assignment 1.1 physics class 11

NCERT Solution Class 11 Physics Chapter 1 Units and Measurement

Q4 :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 (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light. Answer : The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. (a) An atom is a very small object in comparison to a soccer ball. (b) A jet plane moves with a speed greater than that of a bicycle. (c) Mass of Jupiter is very large as compared to the mass of a cricket ball. (d) The air inside this room contains a large number of molecules as compared to that present in a geometry box. (e) A proton is more massive than an electron. (f) Speed of sound is less than the speed of light.

Q5 : 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? Answer : Distance between the Sun and the Earth: = Speed of light x Time taken by light to cover the distance Given that in the new unit, speed of light = 1 unit Time taken, t = 8 min 20 s = 500 s ∴Distance between the Sun and the Earth = 1 x 500 = 500 units

Q7 :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? Answer : Magnification of the microscope = 100 Average width of the hair in the field of view of the microscope = 3.5 mm ∴Actual thickness of the hair is  3.5/100= 0.035 mm.

Q10 : State the number of significant figures in the following: (a) 0.007 m 2 (b) 2.64 x 10 24 kg (c) 0.2370 g cm -3 (d) 6.320 J (e) 6.032 N m -2 (f) 0.0006032 m 2 Answer : (a) Answer: 1 The given quantity is 0.007 m 2 . If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity. (b) Answer: 3 The given quantity is 2.64 x 10 24 kg. Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures. (c) Answer: 4 The given quantity is 0.2370 g cm -3 . For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure. (d) Answer: 4 The given quantity is 6.320 J. For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures. (e) Answer: 4 The given quantity is 6.032 Nm -2 . All zeroes between two non-zero digits are always significant. (f) Answer: 4 The given quantity is 0.0006032 m 2 . If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

Q11 : 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. Answer : Length of sheet, l = 4.234 m Breadth of sheet, b = 1.005 m Thickness of sheet, h = 2.01 cm = 0.0201 m The given table lists the respective significant figures:

Hence, area and volume both must have least significant figures i.e., 3. Surface area of the sheet = 2 (l × b + b × h + h × l) = 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234) = 2(4.25517 + 0.0202005 + 0.0851034) = 2 × 4.36 = 8.72 m 2 Volume of the sheet = l × b × h = 4.234 × 1.005 × 0.0201 = 0.0855 m 3 This number has only 3 significant figures i.e., 8, 5, and 5.

Q12 : The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures? Answer : Mass of grocer’s box = 2.300 kg Mass of gold piece I = 20.15g = 0.02015 kg Mass of gold piece II = 20.17 g = 0.02017 kg (a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg. (b) Difference in masses = 20.17 – 20.15 = 0.02 g In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Q18 : 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). Answer : Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly. On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

Q21 : Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed. Answer : It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ⌊ – ⌋ 10-15 s) are used to measure time intervals in several physical and chemical processes. X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing. The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

Q31 : The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us? Answer : Time taken by quasar light to reach Earth = 3 billion years = 3 x 10 9 years = 3 x 109 x 365 x 24 x 60 x 60 s Speed of light = 3 x 10 8 m/s Distance between the Earth and quasar = (3 x 10 8 ) x (3 x 10 9 x 365 x 24 x 60 x 60) = 283824 x 10 20 m = 2.8 x 10 22 km

NCERT Solutions for Class 11 Physics All Chapters

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 9 Mechanical Properties Of Solids Chapter 10 Mechanical Properties Of Fluids Chapter 11 Thermal Properties Of Matter Chapter 12 Thermodynamics Chapter 13 Kinetic Theory Chapter 14 Oscillations Chapter 15 Waves

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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

• NCERT Solutions
• Chapter 1 Units And Measurement

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements - FREE PDF Download

NCERT for Chapter 1 Units and Measurements Class 11 Solutions by Vedantu, forms the basis for all experimental and theoretical studies in physics. Understanding units and measurements is important as it ensures consistency and accuracy in scientific observations and calculations. This chapter emphasizes the importance of standardized units and accurate measurements in ensuring consistency and precision in scientific observations and calculations. It provides students with the essential tools and techniques to measure physical quantities correctly, which is critical for validating scientific theories and conducting experiments. By practising with Class 11 Physics NCERT Solutions , students can build confidence in their understanding and excel in their studies.

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Glance on Physics Chapter 1 Class 11 - Units and Measurements

The chapter introduces the concept of base units for fundamental physical quantities and derived units for quantities that are combinations of base quantities. 

Various techniques and instruments for measuring physical quantities accurately are discussed. This includes the use of vernier calipers, micrometers, and other precise tools.

The chapter also covers different types of errors—systematic and random—and methods to minimize and account for these errors in measurements.

The chapter discusses the applications of dimensional analysis in checking the correctness of equations, converting units from one system to another, and deriving relationships between physical quantities. Also, it includes significant figures and rules for determining the number of significant figures in a given measurement.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 1 - Units And Measurements, which you can download as PDFs.

There are 17 fully solved questions in class 11th physics chapter 1 Units And Measurements Exercise.

System of Units:

Access NCERT Solutions for Class 11 Physics Chapter 1 – Units and Measurements

a) The Volume of a Cube of 1cm is Equal To ……………… ${{m}^{3}}$. 

We know that, 

$1cm=\frac{1}{100}m$

Volume of a cube of side 1cm would be, 

$V=1cm\times 1cm\times 1cm=1c{{m}^{3}}$

On converting it into unit of ${{m}^{3}}$, we get, 

$1c{{m}^{3}}={{\left( \frac{1}{100}m \right)}^{3}}={{\left( {{10}^{-2}}m \right)}^{3}}$

$\therefore 1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$

Therefore, the volume of a cube of side 1cm is equal to  ${{10}^{-6}}{{m}^{3}}$.

b) The Surface Area of a Solid Cylinder of Radius 2.0cm and Height 10.0cm is Equal To ……………….${{\left( mm \right)}^{2}}$

We know the formula for the total surface area of cylinder of radius r and height h to be,

$S=2\pi r\left( r+h \right)$

We are given:

$r=2cm=20mm$

$h=10cm=100mm$

On substituting the given values into the above expression, we get, 

$S=2\pi \times 20\left( 20+100 \right)=15072m{{m}^{2}}=1.5\times {{10}^{4}}m{{m}^{{{2}^{{}}}}}$

Therefore, the surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to $1.5\times {{10}^{4}}{{\left( mm \right)}^{2}}$. 

c) A Vehicle Moving With a Speed of $18km{{h}^{-1}}$Covers…………………. m in 1s. 

We know the following conversion:

$1km/h=\frac{5}{18}m/s$

$\Rightarrow 18km/h=18\times \frac{5}{18}=5m/s$

Now we have the relation:

$\text{Distance = speed }\times \text{ time }$

Substituting the given values, $\text{Distance = 5}\times \text{1 =5m}$

Therefore, a vehicle moving with a speed of $18km{{h}^{-1}}$covers 5m in 1s.

d) The Relative Density of Lead is 11.3. Its Density Is ………………. $gc{{m}^{-3}}$or………………… $kg{{m}^{-3}}$. 

We know that the relative density of substance could be given by, 

$\text{Relative density = }\frac{density\text{ of substance}}{density\text{ of water}}$

$density\text{ of water = 1kg/}{{\text{m}}^{3}}$

$\text{density of lead = Relative density of lead }\times \text{ density of water = 11}\text{.3}\times \text{1= 11}\text{.3g/c}{{\text{m}}^{3}}$

But we know, 

$1g={{10}^{-3}}kg$

$1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$

$\Rightarrow 1g/c{{m}^{3}}=\frac{{{10}^{-3}}}{{{10}^{-6}}}kg/{{m}^{3}}={{10}^{3}}kg/{{m}^{3}}$

$\therefore 11.3g/c{{m}^{3}}=11.3\times {{10}^{3}}kg/{{m}^{3}}$

Therefore, the relative density of lead is 11.3. Its density is $11.3gc{{m}^{-3}}$or$11.3\times {{10}^{3}}kg{{m}^{-3}}$.

2. Fill ups.

a) $1kg{{m}^{2}}{{s}^{-2}}=..................gc{{m}^{2}}{{s}^{-2}}$

We know that:

$1kg={{10}^{3}}g$

$1{{m}^{2}}={{10}^{4}}c{{m}^{2}}$

$1kg{{m}^{2}}{{s}^{-2}}={{10}^{3}}g\times {{10}^{4}}c{{m}^{2}}\times 1{{s}^{-2}}={{10}^{7}}gc{{m}^{2}}{{s}^{-2}}$

Therefore, $1kg{{m}^{2}}{{s}^{-2}}={{10}^{7}}gc{{m}^{2}}{{s}^{-2}}$

b) $1m=.................ly$

We know that light year is the total distance covered by light in one year. 

$1ly=\text{Speed of light }\times \text{ one year}$

$\Rightarrow 1ly=\left( 3\times {{10}^{8}}m/s \right)\times \left( 365\times 24\times 60\times 60s \right)=9.46\times {{10}^{15}}m$

$\therefore 1m=\frac{1}{9.46\times {{10}^{15}}}=1.057\times {{10}^{-16}}ly$

Therefore, $1m=1.057\times {{10}^{-16}}ly$

c) $3.0m/{{s}^{2}}=.................km/h{{r}^{2}}$ 

$3.0m/{{s}^{2}}=$ ………….$km/h{{r}^{2}}$

We have, $1m={{10}^{-3}}km$

$1hr=3600s$

$\Rightarrow 1{{s}^{2}}={{\left( \frac{1}{3600} \right)}^{2}}h{{r}^{2}}$

Then, 

$3.0m/{{s}^{2}}=\frac{3\times {{10}^{-3}}}{{{\left( \frac{1}{3600}h \right)}^{2}}}km/h{{r}^{2}}$

$\therefore 3.0m/{{s}^{2}}=3.9\times {{10}^{4}}km/h{{r}^{2}}$ 

d)\[6.67\times{{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=.............{{g}^{-1}}c{{m}^{3}}{{s}^{-2}}\] 

$1N=1kgm{{s}^{-2}}$

$1kg={{10}^{-3}}g$

$1{{m}^{3}}={{10}^{6}}c{{m}^{3}}$

\[\Rightarrow 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}=6.67\times {{10}^{-11}}\times \left( 1kgm{{s}^{-2}} \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)\]

\[=6.67\times {{10}^{-11}}\times \left( 1kg\times 1{{m}^{3}}\times 1{{s}^{-2}} \right)\]

\[=6.67\times {{10}^{-11}}\times \left( {{10}^{-3}}{{g}^{-1}} \right)\left( {{10}^{6}}c{{m}^{3}} \right)\left( 1{{s}^{-2}} \right)\]

$\therefore 6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=6.67\times {{10}^{-8}}c{{m}^{3}}{{s}^{-2}}{{g}^{-1}}$

3. A Calorie is a Unit of Heat or Energy and Is Equivalent to 4.2 J Where $1J=1kg{{m}^{2}}{{s}^{-2}}$. Suppose We Employ a System of Units in Which the Unit of Mass Equals $\alpha \text{ kg}$, the Unit of Length Equals $\beta $ m, the Unit of Time is $\gamma \text{ s}$ . Show That a Calorie Has a Magnitude $4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}$ In Terms of the New Unit.

We are given that,

\[1calorie=4.2\left( 1kg \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)\] 

Let the new unit of mass $=\alpha \text{ kg}$. 

So, one kilogram in terms of the new unit, $1\text{ kg}=\frac{1}{\alpha }={{\alpha }^{-1}}$.

One meter in terms of the new unit of length can be written as, $\text{1m}=\frac{1}{\beta }={{\beta }^{-1}}$  or $\text{1}{{\text{m}}^{2}}={{\beta }^{-2}}$.

And, one second in terms of the new unit of time,

$1\text{ s}=\frac{1}{\gamma }={{\gamma }^{-1}}$

$1\text{ }{{\text{s}}^{2}}={{\gamma }^{-2}}$

$1\text{ }{{\text{s}}^{-2}}={{\gamma }^{2}}$

\[\therefore 1calorie=4.2\left( 1{{\alpha }^{-1}} \right)\left( 1{{\beta }^{-2}} \right)\left( 1{{\gamma }^{2}} \right)=4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}\] 

Therefore, the value equivalent to one calorie in the mentioned new unit system is \[4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}\].

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:

The given statement is true because a dimensionless quantity may be large or small, but there should be some standard reference to compare that. 

For example, the coefficient of friction is dimensionless but we could say that the coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

a) Atoms Are Very Small Objects.

Ans: An atom is very small compared to a soccer ball.

b) A Jet Plane Moves With Great Speed.

Ans: A jet plane moves with a speed greater than that of a bicycle.

c)The Mass of Jupiter is Very Large.

Ans: Mass of Jupiter is very large compared to the mass of a cricket ball.

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 that contained by a geometry box.

e) A Proton is Much More Massive than an Electron.

Ans: A proton is more massive than an electron.

f) The Speed of Sound is Much Smaller than the Speed of Light.          

Ans:  Speed of sound is less 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:  

Distance between the Sun and the Earth:

\[\text{x= Speed of light}\times \text{Time taken by light to cover the distance}\] 

It is given that in the new system of units, the speed of light \[c=1\text{ }unit\]. 

Time taken, \[t=8\text{ }min\text{ }20\text{ }s=500\text{ }s\]

Thus, the distance between the Sun and the Earth in this system of units is given by\[x'=c\times t=1\times 500=500\text{ }units\] 

6. Which of the Following is the Most Precise Device for Measuring Length?

Ans: A device which has the minimum least count is considered to be the most precise device to measure length.

a) A Vernier Caliper With 20 Divisions on the Sliding Scale.

Least count of vernier calipers is given by

\[LC=1\text{ }standard\text{ }division\left( SD \right)-1\text{ }vernier\text{ }division\left( VD \right)\] 

$\Rightarrow LC=1-\frac{19}{20}=\frac{1}{20}=0.05cm$

b) A Screw Gauge of Pitch 1 Mm and 100 Divisions on the Circular Scale.

Least count of screw gauge $=\frac{\text{Pitch}}{\text{No of divisions}}$

$\Rightarrow LC=\frac{1 mm}{100}=\frac{0.1 cm}{100}$

$\Rightarrow LC=\frac{1}{1000}=0.001cm$

c) An Optical Instrument that Can Measure Length to Within a Wavelength of Light.

Least count of an optical device $=\text{Wavelength of light}\sim \text{1}{{\text{0}}^{-5}}cm$ 

$\Rightarrow LC=0.00001cm$ 

Hence, it can be inferred that an optical instrument with the minimum least count among the given three options is the most suitable device to measure length.

7. A Student Measures the Thickness of a Human Hair Using 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. Estimate the Thickness of Hair.

We are given that: 

Magnification of the microscope \[=100\] 

Average width of the hair in the field of view of the microscope \[=3.5\text{ }mm\] 

$\therefore $ Actual thickness of the hair would be, $\frac{3.5}{100}=0.035\text{ }mm.$ 

8. Answer the Following:

a) YOu Are Given a Thread and a Meter Scale. How Will You Estimate the Diameter of the Thread?

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

Measure the length that is wounded by the thread using a metre scale. 

The diameter of the thread is given by the relation,

Diameter $=\frac{\text{Length of thread}}{\text{Number of turns}}$ 

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?

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

C) The Mean Diameter of a Thin Brass Rod Is to Be Measured by Vernier Calipers. Why Is a Set of 100 Measurements of the Diameter Expected to Yield a More Reliable Estimate Than a Set of 5 Measurements Only?

A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved will be reduced on increasing the number of measurements.

9. The Photograph of a House Occupies an Area of $1.75c{{m}^{2}}$ On a 35 Mm Slide. the Slide Is Projected Onto a Screen, and the Area of the House on the Screen is $1.55{{m}^{2}}$. What is the Linear Magnification of the Projector-Screen Arrangement?

We are given, 

The area of the house on the $35mm$ slide (area of the object) is given by, 

${{A}_{O}}=1.75c{{m}^{2}}$.

The area of the image of the house that is formed on the screen is given by, ${{A}_{I}}=1.55{{m}^{2}}=1.55\times {{10}^{4}}c{{m}^{2}}$

We know that areal magnification is given by, 

${{m}_{a}}=\frac{{{A}_{I}}}{{{A}_{O}}}$

Substituting the given values, 

${{m}_{a}}=\frac{1.55\times {{10}^{4}}}{1.75}$

Now, we have the expression for Linear magnification as, ${{m}_{l}}=\sqrt{{{m}_{a}}}$ 

$\Rightarrow {{m}_{l}}=\sqrt{\frac{1.55}{1.75}\times {{10}^{4}}}$ 

$\therefore {{m}_{l}}=94.11$

Thus, we found the linear magnification in the given case to be, ${{m}_{l}}=94.11$.

10. State the Number of Significant Figures in the Following:

a) $0.007{{m}^{2}}$

Ans: We know that when the given number is less than one, all zeroes on the right of the decimal point are insignificant and hence for the given value, only 7 is the significant figure. So, the number of significant figures in this case is 1. 

b) $2.64\times {{10}^{24}}kg$

Ans: We know that the power of 10 is considered insignificant and hence, 2, 6 and 4 are the significant figures in the given case. So, the number of significant figures here is 3. 

c) $0.2370gc{{m}^{-3}}$

Ans: For decimal numbers, the trailing zeroes are taken significantly. 2, 3, 7 and 0 are the significant figures. So, the number of significant figures here is 4. 

d) $6.320J$

Ans: All figures present in the given case are significant. So, the number of significant figures here is 4. 

e) $6.032N{{m}^{-2}}$

Ans: Since all the zeros between two non-zero digits are significant, the number of significant figures here is 4. 

f) $0.0006032{{m}^{2}}$

Ans: For a decimal number less than 1, all the zeroes lying to the left of a non-zero number are insignificant. Hence, the number of significant digits here is 4. 

11. The Length, Breadth and Thickness of a Rectangular Sheet of Metal Are 4.234m, 1.005m and 2.01cm Respectively. Give the Area and Volume of the Sheet to Correct Significant Figure. 

Length of sheet, $l=4.234m$; number of significant figures: 4

Breadth of sheet, $b=1.005m$; number of significant figures: 4

Thickness of sheet, $h=2.01cm=0.0201m$; number of significant figures: 3

So, we found that area and volume should have the least significant figure among the given dimensions, i.e., 3. 

Surface area, $A=2\left( l\times b+b\times h+h\times l \right)$

$\Rightarrow A=2\left( 4.234\times 1.005+1.005\times 0.0201+0.0201\times 4.234 \right)=2\left( 4.25517+0.02620+0.08510 \right)$

$\therefore A=8.72{{m}^{2}}$

Volume, $V=l\times b\times h$

$\Rightarrow V=4.234\times 1.005\times 0.0201$

$\therefore V=0.0855{{m}^{3}}$

Therefore, we found the area and volume with 3 significant figures to be $A=8.72{{m}^{2}}$

and $V=0.0855{{m}^{3}}$respectively. 

12. The Mass of a Box Measured by a Grocer's Balance is 2.300 Kg. Two Gold Pieces of Masses 20.15 G and 20.17 G Are Added to the Box. What Is:

a) The Total Mass of the Box?

Mass of grocer’s box $=2.300kg$ 

Mass of gold piece $I=20.15g=0.02015kg$ 

Mass of gold piece $II=20.17g=0.02017kg$ 

Total mass of the box $=2.3+0.02015+0.02017=2.34032kg$ 

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is $2.3kg$.

b) The Difference in the Masses of the Pieces to Correct Significant Figures?       

Difference in masses $=20.17-20.15=0.02g$ 

While subtracting, the final result should retain as many decimal places as there are in the number with the least decimal places.

13. A Famous Relation in Physics Relates ‘Moving Mass’ M to the ‘Rest Mass’ ${{m}_{0}}$of a Particle in Terms of Its Speed v and Speed of Light c. (This Relation First Arise 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:

$m=\frac{{{m}_{0}}}{{{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}}$

We are given the following relation:

 $m=\frac{{{m}_{0}}}{{{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}}$

Dimension of m, ${{M}^{1}}{{L}^{0}}{{T}^{0}}$

Dimension of ${{m}_{0}}$, ${{M}^{1}}{{L}^{0}}{{T}^{0}}$

Dimension of v, ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$

Dimension of ${{v}^{2}}$, ${{M}^{0}}{{L}^{2}}{{T}^{-2}}$

Dimension of c, ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$

For the formula to be dimensionally correct, the dimensions on the LHS should be the same as those on the RHS. In order to satisfy this condition, ${{\left( 1-{{v}^{2}} \right)}^{\frac{1}{2}}}$should be dimensionless and for that we require ${{v}^{2}}$ be divided by ${{c}^{2}}$. So, the dimensionally correct version of the above relation would be,

$m=\frac{{{m}_{0}}}{{{\left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)}^{\frac{1}{2}}}}$

14. The Unit of Length Convenient on the Atomic Scale is Known as an Angstrom and is Denoted By $\overset{{}^\circ }{\mathop{A}}\,:1\overset{{}^\circ }{\mathop{A}}\,={{10}^{-10}}m$. The Size of a Hydrogen Atom Is About 0.5a. What is the Total Atomic Volume In ${{m}^{3}}$of a Mole of Hydrogen Atoms?

Radius of hydrogen atom is given to be, 

$r=0.5\overset{{}^\circ }{\mathop{A}}\,=0.5\times {{10}^{-10}}m$

The expression for the volume is,

$V=\frac{4}{3}\pi {{r}^{3}}$

Now on substituting the given values, 

$V=\frac{4}{3}\pi {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

But we know that 1 mole of hydrogen would contain Avogadro number of hydrogen atoms, so volume of 1 mole of hydrogen atoms would be, 

$V'={{N}_{A}}V=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$

Therefore, we found the required volume to be $3.16\times {{10}^{-7}}{{m}^{3}}$.

15.One Mole of an Ideal Gas at Standard Temperature and Pressure Occupies $22.4L$ (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 $1\overset{{}^\circ }{\mathop{\text{A}}}\,$). Why is This Ratio So Large?

Radius of hydrogen atom, \[r=0.5\overset{{}^\circ }{\mathop{\text{A}}}\,=0.5\times {{10}^{-10}}m\] 

Volume of hydrogen atom, $V=\frac{4}{3}\pi {{r}^{3}}$ 

$\Rightarrow V=\frac{4}{3}\times \frac{22}{7}\times {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

Now, 1 mole of hydrogen contains $6.023\times {{10}^{23}}$ hydrogen atoms. 

Volume of 1 mole of hydrogen atoms,${{V}_{a}}=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$. 

Molar volume of 1 mole of hydrogen atoms at STP, ${{V}_{m}}=22.4L=22.4\times {{10}^{-3}}{{m}^{3}}$ 

So, the required ratio would be, 

$\frac{{{V}_{m}}}{{{V}_{a}}}=\frac{22.4\times {{10}^{-3}}}{3.16\times {{10}^{-7}}}=7.08\times {{10}^{4}}$

Hence, we found that the molar volume is $7.08\times {{10}^{4}}$ times higher than the atomic volume. 

For this reason, the interatomic separation in hydrogen gas is much larger than the size of a hydrogen atom. 

16. Explain This Common Observation Clearly: If You Look Out of the Window of a Fast-Moving Train, the Nearby Trees, Houses Etc. Seems to Move Rapidly in a Direction Opposite to the Train's Motion, but the Distant Objects (hill Tops, the Moon, the Stars Etc.) Seems to Be Stationary. (In Fact, Since You Are Aware That You Are Moving, These Distant Objects Seem to Move With You).

Ans: Line-of-sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc., while sitting in a moving train, they appear to move rapidly in the opposite direction because the line-of-sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc., appear stationary because of the large distance. As a result, the line-of-sight does not change its direction rapidly.

17. The Sun Is a Hot Plasma (ionized Matter) With Its Inner Core at a Temperature Exceeding ${{10}^{7}}K$ and Its Outer Surface at a Temperature of About 6000k. 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$=2.0\times {{10}^{30}}kg$, Radius of the Sun$=7.0\times {{10}^{8}}m$ .

We are given the following:

Mass of the sun, $M=2.0\times {{10}^{30}}kg$

Radius of the sun, $R=7.0\times {{10}^{8}}m$

Now we find the volume of the sun to be, 

$V=\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\pi {{\left( 7.0\times {{10}^{8}} \right)}^{3}}=1437.3\times {{10}^{24}}{{m}^{3}}$

Density of the sun is found to be, 

$\rho =\frac{M}{V}=\frac{2.0\times {{10}^{30}}}{1437.3\times {{10}^{24}}}$

$\therefore \rho \sim 1.4\times {{10}^{3}}kg/{{m}^{3}}$

So, we found the density of the sun to lie in the density range of solids and liquids. 

Clearly, the high intensity is attributed to the intense gravitational attraction of the inner layers on the outer layer of the sun.

Dimensional Formulae of Physical Quantities

Overview of deleted syllabus for cbse class 11 physics units and measurements.

NCERT Class 11 Physics Chapter 1 Solutions on Units And Measurements provided by Vedantu serves as the foundation for all scientific study and experimentation. This chapter introduces students to the fundamental concepts of measuring physical quantities, ensuring consistency and accuracy in their scientific observations and calculations. Understanding units and measurements is critical for conducting experiments, interpreting results, and communicating scientific findings effectively. These concepts are essential for mastering the topic and are often tested in exams. From previous year's question papers, typically around 2-3 questions are asked from this chapter. These questions test students' theoretical concepts as well as their problem-solving skills. 

Other Study Material for CBSE Class 11 Physics Chapter 1

Chapter-specific ncert solutions for class 11 physics.

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

CBSE Class 11 Physics Study Materials

Faqs on ncert solutions for class 11 physics chapter 1 units and measurements.

1. What are the topics that are covered in class 11 physics chapter 2?

The concepts that are covered in chapter 2 are:

Introduction to units and measurement. 

The international system of the units.

SI Base unit

SI Derived unit

Advantages of learning SI units and CGS units  

Significant figures. 

Applications of significant figures

Exact Number

Dimensions of physical quantities. 

Dimensional formulae and dimensional equations. 

Dimensional analysis and its applications.  

Unit conversion and dimensional analysis

Using dimensional analysis to check the correctness of physical equation. 

Homogeneity principle of dimensional analysis. 

Applications of dimensional analysis. 

Limitations of dimensional analysis.

2. What is dimensional analysis?

We quantify the size and shape of things using Dimensional Analysis. It helps us study the nature of objects mathematically. It involves lengths and angles as well as geometrical properties such as flatness and straightness. The basic concept of dimension is that we can add and subtract only those quantities that have the same dimensions. Similarly, two physical quantities are equal if they have the same dimensions.

The main benefits of the dimensional analysis of a problem have reduced the number of variables in the issue by combining dimensional variables to one form of non-dimensional parameters. By far the easiest and most useful method in the analysis of any fluid problem is that of direct mathematical solution.

3. What are gross errors?

This section basically takes into account human oversight and other mistakes while reading, recording and readings. The most common errors, the human error in the measurement fall which is under this category of errors in measurement. For instance, a person taking the reading from the meter of the instrument and he may read 33 as 38. Gross errors could be avoided by using two suitable measures and they are mentioned below:

A proper care should always be taken while reading and recording the data. Also, calculation of error should be done accurately.

By increasing the total number of experimenters we can reduce the gross errors. If each experimenter takes different reading at various points, then by taking an average of more readings we can reduce the gross errors.

4. How Vedantu will help me in exam preparation?

Our NCERT solutions are prepared by our maths experts with various real-life examples. These examples will make you understand the concept quickly and memorise them for a longer time. Solutions provided to the questions are 100% accurate in the exercises which are crisp and concise to the point.

Our solutions are the best study guides, which help you in smart learning and efficient answering of questions. These solutions will also help you in improving a strong conceptual base with all the important concepts in a very easy and understandable language. You will also enjoy learning from our solutions which are really fun and interactive.

5. Are NCERT Solutions important for Class 11 Physics Chapter 2?

Referring to NCERT Solutions is as important as referring to the questions while preparing for your Class 11 Physics Chapter 2. Only reading the questions that are provided in the NCERT is not going to be helpful if students are not able to answer them correctly. NCERT Solutions for Class 11 Physics Chapter 2 available on Vedantu provide students the correct step-by-step solutions for all NCERT questions so that students do not lose any marks in their exams.

6. Do I need to practice all the questions provided in Class 11 Physics Chapter 2 NCERT Solutions?

Questions provided in Class 11 Physics NCERT Solutions for Chapter 2 are to be considered crucial when preparing for your Class 11 Physics exams. The exam question papers always include questions that have been provided in the Physics NCERT for Class 11. Practicing all the questions will only help you increase your understanding of all the concepts taught in Chapter 2 and also the possibility of scoring well in the questions framed from the chapter.

7. How can I understand the Class 11 Physics Chapter 2?

Chapter 2 in Class 11 Physics NCERT is called “Units and Measurement”. This chapter talks in detail about various units used for determining the measurement of different physical quantities, instruments used for such measurements and their accuracy, etc. Students can easily understand this chapter by indulging in regular reading of the chapter and solving the questions provided in the NCERT. For more help, students can also refer to NCERT Solutions for Class 11 Physics Chapter 2 on the official website of Vedantu or download the Vedantu app where these resources are available at free of cost.

8. What is the marks distribution for Class 11 Physics Chapter 2?

Chapter 2 - Units and Measurement in Class 11 Physics is a part of Unit - I along with Chapter 1 - Physical World. According to the marks distribution provided by CBSE for Class 11 Physics, Unit - I consisting of both the chapters, carries a total of 23 marks. Hence, preparation from both chapters should be given equal priority to avoid losing any marks in questions framed from them in the exam.

9. What are the important topics covered in NCERT Solutions for Class 11 Physics Chapter 2?

NCERT Solutions for Class 11 Physics  Chapter 2 - Units and Measurements includes various topics like the International System of Units, Measurement of length, mass, and time, Application of Significant Figures, etc. Among these, the most important topics from the chapter include SI Units, Absolute Errors, Dimensional Analysis, and Significant Figures. Short-answer and numerical-based questions can also be asked from the topic of evaluating errors during the measurement of quantities. 

10. What is the concept of units and measurements in class 11 exercise solutions?

In units and measurements class 11 exercise solutions in Physics Chapter 1 involves quantifying physical quantities in a standardized manner. A unit is a definite magnitude of a quantity, defined and adopted by convention or law, used as a standard for measuring the same kind of quantity. Measurement is the process of determining the size, length, or amount of something, typically using standard units.

11. Which topics are important in units and measurements class 11 NCERT solutions?

Key topics in Physics Class 11 Chapter 1 Exercise Solutions include:

Physical Quantities and Units: Base and derived quantities, SI units, and other systems of units.

Measurement Techniques: Direct and indirect measurement methods.

Significant Figures: Rules for determining significant figures in measurements.

Errors in Measurement: Types of errors (systematic and random) and methods to minimize them.

Dimensional Analysis: Checking the dimensional consistency of equations and converting units.

12. What is the direct method of measurement of length class 11 physics chapter 1 NCERT solutions?

In units and measurements class 11 exercise solutions, the direct method of measurement of length involves using instruments that directly provide the measurement of length, such as rulers, measuring tapes, vernier calipers, and micrometer screw gauges. These tools are used to measure the length of objects by comparing them directly against a standard unit of length.

13. What is the full form of SI unit?

According to physics class 11 chapter 1, the full form of SI unit is "Système International 'Unités", which translates to the International System of Units. It is the modern form of the metric system and the most widely used system of measurement.

14. What are the three main units of measurement in Class 11 Physics Ch 1 NCERT Solutions?

The three main base units and measurements class 11 ncert solutions in the SI system are:

Meter (m) for length

Kilogram (kg) for mass

Second (s) for time

15. What is a measurement of time in units and measurements class 11 solutions?

The measurement of time is the process of quantifying the duration or interval between two events. The standard unit of time in the SI system is the second (s). Time can be measured using various devices such as clocks, stopwatches, and atomic clocks.

NCERT Solutions for Class 11 Physics

NCERT Solutions for Class 11 Physics Chapter 1 Free PDF Download

Ncert solutions for class 11 physics chapter 1 – physical world.

Physics is a subject that studies everything present in the universe. In addition, how everything works and what are their characteristics. But class 11 physics is pretty tough to understand for students. That’s why we prepared these NCERT Solutions for Class 11 Physics Chapter 1 which will guide you through the chapter thoroughly. Also, our video tutorial explains each topic in simple language.

These solutions are pretty easy to understand as our expert teacher has designed them to convey a proper and complete meaning of each topic and equations of the chapter.

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CBSE Class 11 Physics Chapter 1 Physical World NCERT Solutions

The world that we see, feel, and experience around us is the physical world. In this NCERT Solutions for Class 11 Physics Chapter 1, we are going to discuss what is physical world from the viewpoint of physics. Moreover, this NCERT Solution clear all your doubts and queries related to this chapter. Also, it explains each and every topic of the chapter in detail.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 1

1.1 what is physics.

Physics refers to that branch of science that is related to the nature and properties of energy and substance. Likewise, its subject matter includes mechanics, sound, gravity, electricity, heat, light and radiation, magnetism, and the structure of atoms. Besides, this topic discusses how this science work and its origin. And various terms related to physics are discussed in the topic.

• Natural science- the science that relates to nature.

1.2 Scope And Excitement Of Physics

This topic describes how interesting physics is and its scope. Physics is fun or exciting because you can learn the concepts by performing them. Moreover, the scope of physics is very wide that the whole universe comes in it. In addition, the topic describes the scope of physics in gravity, motion, electricity, light, radiation, heat and many others.

1.3 Physics, Technology, and Society

This topic overviews the various scientist and philosopher who contributed to the development of physics and technology. Likewise, the topic contains a chart that contains the name of those scientists and philosophers whom contribution in physics we can’t deny.

1.4 Fundamental Forces in Nature

This topic describes all those forces that work on every object or body. Besides, the topic has a chart that contains the link between physics and technology.

1.4.1 Gravitational Force- It refers to the force that attracts an object towards the earth.

1.4.2 Electromagnetic Force- It refers to the force that every object produce which attract or repel other objects.

1.4.3 Strong Nuclear Force- It refers to the force that binds the protons and electrons with the nucleus of the atom.

1.4.4 Weak Nuclear Force- It refers to that force, which is not strong enough to bind the protons and electrons in an element. Likewise, it only appears in β-decay of the nucleus.

1.4.5 Towards Unification of Forces- This topic defines how various scientists grouped different forces of nature among a common force.

1.5 Nature of Physical Laws

This topic talks about the nature of laws of physics and how different scientist theories are categorized in one group of category. Besides, this topic contains a chart that shows the unification of different forces.

You can download NCERT Solutions for Class 11 Physics Chapter 1 PDF by clicking on the button below.

Solved Questions for You

Question 1: Physics involves the study of the

• Birds and animals
• Nature and natural phenomena

Answer :  Physics is a branch of science which deals with the study of nature and natural phenomena.

Question 2: Which year was declared as International Year of Physics. ?

Answer: The year 2005 was declared as the International Year of Physics.

Question 3: Lightning was discovered by the

Answer: Lightning was discovered by Franklin.

Question 4: What is the full form of GMRT?

• Ground Mobile Receive Terminal
• Geometric Mean Reciprocal Titer
• Giant Metrewave Radio Telescope
• General Maintenance and Repair Technician

Answer: Giant Metrewave Radio Telescope (GMRT).

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NCERT Solutions for Class 11

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NCERT Solutions for Class 11 Physics

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From crucial exams to getting ahead, Class 11 Physics is super important. Our NCERT 11 Physics solutions are online or PDFs you can download, letting you learn your way. Topics cover a lot, so you pick what to study. Whether it’s motion, thermodynamics, or laws of motion, our NCERT 11 Physics solution has your back.

Physics, all about matter, is key for science. Class 11 shapes your future. Our NCERT Physics Class 11 solutions make it simple, reducing stress and boosting understanding. Revise easily with PDF solutions. Be great at concepts for higher classes and tough exams.

Our curated NCERT Physics Class 11 solutions , made by experts, guide you to understand. Access chapters now, download the whole thing, or just a chapter. Our teachers help with problem-solving, showing steps. Key questions and tips help you learn smoothly.

Know Physics, from basics to formulas. Our NCERT Physics Class 11 PDF solutions help you stride into your final school year and beyond with confidence. Boost your skills, tackle the trickiest stuff, and learn more with Infinity Learn’s NCERT Solutions for Class 11 Physics today!

• Chapter 1: Physical World
• Chapter 2: Units and Measurements
• Chapter 3: Motion in a Straight Line
• Chapter 4: Motion in a Plane
• Chapter 5: Laws of Motion
• Chapter 6: Work, Energy, and Power
• Chapter 7: System of Particles and Rotational Motion
• Chapter 8: Gravitation
• Chapter 9: Mechanical Properties of Solids
• Chapter 10: Mechanical Properties of Fluids
• Chapter 11: Thermal Properties of Matter
• Chapter 12: Thermodynamics
• Chapter 13: Kinetic Theory
• Chapter 14: Oscillations
• Chapter 15: Waves

NCERT Solutions Class 11 Physics Chapter Details and Exercises

Ncert solutions for class 11 physics chapter 1: physical world.

We’ll look at how things work and why they work the way they do in this Chapter. You’ll learn more about the world around y and the beginnings and history of science. You’ll also find various aphorisms and their reasons in this Chapter, all of which use Physics concepts in real-life situations. The strong and weak nuclear and gravitational, and electromagnetic forces are described. Listed below are some of the numerous scientists’ theories, observations, and conclusions. You will have a more profound knowledge of the factors influencing our country’s scientific and technological progress. In order to equip students with basic conceptual knowledge, this Chapter expands on the existence of electrons.

The following topics were covered:

Physics-scope and excitement; nature of physical laws; Physics, technology, and society.

Also, access the following resources for Class 11 Physics Chapter 1 The Solid State at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 1 – Introduction
• Class 11 Notes Physics Chapter 1 Physical World

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

Scientists use their senses, such as their ears and eyes, to collect data and make observations. Some are straightforward, such as determining color and texture, whereas others are more difficult and require measurement. It is a basic scientific concept without which scientists would be unable to conduct research. Students will be taught the units of physical quantities as well as how to evaluate them. It will also give you a good idea of the types of errors that can occur when measuring things and large amounts.

Units of measurement; systems of units, SI units, and fundamental and derived units are all required for measurement. Measurements of length, mass, and time; measuring device accuracy and precision; measurement errors; important figures

Dimensions of physical quantities, dimensional analysis, and its applications.

Also, access the following resources for Class 11 Physics Chapter 2 Units and Measurements at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 2 – Units and Measurement
• Units and Measurement Class 11 Notes Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

This Chapter will go over crucial topics like comparing items as point objects and how to draw graphs and calculate values from them. You may easily Learn how to plot an x-t motion graph using the NCERT Solutions from Infinity Learn. By consulting these answers, you can quickly determine how long it takes a bike to travel, how fast a car goes, and how long it takes a bus to travel. Students will Learn the distinction between the magnitude of displacement and the total length of the path traveled. According to CBSE requirements, they will also study average speed, average velocity, instantaneous speed, and velocity.

Motion along a straight line, frame of reference: Speed and velocity, as well as a position-time graph.

Differentiation and integration are covered for characterizing uniform and non-uniform motion, average speed, instantaneous velocity, evenly accelerated motion, velocity-time, and position-time graphs. For uniformly accelerated motion, the following equations apply (graphical treatment).

Also, access the following resources for Class 11 Physics Chapter 3 Motion in a Straight Line at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 3 – Motion in a Straight Line

NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane

This Chapter will go over crucial topics like comparing items as point objects and how to draw graphs and calculate values from them. You may easily Learn how to plot an x-t motion graph using NCERT Solutions from Infinity LearnBy consulting these answers; you can quickly determine how long it takes a bike to travel, how fast a car goes, and how long it takes a bus to travel. Students will Learn the distinction between the magnitude of displacement and the total length of the path traveled according to CBSE requirements; they will also study average speed, average velocity, instantaneous speed, and velocity.

Position and displacement vectors; general vectors and associated notations; equality of vectors; multiplication of vectors by an actual number; Resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors; addition and subtraction of vectors, relative velocity, Unit vector;

Cases of uniform velocity and uniform acceleration in-plane motion homogeneous circular motion, projectile motion

Also, access the following resources for Class 11 Physics Chapter 4 Motion in a Plane at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 4 – Motion in a Plane
• Motion in a Plane Class 11 Notes Physics Chapter 4

NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

This Chapter is critical in mechanics. Usually, problems based on conservation of momentum are asked on the exam. After reading this Chapter, students will understand the laws of motion and how they apply to our daily lives. To assist students in preparing for exams, this Chapter solves various numerical problems based on mechanics. Students with a good comprehension of these concepts will be able to excel in their higher education levels.

Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion.

The law of linear momentum conservation and its applications.

Concurrent forces equilibrium, static and kinetic friction, friction laws, rolling friction, and lubrication

Uniform circular motion dynamics: centripetal force, circular motion examples (vehicle on a level circular road, vehicle on a banked road).

Also, access the following resources for Class 11 Physics Chapter 5 Laws of Motion at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 5 – Law of Motion

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy, and Power

The Work-Energy Theorem is an important concept in this Chapter. Because this Chapter will be studied again in future sessions, it is critical that students memorize it completely. To ensure that students comprehend the ideas, the labour that a person performs, and the energy required to execute that work is quickly explained with appropriate examples. The CBSE board’s mark weightage is used to answer problems involving determining energy and power in a step-by-step manner. They will have no trouble learning the Chapter if they apply these ideas regularly.

The following subjects were discussed:

Kinetic energy, work-energy theorem, and power are terms used to describe the work done by a constant force and a variable force.

Non-conservative forces: motion in a vertical circle; elastic and inelastic collisions in one and two dimensions; conservative forces: conservation of mechanical energy (kinetic and potential energies).

Also, access the following resources for Class 11 Physics Chapter 6 Work, Energy, and Power at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 6 – Work, Energy and Power
• Work, Energy and Power Class 11 Notes Physics Chapter 6

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion

This Chapter will help you understand how extended bodies, such as a particle system, move. The central mass of a system of particles is the main topic of this Chapter. The concept of motion and its utility are briefly defined in the CBSE board’s syllabus. Students are strongly urged to use excellent study materials when answering the textbook’s practice questions because this Chapter covers many topics.

Momentum conservation, centre of mass motion, and the centre of mass of a two-particle system A rigid body’s centre of mass; a uniform rod’s centre of mass. Force moment, torque, angular momentum, the law of conservation of angular momentum, and its applications

Rigid body equilibrium, rigid body rotation, rotational motion equations, and comparison of linear and rotational motions

Inertia moment, gyration radius, and moment of inertia values for simple geometrical objects (no derivation). Theorems of parallel and perpendicular axes and their applications are stated.

Also, access the following resources for Class 11 Physics Chapter 7 System of Particles and Rotational Motion at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 7 – Systems of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

All material objects are attracted to the earth, as we already know. Things we throw up end up on the ground. It is more difficult to climb a hill than it is to descend one. Gravitation is a fairly common idea and anunderpins most of the concepts we’ll be studying in the future. In order to understand the more advanced issues in this Chapter, students must first understand the difference between gravity and gravitation. Other concepts explored in this Chapter include the potential energy differential between two points, acceleration due to gravity, and interplanetary motion.

The universal law of gravitation, Kepler’s rules of planetary motion. Gravitational acceleration and how it varies with altitude and depth.

Geo-stationary satellites, gravitational potential energy, gravitational potential escape velocity, orbital velocity of a spacecraft

Also, access the following resources for Class 11 Physics Chapter 8 Gravitation at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 8 – Gravitation
• Gravitation Class 11 Notes Physics Chapter 8

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

All material objects are attracted to the earth, as we already know. Things we throw up end up on the ground. It is more difficult to climb a hill than it is to descend one. Gravitation is a fairly common idea and underpins most of the concepts we’ll be studying in the future. To understand the more advanced issues in this Chapter, students must first understand the difference between gravity and gravitation. Other concepts explored in this Chapter include the potential energy differential between two points, acceleration due to gravity, and interplanetary motion.

Geostationary satellites, gravitational potential energy, gravitational potential escape velocity, orbital velocity of a spacecraft

Also, access the following resources for Class 11 Physics Chapter 9 Mechanical Properties of Solids at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 9 – Mechanical Properties of Solids

NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

In this part, we’ll look at the physical properties of liquids and gases. Fluids are defined as substances that can flow. This is an important distinction between solids and liquids, and gases. Other important concepts covered are Bernoulli’s principle, Reynold’s number, streamline flow, viscosity, and surface tension. The ability to flow is the most basic characteristic of a fluid. It cannot change its shape. Students will have a good understanding of fluid mechanical properties, which will aid their performance on the final exam.

Pascal’s l, its applications (hydraulic lift and hydraulic brakes), and the influence of gravity on fluid pressure are all discussed.

Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem, and applications, Surface energy, and surface tension, angle of contact, excess pressure across a curved surface, application of surface tension ideas to drops, bubbles, and capillary rise.

Also, access the following resources for Class 11 Physics Chapter 10 Mechanical Properties of Fluids at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 10 – Mechanical Properties of Fluids
• Mechanical Properties of Fluid | Class 11 Physics

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

We’re all familiar with the concepts of heat and temperature. The procedure for determining the temperature is covered in this Chapter. Students will also obtain a solid comprehension of Newton’s Law of Cooling, which is an important concept for many competitive assessments. Temperature is the hotness of a body that can be measured with a thermometer. This Chapter walks students through the various thermal properties of matter in a step-by-step format to help them grasp the concepts.

Heat, temperature, and thermal expansion of solids, liquids, and gases, as well as anomalous expansion of water; specific heat capacity. Change of state – latent heat capacity; Cp, Cv – calorimetry

Heat transfer-conduction, convection radiation, thermal conductivity, qualitative conceptions of Blackbody radiation, Wein’s displacement law, Stefan’s law, and the Greenhouse effect are alf heat transfer-conduction.

Also, access the following resources for Class 11 Physics Chapter 11 Thermal Properties of Matter at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 11 – Thermal Properties of Matter
• Thermal Properties Of Matter Questions for CBSE Class 11th

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

The study of heat and temperature, as well as their conversion to various forms of energy, is known as thermodynamics. Most exam questions are located in this Chapter, one of the most popular themes among question paper writers. In order to get great grades, students must understand all of these principles. This Chapter discusses, among other things, thermodynamic rules, specific heat capacity, various thermodynamic processes, and the Carnot engine.

The zeroth law of thermodynamics, thermal equilibrium, heat, work, and internal energy are discussed—isothermal and adiabatic processes, as well as the first law of thermodynamics.

Reversible and irreversible processes, heat engines, and refrigerators are all examples of the second law of thermodynamics.

Also, access the following resources for Class 11 Physics Chapter 12 Thermodynamics at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 12 – Thermodynamics
• Thermodynamics Questions for CBSE Class 11th

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

This Chapter is more important when it comes to exam preparation. Unlike solids and liquids, the properties of gases are easy to comprehend. The molecular nature of matter, the behaviour of gases, the kinetic theory of an ideal gas, the law of energy equipartition, specific heat capacity, and the mean free path are all covered in this Chapter. The concerns are also addressed systematically and CBSE-compliantly, making it easy for Class 11 students to earn decent scores.

Work done in compressing a gas, equation of state of a perfect gas.

Assumptions and the concept of pressure in the kinetic theory of gases. Degrees of freedom, the law of equipartition of energy (statement only) and application to specific heat capacities of gases; the concept of mean free path, Avogadro’s number; kinetic interpretation of temperature; RMS speed of gas molecules; degrees of freedom, the law of equipartition of energy (statement only) and application to specific heat capacities of gases; the concept of mean free path, Avogadro’s number

Also, access the following resources for Class 11 Physics Chapter 13 Kinetic Theory at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 13 – Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Oscillations are a basic physics concept. Because the principles provided in this Chapter will be revisited in the following lectures, it is vital that students comprehend them. This Chapter covers periodic and oscillatory motions, basic harmonic motion, uniform circular motion, and other concepts. It is critical for students to use adequate study materials in order to master these topics. If students study the Chapter daily, they will be better prepared to face the more challenging questions on the exam.

Period, frequency, displacement as a function of time, and periodic functions are all examples of periodic motion.

Phase; oscillations of a loaded spring-restoring force and force constant; energy in S.H.M. Simple pendulum derivation of expression for its period; kinetic and potential energies Resonance, free, forced, and damped oscillations (qualitative notions only).

Also, access the following resources for Class 11 Physics Chapter 14 Oscillations at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 14 – Oscillations
• Oscillations Class 11 Notes Physics Chapter 14
• Important Topic of Physics – Oscillation

NCERT Solutions for Class 11 Physics Chapter 15 Waves

This Chapter includes in-depth questions on key subjects, including wave dynamics and other relevant topics. This Chapter’s questions will walk you through concepts like wave types, sound speed in the air, string tension, and sound speed in air dependence. Important formulas and concepts are highlighted in the solutions to make it easier for students to revise them. Numerous wave characteristics are discussed in this Chapter, including amplitude, frequency, phase, wavelength, wave displacement, and resonance.

Transverse and longitudinal waves, traveling wave speed, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect

Also, access the following resources for Class 11 Physics Chapter 15 Waves at Infinity Learn

• NCERT Exemplar Solutions for Class 11 Physics Chapter 15 – Waves

Because Infinity Learn is a learning organization, we understand the significance of study materials and how they influence academic performance. As a result, we developed Class 11 NCERT Solutions, a learning resource aimed at helping students study and achieve their objectives. Because each Chapter covers a huge number of topics, students must study them frequently to ensure that they understand them. It will help students understand how to approach difficult questions on the final test in the right way. NCERT Class 11 Books might help students better understand the syllabus.

CBSE Marking Scheme 2023-24

The CBSE board has thoughtfully organized the academic sessions into well-defined segments, strategically covering the entire syllabus. This meticulous approach is designed to empower students with effective learning experiences. The subject matter experts have ingeniously connected various topics and concepts, establishing a strong foundation for the learners. This interconnectedness not only aids in comprehension but also highlights the beautiful harmony within the subject.

The culmination of these sessions is the eagerly awaited board examinations, masterminded by the CBSE board itself. These examinations, held at the culmination of the academic year, serve as the ultimate assessment of the student’s grasp of the subjects. These exams are structured in accordance with the thoughtfully divided syllabus, ensuring that students are evaluated on the complete spectrum of what they have learned. This approach adds an extra layer of clarity and coherence to the entire academic process.

The significance of this approach lies in its ability to enhance the overall quality of education. By conducting these comprehensive assessments, students are encouraged to engage more profoundly with the subjects throughout the year. The anticipation of the year-end examinations acts as a guiding light, propelling students toward consistent learning and preparation.

Why Should One Opt for Infinity Learn NCERT Solutions?

Opting for Infinity Learn NCERT Solutions offers a multitude of benefits that can significantly enhance your learning experience and academic performance. Here are some compelling reasons why you should consider choosing Infinity Learn’s NCERT Solutions:

• Comprehensive Coverage: Infinity Learn’s NCERT Solutions cover the entire NCERT curriculum for various subjects and classes. This ensures that you have access to accurate and detailed explanations for every topic, chapter, and concept, leaving no gaps in your understanding.
• Clarity and Simplicity: The solutions provided by Infinity Learn are designed to be clear, concise, and easy to understand. Complex concepts are broken down into simple explanations, making it easier for you to grasp and apply the knowledge effectively.
• Expertly Crafted: Infinity Learn’s NCERT Solutions are crafted by subject-matter experts who have a deep understanding of the curriculum and the intricacies of each topic. This ensures that you receive accurate and reliable information that aligns with the NCERT syllabus.
• Step-by-Step Approach: The solutions follow a step-by-step approach, guiding you through each problem or concept methodically. This approach not only helps you solve problems but also teaches you the thought process behind it, enhancing your problem-solving skills.
• Variety of Questions: Infinity Learn’s solutions cover a wide range of questions, including examples from the NCERT textbook as well as additional practice questions. This variety prepares you for different types of questions that might appear in exams.
• Time Management: By providing efficient solutions, Infinity Learn helps you save time in understanding concepts and solving problems. This gives you more time to focus on revision, practice, and exploring additional study materials.
• Visual Aids: Some solutions may include diagrams, graphs, and illustrations to enhance your understanding of certain concepts. These visual aids make learning more engaging and help in retaining information better.
• Concept Reinforcement: Infinity Learn’s solutions aim to reinforce the core concepts of each topic. This helps in building a strong foundation, which is crucial for understanding more complex concepts in the future.
• Exam Preparation: The solutions are structured in a way that prepares you for exams. They not only provide answers but also offer insights into the right approach to answering questions and highlighting key points that examiners might look for.
• Self-Assessment: Most solutions include exercises and practice questions. By attempting these questions and comparing your answers with the provided solutions, you can assess your own understanding and identify areas that need more attention.
• Convenient Accessibility: Infinity Learn’s NCERT Solutions are often available online, making them accessible anytime, anywhere. This convenience allows you to study and practice at your own pace and convenience.
• Supplement to Classroom Learning: These solutions can serve as a supplement to your classroom learning, helping you reinforce what you’ve learned in school and clarifying any doubts you might have.

In conclusion, Infinity Learn’s NCERT Solutions offer a wealth of advantages, from comprehensive coverage to expert guidance and efficient learning. They can be a valuable tool to complement your studies, enhance your understanding, and boost your academic success.

Vijayi Bhava Course | Sample Video for Class 11 Physics

Frequently Asked Questions on NCERT Solutions for Class 11 Physics

Why should i consider using ncert solutions for class 11 physics.

NCERT Solutions for Class 11 Physics serve as a valuable resource to tackle textbook problems effectively. The challenges posed by Class 11 Physics chapters can be daunting initially, given the vast syllabus and introduction to new concepts. To navigate this, students can turn to NCERT Solutions available on e-learning platforms like Infinity Learn for comprehensive clarity. These solutions play a crucial role in resolving doubts related to any question, thereby aiding students in better exam preparation. Since Class 11 Physics content holds significant weightage in competitive exams like NEET and JEE, it's vital for students to focus on mastering the subject. By opting for NCERT Solutions, students can enrich their understanding and gain the proficiency needed to solve problems accurately.

Which chapter should I begin with if I'm finding Class 11 Physics challenging?

Within the NCERT Solutions for Class 11 Physics, you'll discover 15 diverse chapters, each with its level of complexity tailored to individual students. If you're seeking a starting point, consider Chapter 1: Physical World, which offers a straightforward introduction. Additionally, Chapter 2: Units and Measurements is both manageable and essential. Infinity Learn's solutions provide clear and accessible explanations for all chapters, offering substantial support to students. Initiating your journey with these chapters can boost your confidence and provide a solid foundation for progressing through the rest of the material with greater ease.

Which chapter is important in Class 11 Physics?

In the realm of Physics, the significance of each chapter is uniform and noteworthy. Despite varying levels of complexity, these chapters collectively contribute to your academic achievement and lay the groundwork for future endeavors. Adhering to the NCERT-prescribed syllabus is essential, as it aligns with examination content. Infinity Learn’s NCERT solutions for Class 11 Physics echo this approach, furnishing students with confidence in their studies. This assurance stems from the alignment with curriculum, ensuring a holistic understanding of the subject matter.

Is Class 11 Physics Challenging?

No subject becomes challenging when approached with proper guidance and consistent practice. Within the Class 11 curriculum, Physics might present a slightly more intricate landscape compared to other subjects. It demands a firm grasp of concepts and a sturdy foundation to conquer its intricacies. Infinity Learn's Class 11 Physics solutions serve as a valuable resource for demystifying these complex topics. These solutions present the material in a clear, engaging manner, ensuring that students can readily comprehend and retain these fundamental principles for a lifetime.

Which chapter in the NCERT Solutions for Class 11 Physics is considered less challenging?

When students dedicate consistent effort to understand all concepts, no chapter becomes overly difficult. The extensive syllabus might initially seem overwhelming, but as you progress, you'll find comfort in the subject. Should any uncertainties arise, you can seek guidance from experienced faculty members at Infinity Learn. They have years of expertise in this domain. Furthermore, you can freely access the solutions PDF on the Infinity Learn website

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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements in Hindi and English Medium updated for session 2024-25 with 11th Physics chapter 1 MCQ explanation and answers. Preparing for the important topics in CBSE Class 11 Physics from the NCERT textbook requires a systematic and focused approach. Tiwari Academy provides solutions as well as notes based on chapter 1 of 11th Physics.

Class 11 Physics Chapter 1 Units and Measurements Question Answers

• Class 11 Physics Chapter 1 NCERT Solutions
• Class 11 Physics Chapter 1 in Hindi Medium
• Class 11 Physics Chapter 1 NCERT Book
• Class 11 Physics all Chapters NCERT Solutions
• Class 11 all Subjects NCERT Solutions

Light year is unit of

In scientific usage, a light-year is a unit of length that is equal to the distance light travels in one year in the vacuum of space about 5,878,000,000,000 miles. But non-scientist apply the term far more freely, using light-year for any extremely large measure of comparison-be it distance or time.

• View Answer

Physical quantities are

The quantities which can be describe the law of physics is known as the physical quantity. A physical quantity is one that can measured length, mass, time, pressure, temperature, current and resistance are the physical quantities.

Which of the following has the same dimensions?

The dimensional formula for impulse is [MLT-1] The dimensional formula for momentum is [MLT-1] which is same as the dimensional formula for impulse.

What is the value of SI unit? when the atmospheric pressure is 10⁶ dyne/cm²

Pressure = F/A Taking dimensions on both sides [P]=[MLT⁻²]/[L⁻²]= [ML⁻1T⁻²] 10⁶*[1g]/[1kg]¹ = [1cm]/[1m]⁻¹=10⁵ newton/m²

How many units there are are independent quantities. Take three physical quantities: mass, length and time. These quantities are independent of each other. Therefore, measuring all quantities requires three separate units. Therefore, it becomes important to establish a system of units. Historically, the options for each unit have changed. For example, meters were originally defined in terms of the distance between the North Pole and the equator. This distance is almost equal to 107 m. Until recently, the world standard meter was the distance between two scratches on a platinum-iridium alloy rod, which was kept at the International Bureau of Weights and Measures in France. The number of light wavelengths for the different spectral lines of the Kr-86 isotope has been specified in France.

Which one of the following is same as the dimensions of kinetic energy?

The relation between work and energy is that the work force does on an object equals the change in energy , not just kinetic energy but any form of energy. The box cannot have kinetic energy if it is stationary but it acquires kinetic energy when a force does work on it continuous to be in motion.

Electron volt is a unit of

Electron volt is the unit of energy which is commonly used in atomic and nuclear physics, equal to the energy gained by an electron. When the electrical potential at the electron increases by one volt.

In SI system the fundamental units are

Nothing to explain it because it is the universally accepted but we can explain is at telling the unit of that fundamental quantity.

What is the unit of solid angle?

A solid, omega made up of all the lines from a closed curve meeting at a vertex, is defined by the surface area of a sphere subtended by the lines and by the radius of that sphere, The dimensionless unit of solid angle is the steradian, with 4pie steradians in a sphere.

The concept of length involves the comparison of two objects. Suppose we put a rubber string on a standard meter scale. Make it 50 cm long. Now pull the rope from both ends and measure the length. Now take a new length of 60 cm. Since the final reading is greater than the initial reading, the length of the string is said to have increased. Therefore, it proves that the final reading is more than the initial reading. The next unit used to measure very long distances is the distance traveled by light.

It’s short for Parallax Seconds. It is the distance of one diagonal astronomical unit of one angular second. In other words, the distance corresponding to one second of parallax per year. The annual parallax 0 is the angle of the semi-major axis of the Earth’s orbit seen from the star perpendicular to the direction of the star. 1 Parsec = 3.26 light years

Is the NCERT solution important for Class 11 Physics Chapter 1?

When preparing for Class 11 Physics Chapter 1, it is just as important to refer to the NCERT solution as to the problem. The answers and explanations are given here in simplified manner. If students cannot answer these questions correctly, it is not enough to read the questions provided in the NCERT. 11 Physics NCERT Solutions Chapter 1 available on Tiwari Academy provides students with correct step by step solutions to all NCERT questions. In this way students do not lose any marks in the exam.

How to understand chapter 1 of the course of physics 11th class?

Chapter 1 of grade 11 NCERT Physics is called Units and Measurements. This chapter details the different units used to determine the measurement of different physical quantities, the instruments used for these measurements and their accuracy, etc. Class 11 Physics chapter 1 is the base of other all the chapters. Students can easily understand chapter 1 by engaging in regular reading of the chapter and solving the problems provided in the exercises and additional exercises. For more help, students can also check out the NCERT Solutions for Class 11 Physics Chapter 1 on website or download the Tiwari Academy app, which is available for free.

What important topics are covered in Chapter 1 of Class 11 Physics?

NCERT Class 11 Physics Solutions Chapter 1 Units and Measurements covers a variety of topics such as the International System of Units, measurement of length, mass and time, application of significant figures, etc. Of these, the most important topics in this chapter include SI units, absolute error, dimensional analysis, and significant figures. Most of the concepts given here can be understand by solving the numerical questions. Short-answer and number-based questions can be asked on the topic of evaluating errors in quantitative measurement processes.

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