics laws (think Newton's laws!)

Compare experiments and results

Solve problems and predict future behavior

14.  Are these Class 11 Physics Chapter 1 Units and Measurements Notes based on the latest CBSE syllabus?

Yes, Class 11 Physics Chapter 1 Units and Measurements Notes are aligned with the latest CBSE syllabus, ensuring that students cover all the topics prescribed by the board.

15. Are Unit and Dimensions Class 11 Notes suitable for self-study?

Yes, Unit and Dimensions Class 11 Notes are well-suited for self-study purposes. These notes are designed to provide comprehensive coverage of the topic, including explanations of fundamental concepts, examples, and practice problems. They are structured in a way that allows students to study at their own pace and reinforce their understanding through self-assessment. Additionally, the notes may include tips and strategies for effective learning and problem-solving. Whether you are studying independently or preparing for exams, these notes serve as a valuable resource for enhancing your understanding of units and measurements in physics.

CBSE Study Materials for Class 11

• Units And Dimensions
• Measurement And Units
• Active page

Units And Dimensions of Class 11

Physical quantity.

Any meaningful term which can be measured is a physical quantity. For example length, velocity, time etc. are physical quantity. But handsomeness, beauty are not physical quantity. Why measurement is needed?: Physics is an experimental science and experiments involve measurement of different physical quantities in which laws of physics are expressed. Without measuring results of experiments, it would not be possible for scientists to communicate their results to one another or to compare the results of experiments from different laboratories.

Units of Measurement

To measure a physical quantity we need some standard unit of that quantity. For example, if a measurement of length is quoted as 5 meters, it means that the measured length is 5 times as long as the value accepted for a standard length defined to be “one meter”.

Any set of standards of units must fulfill the following two conditions

• It must be accessible.
• It must be invariable with the passage of time

Two more auxiliary conditions

• It is necessary to have wide unlimited agreement about those standards.
• It is inter convertible to different units of same quantity.

A measurement consists of two parts, one is numeric and the other is standard chosen. For example, 5 meter of length implies 5 times the “standard meter”. It is not necessary to establish a measurement standard for every physical quantity. Some quantities can be regarded as fundamental and the standard for other quantities can be derived from the fundamental ones. For example, in mechanics length, mass and time are regarded as fundamental quantities and the standard for speed (= length / time) can be derived from fundamental quantities length and time.

Two Supplementary Units

• Two other system of units compete with the international system- One is Gaussian System in terms of which much of the literature of physics is expressed. In India this system is not in use.
• The other is the British system- This system is still in daily use in United states and United Kingdom. But SI units are standard units worldwide.
• C.G.S. Unit- In this system of unit, centimeter, gram and seconds are units of length, mass and time respectively.
• Conversion of One System of Units to another System- The basic formula is n 1 u 1 = n 2 u 2 where n 1 and n 2 are numbers.

Q1. How many dyne−centimeter are equal to 1 N −m ?

Ans. 1N -m = (1 kg)(1m 2 ) (1s -2 )

1 dyne - centimeter = (1kg)(1cm 2 )(1 s -2 )

1N -M / 1 dyne -cm = (1000g/1g)(100cm 2 / 1cm)

= 1000 x 10000

1N -m = 107 dyne -cm

Measurement of Length

Depending upon the range of length, there are three main methods for measuring length:

Direct method

Indirect or mathematical method.

• Chemical Mehtod

The simplest method measuring the length of a straight line is by means of a meter scale. But there exist some limitations in the accuracy of the result:

• The dividing lines have a finite thickness.
• Naked eye cannot correctly estimate less than 0.5 mm

For greater accuracy devices like

• Vernier calliper
• micrometer scale (screw gauge)

Vernier Calliper

It consists of a main scale graduated in cm/mm over which an auxiliary scale (or Vernier scale) can slide along the length. The division of the Vernier scale being slightly shorter than the divisions of the main scale.

Vernier Calliper Least count

The least count or the Vernier constant (V.C.) is the minimum value of correct estimation of length without eye estimation. The difference between the values of one main scale division and one vernier scale division is known as vernier constant if N division of vernier scale coincides with (N−1) divisions of main scale, then vernier constant,

n.V.S.D. = (n −1 ) M.S.D.

1.V.S.D. = (n -1 / n) M.S.D., then

LC = 1.M.S.D. − 1.V.S.D. = 1.M.S.D. (n -1 / n) M.S.D.

=1/n M.S.D.

Vernier Calliper Least Count Formula = Smallest Division on main scale / No. of division on vernier scale

Vernier Calliper Reading

Let one main scale division be 1 mm and 10 vernier scale divisions coincide with 9 main scale divisions

1.V.S.D. = 9/10 M.S.D.=0.9 mm

Vernier constant = 1.M.S.D − 1.V.S.D. = 1 mm − 0.9 mm

= 0.1 mm = 0.01 cm

The reading with vernier scale is read as given below :

• Firstly take the main scale reading (N) before on the left of the zero of the vernier scale.
• Find the number (n) of vernier division which just coincides with any of the main scale division. Multiply this number (n) with vernier constant (V.C.)
• Total reading = (N + n × V.C.)

Caution: The main scale reading with which the Vernier scale division coincides has no connection with reading

Suppose If we have to measure a length AB, the end A is coincided with the zero of the vernier scale as shown in fig. Its enlarged view is given in fig.

1.0 cm < Length AB < 1.1. cm

Let 5th division of vernier scale coincide with 1.6 cm of main scale. From diagram it is clear that the distance between 4th division of vernier scale and 1.5 cm of main scale is equal to one V.C. and distance between zero mark of vernier scale and 1.0 cm mark on the main scale is equal to 5 times the vernier constant.

AB = 1.0 + 5 × L.C. = 1.0 + 5 × 0.01 = 1.05 cm.

Q2. In travelling microscope the vernier scale used has the following data. 1 M.S.D. = 0.5 mm, 50 V.S.D. = 49 M.S.D. and the actual reading for distance travelled by travelling microscope is 2.4 cm with 8th division coinciding with a main scale graduation . Estimate the distance travelled.

Ans. In this case vernier constant = 1.M.S.D. − 1.V.S.D.

= 1.M.S.D. − 49/50 M.S.D.

= 1/50 M.S.D

=1/50 x 0.5 mm

= 5/10 x 1/50 = 0.01mm = 0.001cm

Distance travelled = 2.4 + 8 × 0.001 cm

Q3. The Vernier scale used in Fortin’s barometer has 20 divisions coinciding with the 19 main scale divisions. If the height of the mercury level measured is 5 mm and 15th division of vernier scale is coinciding with the main scale division. Then calculate the exact height of the mercury level (given that 1.M.S.D. = 1mm)

Ans. 20 V.S.D. = 19 M.S.D. (Given)

1.V.S.D. = 19/20 M.S.D

V.C. = 1. M.S.D. − 1.V.S.D = (1 - 19/20M.S.D.)

= 1/20 M.S.D.

= 1/20 x 1 mm

Height of mercury level = 5 + 0.05 × 15

If the zero marking of main scale and Vernier scale do not coincide, necessary correction has to be made for this error which is known as zero error of the instrument. If the zero of the vernier scale is to the right of the zero of the main scale the zero error is said to be positive and the correction will be negative and vice versa.

Q1. Consider the following data:

10 main scale divisions = 1cm, 10 vernier division = 9 main scale divisions, zero of Vernier scale is to the right of the zero marking of the main scale with 6th Vernier division coinciding with a main scale division and the actual reading for length measurement is 4.3 cm with 2nd Vernier divisions coinciding with a main scale graduation. Estimate the length.

Ans. In this case, vernier constant = 1mm/100

Zero error = 6 × 0.1 = + 0.6 mm

Correction = −0.6 mm

Actual length = (4.3 + 2 × 0.01) + correction

= 4.32 − 0.06 = 4.26 cm.

Screw Gauge (or Micrometer Screw)

In general Vernier Callipers can measure accurately upto 0.02 mm and for greater accuracy micrometer screw devices, e.g. screw gauge, spherometer are used. These consist of accurately cut screw which can be moved in a closely fitting fixed nut by turning it axially. The instrument is provided with two scales:

• The main scale or pitch scale M graduated along the axis of the screw.
• The cap−scale or head scale H round the edge of the screw head.

Constants of the screw gauge

• Pitch: The translational motion of the screw is directly proportional to the total rotation of the head. The pitch of the instrument is the distance between two consecutive threads of the screw which is equal to the distance moved by the screw due to one complete rotation of the cap. Thus if 10 rotations of cap ≡5 mm, then pitch = 0.5 mm

In general, pitch = Distance travelled by screw on main scale / No. of rotation taken by the cap to travel that much distance

• Least count: In this case also, the minimum (or least) measurement (or count) of length is equal to one division on the main scale which is equal to pitch divided by the total cap divisions. Thus in the aforesaid Illustration, if the total cap division is 100, then least count = 0.5 mm/100

In general, In case of circular scale,

Least count = Pitch / Number of division on circular scale

If pitch is 1 mm and there are 100 divisions on circular scale, then

Least count = 1mm/100 = 0.01 mm = 0.001 cm

= 0.00001 m = 10−5 m = 10 μm

Since least count is of the order of 10 μm, So the screw is called a micrometer screw. Screw gauge and the spherometer which work on the principle of micrometer screw, consist essentially of the following two scales.

• Linear or Pitch scale: It is a scale running parallel to the axis of the screw.
• Circular of Head scale: It is marked on the circumference of the circular disc or the cap attached to the screw.
• Zero Error: In a perfect instrument the zero of the head scale coincides with the line of graduation along the screw axis with no zero−error, otherwise the instrument is said to have zero−error which is equal to the cap reading with the gap closed. This error is positive when zero line or reference line of the cap lies below the line of graduation and vice−versa. The corresponding corrections will be just opposite.

Q1. A screw gauge has 100 divisions on its circular scale. Circular scale travels one division on linear scale in one rotation and 10 divisions on linear scale of screw gauge is equal to 5 mm. What is the least count of a screw gauge.

Ans. Pitch = 1 division on linear scale / 1 rotation = 1 div

10 division = 5 mm

1 division = 0.5 mm

pitch = 0.5 mm

least count = Pitch / No. of divisions on circular scale

= 0.5 mm /100

Q2. The screw gauge mentioned in above illustration is used to measure thickness of a coin. The reading of the linear scale is 4th div and 25th division of circular scale is coinciding with it. What is the value of thickness of the coin.

Ans. Reading = Linear scale Reading + Least count × circular scale reading

= 4th division on linear scale + 0.005 mm × 25

= 4 × 0.5 mm + 0.125 mm

= 2 mm + 0.125 mm

Q3. A spherometer has 250 equal divisions marked along the periphery of its disc and one full rotation of the disc advances it on the main scale by 0.0625 cm. The least count of the spherometer is

(A) 2.5x10 −2 cm

(B) 25 x10 −3 cm

(C) 2.5x10 −4 cm

(D) none of the above

Ans. Correct option is (C)

Least count = 0.0625/250 cm = 2.5 x 10 −4 cm

This method involves measurement of long distances. Main methods of this category are −

Reflection method: Suppose we want to measure the distance of a multi story building from a destination point P. If a shot be fired from P, the sound of shot travels a distance x towards the building, gets reflected from the building. The reflected sound travels the distance x to the point of P, when an echo of the shot is heard.

Let t = time interval between the firing of the shot and echo sound.

v = velocity of sound in air.

Distance = velocity x time

x + x = (v) (t)

⇒ x = (v) (t/2)

As v is known, x can be calculated by measuring the time t.

Q8. A rock is at the bottom of a very deep river. An ultrasonic signal is sent towards rock and received back after reflection from rock in 4 seconds. If the velocity of ultrasonic wave in water is 1.45 km/s, find the depth of river.

Ans. Here x = ?

v = 1.45 km/s = 1450 m/sec.

so, x = v x t / 2 = 1450 x 4 / 2 = 2900 m.

• Parallel method: This method is used for measuring the distance of nearby stars. If we have to measure the distance D of a faraway star S by this method. We observe this star from two different positions A and B on the earth, separated by a distance AB = b at the same time as shown in the figure. Let ∠ASB =θ, the angle θ is called parallatic angle. As the star is very far away, b/D << 1 and θ is very small.

Here we can take AB as an arc of length b of a circle with center at S and the distance D as the radius AS=BS so that AB = b = Dθ where θ is in radians.

D = b/θ

Knowing b and measuring θ, we can calculate D.

• Copernicus method: This method is used to measure the relative distances of the planets from the Sun.
• For Interior Planets: The angle formed at earth between the earth−planet direction and the earth−sun direction is called the planet’s elongation. This is the angular distance of the planet from the sun as observed from earth. When the elongation attains its maximum value ε as in the figure, the planet appears farthest from Sun.

• For Exterior Planets: This method is a consequence of Kepler’s 3rd law of planetary motion. For two planets P 1 and P 2 we have,

a 3 2 / a 3 1 = T 2 2 / T 2 1

where a 1 and a 2 are semi−major axes, of respective orbits. The period can be ascertained by direct observation. Therefore if a 1 is measured, a 2 can be calculated.

Chemical Method

This method is used to measure the distance of the order of 10−10 m. Let us calculate the size of an atom.

Let m = mass of substance,

V = volume occupied by substance &

ρ = density of the substance

v = m /ρ (1)

Let M be the atomic weight of the substance and N be the Avogadro number.

No. of atoms in mass m of the substance = Nm / M

If r = radius of each atom then V = volume of each atom = 4/3 πr 3

Volume of all the atoms in substance = (4/3 πr 3 x Nm)/M.

According to Avagordo’s hypothesis,

Volume of all the atoms = (2/3) x volume of substance

4/3 πr 3 x Nm/M = (2/3) m/ρ

r = (M/2πNρ) 1/3

Frequently Asked Question (FAQs)

Q1. What is Vernier Calliper?

Ans. A measuring device is used for the measurement of linear dimensions. It is also used to measure the diameters of round objects with the help of the measuring jaws.

Q2. What is the least count of Vernier callipers?

Ans. The zero mark of the Vernier scale has gone seven-tenths of a millimeter past the 11.0 mm mark, giving a total reading of 11.7 mm. The least count of the Vernier caliper is 0.1 mm. 0.1 mm is the smallest scale reading that can be made without estimation.

Q3. What is the atomic mass unit (AMU) of 1 Å?

Ans. 1 A.U. = 1.496 × 10 11 m and 1 Å = 10 10 m.

Q4. Are S.I. units Coherent? Why?

Ans. Yes, because all the derived units in this system can be obtained by multiplying or dividing a certain set of basic units.

• Introduction
• Scope And Excitement Of Physics
• Technology And Society
• Fundamental Forces In Nature
• Conservation Laws
• Measurement Of Mass
• Errors In Measurement
• Physical Quantities

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NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

June 20, 2022 by Sastry CBSE

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements are part of NCERT Exemplar Class 11 Physics . Here we have given NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements.

Single Correct Answer Type Q1. The number of significant figures in 0.06900 is (a) 5 (b) 4 (c) 2 (d) 3 Sol: (b) Key concept: 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. The reverse is also true. The following rules are observed in counting the number of significant figures in a given measured quantity. 1. All non-zero digits are significant. 2. A zero becomes significant figure if it appears between two non¬zero digits. 3. Leading zeros or the zeros placed to the left of the number are never significant. 4. Trailing zeros or the zeros placed to the right of the number are significant. 5. In exponential notation, the numerical portion gives the number of significant figures. Leading zeros or the zeros placed to the left of the number are never

Q3. The mass and volume of a body are 4.237 g and 2.5 cm 3 , respectively. The density of the material of the body in correct significant figures is (a) 1. 6048 g cm -3 (b) 1.69 g cm -3 (c) 1.7 g cm 3                                             (d) 1.695 g cm -3

After rounding off the number, we get density =1.7

Q4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give (a) 2.75 and 2.74 (b) 2.74 and 2.73 (c) 2.75 and 2.73 (d) 2.74 and 2.74 Sol: (d) Key concept: While rounding off measurements, we use the following rules by convention: 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged. 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one. 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one. 4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even. 5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd. Units and Measurements Let us round off 2.745 to 3 significant figures. Here the digit to be dropped is 5, then preceding digit is left unchanged, if it is even. Hence on rounding off 2.745, it would be 2.74. Now consider 2.737, here also the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. Hence on rounding off 2.735 to 3 significant figures, it would be 2.74.

Q9. Which of the following measurements is most precise? (a) 5.00 mm (b) 5.00 cm (c) 5.00 m (d) 5.00 km Sol: (a) Key concept: Precision is the degree to which several measurements provide answers very close to each other. It is an indicator of the scatter in the data. The lesser the scatter, higher the precision. Let us first check the units. In all the options magnitude is same but units of measurement are different. As here 5.00 mm has the smallest unit. All given measurements are correct upto two decimal places. However, the absolute error in (a) is 0.01 mm which is least of all the four. So it is most precise.

Q11. Young’s modulus of steel is 1.9 x 10 11 N/m 2 . When expressed in CGS units of dyne/cm 2 , it will be equal to (1 N = 10 5 dyne, 1 m 2 = 10 4 cm 2 )                     . (a) 1.9 xlO 10                                              (b) 1.9×10 12 (c) 1.9 xlO 12                                             (d) 1.9 xlO 13

More Than One Correct Answer Type Q13. On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?

Q14. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity? (a) (P-Q)/R           (b) PQ-R (c) PQ/R                                   (d) (PR-Q 2 )/R (e)(R + Q)/P Sol: (a, e) Key concept: Principle of Homogeneity of dimensions: It states that in a correct equation, the dimensions of each term added or subtracted must be same. Every correct equation must have same dimensions on both sides of the equation. According to the problem P, Q and R are having different dimensions, since, sum and difference of physical dimensions, are meaningless, i.e., (P – Q) and (R + Q) are not meaningful. So in option (b) and (c), PQ may have the same dimensions as those of R and in option (d) PR and Q 2 may have same dimensions as those of R. Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

Q15. Photon is a quantum of radiation with energy E = hv , where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of (a) Linear impulse (b) Angular impulse (c) Linear momentum                           (d) Angular momentum

Q16. If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities? (a) Mass of electron (m e )              (b) Universal gravitational constant (G) (c) Charge of electron (e)              (d) Mass of proton (m p )

Q17. Which of the following ratios express pressure? (a) Force/Area                                        (b) Energy/Volume (c) Energy/Area                                      (d) Force/Volume

Sol:  (a, b) Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.

Very Short Answer Type Questions

Q19. Why do we have different units for the same physical quantity? Sol:  Magnitude of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required. For example: 1.Mass ranges from 10 -30 kg (for an electron) to 10 53 kg (for the known universe). We need different units to measure them like miligram, gram, kilogram etc. 2.The length of a pen can be easily measured in cm, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.

Q21. Name the device used for measuring the mass of atoms and molecules. Sol: A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.

Q22. Express unified atomic mass unit in kg. Sol: The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

Q24. Why length, mass and time are chosen as base quantities in mechanics? Sol: Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because (i) Length, mass and time cannot be derived from one another, that is these quantities are independent. (ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

Short Answer Type Questions 25. (a) The earth-moon distance is about 60 earth radius. What will be the . diameter of the earth (approximately in degrees) as seen from the moon? (b) Moon is seen to be of (1/2)° diameter from the earth. What must be the relative size compared to the earth? (c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Q26. Which of the following time measuring devices is most precise? (a) A wallclock                                         (b) A stop watch (c) A digital watch                                   (d) An atomic clock Given reason for your answer. Sol: Option (d) is correct because a clock can measure time correctly up to one second. A stop watch can measure time correctly up to a fraction of a second. A digital watch can measure time up to a fraction of second whereas an atomic clock is the most accurate timekeeper and is based on characteristic frequencies of radiation emitted by certain atoms having precision of about 1 second in 300,000 years. So an atomic clock can measure time most precisely as precision of this clock is about 1 s in 10 13 s.

Q27. The distance of a galaxy is of the order of 10 25 Calculate the order of magnitude of time taken by light to reach us from the galaxy. Sol: According to the problem, distance of the galaxy = 10 25 m. Speed of light = 3 x 10 8 m/s Hence, time taken by light to reach us from galaxy is

Q29. During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon. Sol: Key point: In geometry, a solid angle (symbol: Ω or w) is the two­dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

Q32. Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π/6 at the centre.

Q35. Time for 20 oscillations of a pendulum is measured as t 1 =39.6 s; t 2 = 39.9 s and t 3 = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement? Sol: According to the problem, time for 20 oscillations of a pendulum, t 1 = 39.6 s, t 2 = 39.9 s and t 3 = 39.5 s It is quite obvious from these observations that the least count of the watch is 0.1 s. As measurements have only one decimal place. Precision in the measurement = Least count of the measuring instrument= 0.1 s Precision in 20 oscillations = 0.1

Long Answer Type Questions

Q40. If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities. Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal, We know that, dimensions of

Q42. In an experiment to estimate ‘the size of a molecule of oleic acid, 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions.

• Why do we dissolve oleic acid in alcohol?
• What is the role of lycopodium powder?
• What would be the volume of oleic acid in each mL of solution prepared?
• How will you calculate the volume of n drops of this solution of oleic
• What will be the volume of oleic acid in one drop of this solution?

Sol: (a) Since Oleic acid does not dissolve in water, hence it is dissolved in alcohol.

(b)Lycopodium powder spreads on the entire surface of water when it is sprinkled evenly. When a drop of prepared solution of oleic acid and alcohol is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can thus be able to measure the area over which oleic acid spreads.

(c)Since 20 mL (1 mL oleic acid + 19 mL alcohol) contains 1 mL of oleic acid, oleic acid in each mL of the solution =1/20 mL. Further, as this 1 mL is diluted to 20 mL by adding alcohol. In each mL of solution prepared, volume of oleic acid = 1/20 mL x 1/20  = 1/400 mL

(d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette (used to make solution in the form of countable drops) and measuring cylinder and measuring the number of drops.

(e) As 1 mL of solution contains n number of drops, then the volume of oleic acid in one drop will be = 1/(400)n mL

Q43. (a) How many astronomical units (AU) make 1 parsec? (b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute. (c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate at what size it will appear when seen through the same telescope.

NCERT Exemplar Class 11 Physics Solutions

• Chapter 1 Units and Measurements
• Chapter 2 Motion in a Straight Line
• Chapter 3 Motion in a Plane
• Chapter 4 Laws of Motion
• Chapter 5 Work, Energy and Power
• Chapter 6 System of Particles and Rotational Motion
• Chapter 7 Gravitation
• Chapter 8 Mechanical Properties of Solids
• Chapter 9 Mechanical Properties of Fluids
• Chapter 10 Thermal Properties of Matter
• Chapter 11 Thermodynamics
• Chapter 12 Kinetic Theory
• Chapter 13 Oscillations
• Chapter 14 Waves

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NCERT Class 11 Physics Chapter 1 Units of Measurement

NCERT Class 11 Physics Chapter 1 Units of Measurement   Solutions, NCERT Class 11 Physics Chapter 1 Units of Measurement Notes to each chapter is provided in the list so that you can easily browse throughout different chapters  NCERT Class 11 Physics Chapter 1 Units of Measurement Question Answer and select needs one.

Also, you can read the SCERT book online in these sections NCERT Class 11 Physics Chapter 1 Units of Measurement Solutions by Expert Teachers as per SCERT ( CBSE ) Book guidelines. These solutions are part of SCERT  All Subject Solutions . Here we have given NCERT Class 11 Physics Chapter 1 Units of Measurement Solutions for All Subjects, You can practice these here.

Units of Measurement

Note: In stating numerical answers, take care of significant figures. 

1. Fill in the blanks:

(a) The volume of a cube of 1 cm is equal to …..m 3 .

Ans: 10 – 6 m 3 . 

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to …(mm) 2 .

Ans: 1.26 × 10 4 mm 2 .

(c) A vehicle moving with a speed of 18 km h –1 covers….m in 1 s.

Ans: 5 ms -1 .

(d) The relative density of lead is 11.3. Its density is ….g cm –3 or ….kg m –3 .

Ans: 1.13 × 10 4 kgm -3 . 

2. Fill in the blanks by suitable conversion of units.  

(a) 1 kg m 2 s –2 = ….g cm 2 s –2 .

(b) 1 m = ….. ly.

(c) 3.0 m s –2 = …. km h-2

= 3 × 3600 × 10 -3 kmh -2

= 3.888 × 10 4 km h -2

(d) G = 6.67 × 10 –11 N m 2 (kg) –2 = …. (cm)3 s –2 g –1 .

Ans: G = 6.67 × 10-11 Nm2 kh-2

= 6.67 × 10-11 × 1 kg-1 m3 s-2 

= 6.67 × 10 -8 cm -3 s -3 g -1 .

3. A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α –1 β –2 𝝲 2 in terms of the new units.

Ans: 1 cal. = 4.2 kg m 2 S -2

4. Explain this statement clearly:  

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:  

(a) Atoms are very small objects.

Ans: Atoms are very small objects compared to chalk.

(b) A jet plane moves with great speed.

Ans: A jet plane moves with higher speed as compared to an aeroplane.

(c) The mass of Jupiter is very large.

Ans: The mass of Jupiter is very large compared to Mercury. 

(d) The air inside this room contains a large number of molecules.

Ans: The air inside this room contains a large number of molecules as compared to a car tire.

(e) A proton is much more massive than an electron.

Ans: These statements need not be reframed.

(f) The speed of sound is much smaller than the speed of light.

5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?  

Ans: The time sun lights taken to reach the earth t = 8 min 20s

Now, convert 8 min 20 second into seconds

t = 8 × 60 + 20

Velocity of light in vacuum, c = 1 new unit of length s -1

Distance = speed x time ( c × t)

Distance = 1 new unit of length s -1 × 500.

= 500 new units of length.

6. Which of the following is the most precise device for measuring length: 

(a) a vernier callipers with 20 divisions on the sliding scale.

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale.

(c) an optical instrument that can measure length to within a wavelength of light? 

Ans: An optical instrument which can measure length within a wavelength of light is capable of extremely high precision. Therefore, option (c) is the correct answer. 

7. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair? 

Ans: Magnification = 100

Average width of the hair in the field of view = 3.5 mm

The estimate on the thickness of hair = Average width in field of view/Magnification 

8. Answer the following:

(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?  

Ans: Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other.

The diameter of the thread is given by the relation,

Diameter = Length of rod/Number of turns.

(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

Ans: Increasing the number of divisions of the circular scale will increase its accuracy to a certain extent only. 

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only? 

Ans: Because random errors involved in the former are very less as compared to the latter.

9. The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is 1.55 m2 . What is the linear magnification of the projector-screen arrangement?  

Ans: Area of the house on the screen  A = 1.75 cm²

= 1.55.cm 2 = 1.55 × 10 4 cm 2 .

Area of the house on the screen A′= 1.55 m²

= 1.55 × 10 4 / 1.75 

= 8.857 × 10 4 cm 2 .

Linear magnification

= √areal magnification

= √ 8.857 × 10 3

= 94.1 cm 2 .

10. State the number of significant figures in the following:  

(a) 0.007 m 2 .

(b) 2.64 × 10 4 kg.

(c) 0.2370 g cm –3 .

(d) 6.320 J.

(e) 6.032 N m –2 .

(f) 0.0006032 m 2 .  

11. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.  

Ans: The area (AAA) of the rectangular sheet is given by: 

Area = (4.234 × 1.005) x 2

= 8.51034 = 8.5 m 2 .

Volume The volume (VVV) of the rectangular sheet is given by:

V = l × b × t 

Volume =  (4.234 × 1.005) × (2.01 × 10 -2 )

= 8.55289 × 10 -2

= 0.0855m 3 .

12. The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is

(a) the total mass of the box.

Ans: Total mass of the box.

Mass of the box: 2. 30 kg.

First convert the masses of the gold pieces to kilograms:

20.15 g = 0.02015 kg.

20.17 g = 0.02017 kg.

Next, we add the masses:

(2.3kg + 0.02015 kg + 0.02017 kg)

= 2.3432 kg.

(b) the difference in the masses of the pieces to correct significant figures?  

Ans: Difference in the Masses of the Pieces.

The masses of two gold pieces are:

First piece: 20.15g

Second piece: 20.17g

The difference is:

20.17g – 20.15 g

13. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

Guess where to put the missing c.  

Ans: On rearranging, we have 

Since, the left hand side is dimensionless, so the right hand side should also be dimensionless. This will be so, if

14. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?

Ans: Given that:

r = 0.5 A = 0.5 × 10-10 m

Volume of hydrogen atom = (4/3) 𝜋r 3

= 0.524 × 10 -30 m 3

1 mole of hydrogen contains 6.023 × 10 23 hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 × 10 23 × 0.524 × 10 -30  

= 3.16 × 10 -7 m3. 

15. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of a hydrogen molecule to be about 1 Å). Why is this ratio so large?

Ans: Molar volume of ideal Gas:

1 mole of an ideal gas occupies 22.4 litres.

Size of hydrogen molecules: 1 Å = 1 × 10 -10 metres. 

V = 4/3 𝜋r 3 (r = 0.5 × 10 -10 metres)

V = 4/3 𝜋 (0.5 × 10 -10 ) 3

= 4/3 𝜋 (1.25 × 10 -31 )

= 5 × 10 -31 m 3

(convert to litres 1 m 3 = 1000 litres)

5 × 10 -31 m 3 = 5 × 10 -28 litres

Avogadro’s number = 6.022 × 10 23

Volume = 6.022 × 10 23 × 5 × 10 -28 litres

= 0.030 litres

Ratio = 22.4 / 0.030 

= 747.   

This high ratio is because of intermolecular spaces in gas being much larger than the size of the molecule. Thus, even though individual hydrogen molecules are very small, the gas volume reflects the much larger space between molecules in the gas phase.

16. Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Ans: Due to the motion of the train the nearby objects appear to move quickly in the opposite direction to the train’s motion. This is because as the train moves forward, your line of sight to these nearby objects changes rapidly. On the other hand the sight of a distant star or a cliff does not change its direction because they are so far away that their relative motion due to the train’s movement is imperceptible during the short duration of observation. Due to this the distant cliff or the star appears stationary.

17. The Sun is a hot plasma (ionised matter) with its inner core at a temperature exceeding 10 7 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×10 30 kg, radius of the Sun = 7.0 × 10 8 m.

M = 2 × 10 30 kg

r = 7 × 10 8 kg

Volume of sun = 4/3 𝜋r 3

= 4/3 × 22/7 × ( 7 × 10 8 ) 3

= 1.437 × 10 27 m 3 .

Now V = 1.437 × 10 27 .

As, r = M/V

= 2 × 10 30 / 1.437 × 10 27 .

= 1391.8 kg m -3 .

= 1.4 × 10 3 kg m -3 .

This is higher than the density of gases under standard conditions. Therefore, the mass density of the Sun is in the range of densities of solids and liquids, not gases. 

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NCERT Solutions for Class 11 Physics Chapter 2 Free PDF Download

Ncert solutions for class 11 physics chapter 2 – units and measurement.

Class 11 physics is pretty tough and difficult to understand by students. Moreover, there are certain topics that completely bounce off over students head. So to help them know the topic we have prepared NCERT Solutions for Class 11 Physics Chapter 2 that will help them with the topic in which they face difficulty.

Besides, these NCERT Solutions are prepared by our panel of professionals who have developed it after thorough research. In addition, they are in easy language so that students can easily get their meaning.

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CBSE Class 11 Physics Chapter 15 Units and Measurement NCERT Solutions

Unit refers to the international standards that are accepted worldwide. While on the other hand, measurement refers to the use of the unit for measuring the quantity. In this NCERT Solutions for Class 11 Physics Chapter 2, we will discuss all the topic of the chapter in brief and in an easily understandable language so that students can get what the content of the chapter.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 15

2.1 introduction.

This topic discusses what is unit and measurement which we have discussed above.

2.2 The International System of Units

This topic defines that earlier scientist of different countries uses the different unit for measurement. But with time and international usage, they started using SI (Standard Unit).

2.3 Measurement of Length

This topic overview the different unit of measuring the length that is used worldwide.

2.3.1 Measurement of Large Distances- This topic defines the use of parallax method for measuring long distances. Besides, the topic provides various examples of using large distances.

2.3.2 Estimation of very Small Distances: Size of a Molecule- This topic defines the use of the small unit by which we can measure the distance of the molecule.

2.3.3 Range of Lengths- this topic defines the use of such unit that helps to measure the lengths of objects that are spread in this wide universe.

2.4 Measurement of Mass

The weight of any object is its mass and this topic discusses the various unit of measurement of mass.

2.4.1 Range of Masses- It refers to all those objects that are spread over the universe that has a fixed mass and can be measured using these units.

2.5 Measurement of Time

This topic defines the atomic standard of time that is used for measuring the atomic clock and cesium clock.

2.6 Accuracy, Precision of Instruments and Errors in Measurement

This topic firstly defines what are errors and the mistakes we do which causes the errors.

• Systematic Errors
• Instrumental errors
• Imperfection in experimental technique or procedure
• Personal errors
• Random Errors
• Least Count Errors

2.6.1 Absolute Error, Relative Error, and Percentage Error- This topic defines all the three errors and how they affect the results.

2.6.2 Combination of Errors- This topic defines the error that we do while performing several measurements

• An error of a sum or a difference
• The error of a product or a quotient
• Error in case of a measured quantity raised to a power

2.7 Significant Figures

This refers to the first uncertain digit plus the reliable digit. Moreover, this topic several other points related to significant figures.

2.7.1 Rules of Arithmetic Operations with Significant Figures- The topic define the rules related to the arithmetic operations.

2.7.2 Rounding off the Uncertain Digits- This topic defines the rule of rounding off of unclear digits.

2.7.3 Rules of Determining the Uncertainty in the Results of Arithmetic Calculations- This topic defines the rules that help to govern the doubt of the results of mathematics calculations.

2.8 Dimensions of Physical Quantities

This topic defines the scope of physical quantities.

2.9 Dimensional Formulae and Dimensional Equations

This topic describes the dime national equations and formula that we use to equitize a physical quantity.

2.10 Dimensional Analysis and its Applications

This topic defines the various applications by which we can analyze the physical quantities of a dimension.

2.10.1 Checking the Dimensional Consistency of Equations- This topic checks the consistency of dimensional equations.

2.10.2 Deducing Relation among the physical Quantities- This topic defines how we can use the relation of the physical quantities for reasoning.

You can download NCERT Solutions for Class 11 Physics Chapter 2 PDF by clicking on the button below.

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NCERT Solutions for Class 11

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• NCERT Solutions for Class 11 Physics Chapter 6
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• NCERT Solutions for Class 11 Biology Chapter 19 Free PDF Download

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Units and Measurements Assignments/DPPs

The chapter “Units and Measurements” in Class 11 physics forms the cornerstone for all subsequent scientific endeavors. This crucial chapter lays the groundwork for comprehending and interpreting physical quantities and measurements, a skill vital for success in the Joint Entrance Examination (JEE) and beyond. In this journey of scientific exploration, we delve into the world of units, measurements,

Overview : Units and Measurements for Class 11 and JEE

Understanding the Fundamentals:

• Physical Quantity:  Any measurable characteristic of a physical system (e.g., length, mass, time, temperature).
• Units:  Standardized references used to express the magnitude of a physical quantity. They ensure consistency and allow comparison of measurements across different contexts.
• Measurement:  The process of assigning a numerical value to a physical quantity using a chosen unit.

The International System of Units (SI):

The SI serves as the modern and globally accepted system of units. It comprises seven fundamental units:

• Length:  Meter (m)
• Mass:  Kilogram (kg)
• Time:  Second (s)
• Electric Current:  Ampere (A)
• Thermodynamic Temperature:  Kelvin (K)
• Amount of Substance:  Mole (mol)
• Luminous Intensity:  Candela (cd)

Derived Units:

Through combinations of fundamental units, we can derive numerous derived units to measure various physical quantities. Examples include:

• Area:  m² (square meter)
• Volume:  m³ (cubic meter)
• Speed:  m/s (meter per second)
• Acceleration:  m/s² (meter per square second)
• Force:  N (Newton) = kg * m/s²

Essential Formulas and Relationships:

• Converting between units:  Utilize conversion factors to convert from one unit to another. (e.g., 1 km = 1000 m)
• Dimensional analysis:  Check the consistency of physical quantities in equations by analyzing their dimensions (units raised to powers).

Question Types in the JEE:

The JEE assesses your understanding of Units and Measurements through various question types, including:

• Unit conversions:  Converting numerical values between different units of the same quantity.
• Dimensional analysis:  Checking the dimensional homogeneity of equations and identifying inconsistencies.
• Estimation problems:  Implementing appropriate units and estimating numerical values based on physical reasoning.
• Interpreting graphs and data:  Analyzing data presented in graphs and tables, considering their units and scales.
• Word problems:  Applying your knowledge of units and measurements to solve real-world scenarios involving various physical quantities.

DPPs for Units and Measurements

Mastering Units and Measurements with PRERNA EDUCATION:

• Targeted assignments :  Enhance your understanding by practicing dedicated exercises from PRERNA EDUCATION , focusing on specific unit conversions, dimensional analysis techniques, and real-world applications.
• Daily Practice Problems (DPPs):  Sharpen your problem-solving skills and build speed by tackling daily practice problems (DPPs) encompassing various difficulty levels and diverse problem formats.
• Expert guidance:  Don’t hesitate to seek clarification from teachers or utilize online resources provided by PRERNA EDUCATION for additional insights and explanations.
• Focus on unit awareness:  Be mindful of units throughout your problem-solving process. Always include units in your calculations and final answers.
• Practice dimensional analysis:  Utilize dimensional analysis to check the consistency of equations and identify potential errors.

Embrace the Exploration:

A solid foundation in Units and Measurements unlocks the doors to exploration in all scientific disciplines. By leveraging the resources offered by PRERNA EDUCATION, consistently practicing with diverse problems, and fostering both conceptual understanding and unit awareness, you can excel in this crucial chapter and gain valuable skills for your journey in Class 11 physics and beyond. Remember, a curious mind, dedication to practice, and a strong foundation in fundamental concepts form the key to success in navigating the fascinating world of science and measurements.

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