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CBSE Sample Papers for Class 11 Physics with Solutions 2023-2024

Solved CBSE Sample Paper Class 11 Physics 2023-2024 with Solutions: Solving Pre Board CBSE Sample Papers for Class 11 Physics with Solutions Answers 2023-2024 Pdf Download helps to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for CBSE Class 11 Physics board exam and Score More marks. Here we have given CBSE Class 11 Physics Sample Papers 2023.

Board  – Central Board of Secondary Education (CBSE) Subject – CBSE Class 11 Physics Year of Examination – 2023-2024

CBSE Sample Paper Class 11 Physics 2023-2024 with Solutions

According to the new CBSE Exam Pattern, MCQ Questions for Class 11 Physics also carry marks in the board exam. Click on the link below to access the CBSE Class 11 Physics Sample Paper 2023 Solved.

CBSE Class 11 Physics Sample Papers 2023 with Solutions

• CBSE Sample Papers for Class 11 Physics Set 1
• CBSE Sample Papers for Class 11 Physics Set 2
• CBSE Sample Papers for Class 11 Physics Set 3
• CBSE Sample Papers for Class 11 Physics Set 4
• CBSE Sample Papers for Class 11 Physics Set 5

Physics Class 11 Sample Paper Marking Scheme Question Paper Design 2023-24

Note: The above template is only a sample. Suitable internal variations may be made for generating similar templates keeping the overall weightage to different forms of questions and typologies of questions the same.

We hope these CBSE Sample Papers for Class 11 Physics with Solutions 2023-2024 will help in self-evaluation. Stay tuned for further updates on the CBSE Sample Paper of Physics Class 11 2023 for their exam preparation.

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Class 11 Physics Sample Paper

Please refer to the Class 11 Physics Sample Paper below. These CBSE Sample Papers for Class 11 Physics have been prepared based on the latest guidelines and examination patterns issued for the current academic year. We have provided Term 1 and Term 2 sample papers with solutions. You can click on the links below and access the free latest CBSE Sample Papers for Physics for Standard 11. All guess papers for Physics Class 11 have been prepared by expert teachers. You will be able to understand the type of questions which are expected to come in the examinations and get better marks.

CBSE Sample Papers for Class 11 Physics

Class 11 Physics Sample Paper Term 2 With Solutions Set A

SECTION – A

1. The bulk modulus of elasticity of solids, liquids and gases are B s , B l , B g arrange them according to their magnitude. Give reason. Answer :  Under constant stress, magnitude of bulk modulus of elasticity of material is inversely proportional to volumetric strain. B ∝ 1 / V/ΔV Since, gases has higher compressibility than liquid and solids have the least, therefore B s > B l > B g .

2. When a stretched wire and a tuning fork are sounded together, 5 beats per second are produced, when length of wire is 95 cm or 100 cm, find frequency of the fork. Answer :  Let the frequency of tuning fork be n. Suppose ∫ 1 = 95 cm has frequency n 1 and ∫ 2  = 100 cm has frequency n 2 ( < n 1 )

OR From Doppler effect, beats, transverse waves and longitudinal waves which of the following does the intensity of sound vary with time? Answer :  Beats is the phenomenon of regular variation in the intensity of sound with time when two sources of nearly equal frequencies are sounded together. One minimum of sound followed by one maximum constitute one beat.

3. Refer to the plot of temperature versus time showing the changes in the state of ice on heating (not to scale).

What represents AB in the given figure? Answer :  AB represents ice and water are not in the thermal equilibrium.

4. The time period of a simple pendulum is T. If the length of the pendulum is made 9 times and mass of the bob is made 4 times then find the value of time period. Answer :  The time period of the simple pendulum is

  OR A body of mass 500 g is attached to a horizontal spring of spring constant 8π 2 N m –1 . If the body is pulled to a distance of 10 cm from its mean position, then find its frequency of oscillation. Answer : 

5. Why greenhouse are provided with glass roofs and glass doors? Answer :  Because glass is athermanous.

Assertion & Reason  Question :

two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false

6. Assertion (A) : The speed of sound in a gas is not affected by change in pressure provided the temperature of the gas remains constant. Reason (R): The speed of sound is inversely proportional to the square root of the density of the gas.  

7. Assertion (A): Coefficient of thermal conductivity of a metal rod is a function of length of the rod. Reason (R): Longer the rod, larger is the amount of heat conducted.   

SECTION – B

Case Study Based Question :

8. The appearance of sound to a human ear is characterised by three parameters (a) pitch (b) loudness and (c) quality. Pitch is related to frequency, loudness is related to intensity and quality is related to the waveform of the sound wave. The pitch of a sound corresponds to frequency. Frequency is an objective property which can be assigned a unique value. However a person’s perception of frequency is subjective. The brain interprets frequency primarily in terms of a subjective quality called Pitch. The loudness that we sense is related to the intensity of sound . The intensity of sound depends on pressure variations within the sound wave, it depends on the amplitude. Sound “quality” or “timbre” describes those characteristics of sound which allows the ear to distinguish sounds which have the same pitch and loudness. Intensity of sound waves : Intensity of a wave is given by I = 1/2 pA 2 ω 2 v Sound intensity in decibels : Physiological sensation of loudness is closely related to the intensity of wave. There is a wide range of intensities that the human ear can detect. At frequency f = 1000 Hz, lowest audible intensity is 10 –12 W/m 2 and intensity of 1 W/m 2 causes pain.

(i) If intensity of sound is increased by a factor of 1000, what will be increase in decibels? (a) 30 dB (b) 40 dB (c) 50 dB (d) 60 dB  

(ii) The unit of sound is (a) decible (b) hertz (c) meter (d) meter 2  

(iii) At what temperature will the speed of sound be double its value at 0°C? (a) 1090 K (b) 1010 K (c) 1082 K (d) 1092 K

(iv) When sound wave is refracted from air to water which of the following quantities remains unchanged? (a) Wavelength (b) Wave number (c) Wave velocity (d) Frequency

(v) Out of the following gases under similar condition of temperature and pressure, the velocity of sound will be maximum in (a) hydrogen (b) nitrogen (c) oxygen (d) chlorine

SECTION – C

All questions are compulsory. In case of internal choices, attempt anyone.

9. A wire of length l and radius r has a weight W and the Young’s modulus of elasticity Y. It is suspended vertically from a fixed point. Calculate the increase in length of wire produced due to its own weight. Answer :  Since the tension is non-uniform along the wire, we have variable stress in wire. Let us consider a small element of length dx at a distance x from bottom end P as shown figure.

OR The Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then what will be the ratio of the weights added to the steel and brass wires ? Answer :  

Let L and A be length and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level (i.e. same elongation is produced in both wires), let weights W s and W b be added to steel and brass wires respectively. Then By definition of Young’s modulus, the elongation produced in the steel wire is

10. An ideal gas is expanding such that PT 2 = constant. Find the coefficient of volume expansion of the gas. Answer :  

11. A sample of 0.2 g of water at 100°C and normal pressure (1.013 × 10 5 N m –2 ) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, find the change in internal energy of the sample. Answer :  Using first law of thermodynamics,

OR At constant pressure heat is supplied to a diatomic gas. Find ratio of ΔQ : ΔU : ΔW. Answer :  

12. A gas has two specific heats whereas a liquid and a solid have only one. Why ? Answer :  When solids and liquids are heated, there is only a slight change in their volume and as such they possess only one specific heat, i.e., specific heat at constant volume. But in case of gases, pressure and volume both change and as such they possess two principal specific heats; one at constant pressure and one at constant volume.

SECTION – D

All questions are compulsory. In case of internal choices, attempt any one.

13. Two bodies are combined to form a single composite body having specific heats s 1 and s 2 What is the specific heat of the composite body? Answer : Let m 1 and m 2 are the masses of two bodies. Heat capacity of the composite body

14. (a) State two conditions of a reversible process. (b) What is the quasi-static process? Answer :  (a) : The process should take place very slowly, so that it satisfies the following conditions : (i) The system should be in thermal equilibrium. (i) The system should be in chemical equilibrium. (b) A process in which the system passes through a continuous sequence of equilibrium states is said to be quasi-static process, i.e., apparently static. OR A gas is taken through the cycle A → B → C → A, as shown. What is the net work done by the gas?

Answer :  In a cyclic process, work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if the cycle is anticlockwise.

SECTION – E

15. (a) What is the phase difference between two simple harmonic motions represented by x 1 = A sin  {ωt + π/6} x 2 = Acos(ωt)? (b) The maximum velocity of particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m s –1 . Find the time period of oscillation. Answer : 

OR (a) A particle executing a simple harmonic motion has a period of 6 sec. Find the time taken by the particle to move from the mean position to half the amplitude, starting from the mean position. (b) The equation of SHM of a particle is given as 2 d 2 x/dt 2 + 32x = 0 where x is the displacement from the mean position of rest. What is the period of its oscillation in seconds ? Answer :  (a) Since the motion is started from the mean position, therefore displacement x of a particle executing SHM at any time t from its mean position is given by

16. State and prove Bernoulli’s theorem. Answer :   Bernoulli’s principle : Bernoulli’s principle states that the sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline.  Mathematically, it can be expressed as P + 1/2  pv 2  + pgh = constant Proof: Consider a non-viscous and incompressible fluid flowing steadily between the sections A and B of a pipe of varying cross-section. Let a 1 be the area of cross-section at A, v 1 the fluid velocity, P 1 the fluid pressure, and h 1 the mean height above the ground level. Let a 2 , v 2 , P 2 and h 2  be the values of the corresponding quantities at B.

Let r be the density of the fluid. As the fluid is imcompressible, so whatever mass of fluid enters the pipe at section A in time Δt, an equal mass of fluid flows out at section B in time Δt. This mass is given by

This proves Bernoulli’s principle according to which the total energy per unit volume remains constant. OR (a) State Pascal’s law of fluid pressure. Explain the working of hydraulic lift with suitable diagram. (b) Why air bubbles in a liquid moves in upward direction? Answer :  (a) Pascal’s Law : If gravity effect is neglected the pressure in a fluid at rest is same at all points.

Hydraulic lift : It is used to lift the heavy loads. Its working is based on Pascal’s law. A simple hydraulic lift is shown in figure. Here C and D are two cylinders of different areas of cross-section. They are connected to each other with a pipe E. Each cylinder is provided with airtight frictionless piston. Let a, and A be the areas of crosssection of the pistons in C and D respectively, where a < < A. The cylinders are filled with an incompressible liquid.  Let a downward force f be applied on the piston of C. Then the pressure exerted on the liquid, P = f/a. According to Pascal’s law, this pressure it transmitted equally to piston of cylinder D.  ∴ Upward force acting on the piston of cylinder  D will be, F + PA + f/a A + f A/a (b) The fluids move from higher pressure to lower pressure and a fluid pressure increases with depth. Hence  pressure at the top is less than that at the bottom and so the air bubble will rise from bottom to top.  

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CBSE Sample Papers for Class 11 Physics 2024

CBSE Sample Papers for Class 11 Physics : Solving Class 11 Physics sample papers will provide an overview of the annual exam and provide the aspirants with a variety of questions for preparation. After completing the syllabus, students must revise the sample paper. Physics sample papers will include both theory and numerical. The sample papers will include the complete curriculum of the subject.

Physics is a challenging yet interesting subject. Students can score well with constant practice of numerical and retaining and understanding the theoretical part. Solving sample papers constantly helps the students to keep a check on their progress. Note the performance in every sample paper and work on the areas that need improvement. Keep reading to know more about the CBSE Class 11 Physics sample papers. 

Latest Update: — Students can check out previous Class 11 sample question papers for Physics in PDF , along with the Marking Scheme to get an idea of 2024 paper. Check out the table added below for direct PDF links .

Class 11 Physics Sample Papers 2024

The students can access the sample papers online. The article will include five sets of physics sample papers to guide the students with a variety of questions. The sample papers cover all the units of the curriculum. The online availability of the papers increases accessibility, and all students, irrespective of their location, can download the same for preparation. Sample papers are extremely useful in the last days of preparation.

The CBSE Class 11 Physics sample papers haven’t been released by the official board yet. Once it is available, we will provide the same here. Keep a check on the page to stay updated.

What Units Do Class 11 Physics Cover?

The sample papers are prepared as per the syllabus specified by CBSE. Aspirants can solve the paper after completing the syllabus to test their progress. Questions from all the topics of the curriculum are included in the sample paper for comprehensive preparation. The students must treat the sample papers as a test and solve accordingly to know the level of preparation. Check below the Physics curriculum that will be covered in Class 11 Physics.

CBSE Class 11 Physics Curriculum

CBSE Sample Papers for Class 11 Physics: Benefits

The difficulty level of the exam increases as one reaches the higher classes. Maths and Physics are some subjects that require frequent practice. CBSE sample papers allow the students to prepare the units of the curriculum and analyze the areas of improvement. The students must download the PDF of the sample paper and attempt all the questions. Check below the benefits of CBSE sample papers for class 11 physics-

• The students can download any or all of the five sets of physics sample papers.
• The sample papers are similar to the annual exam and give an overview of the same for the understanding of the students.
• Aspirants unable to solve a question in the sample paper help them to analyze the weak areas. The students can use the result of the sample paper to know the time they should give to each topic.
• Solving sample paper within the specified time helps students to work on their speed.
• It boosts the confidence of the students.
• Students must solve as many sample papers as possible to know the frequently asked questions. 

FAQs On CBSE Sample Papers for Class 11 Physics 2024

Ans: Aspirants can access the Class 11 CBSE sample papers of Physics via online mode. 

Ans: Yes, sample papers of CBSE are of great use. Students can attempt the sample papers to know the type of questions asked in the annual exam.

Ans: Students can check the syllabus for the CBSE Class 11 exam on the official website of the Central Board of Secondary Education.

Ans: The students can download the CBSE sample papers for Physics through the links available in this article.

Ans: Yes, Aspirants can download the CBSE sample papers for Class 11 available in PDF format.

Also check:

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CBSE Sample Paper Physics Class 11 2023-24 PDF (Free Download)

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Those students who were looking for CBSE Sample Paper Physics Class 11 can now download the sample questions papers for 2023-24 from the given link here. Apart from the official sample paper of Physics class 11 CBSE 2023-24, here we have set wise question papers for Physics.

Going through these Physics papers, a student can understand the questions formatting and be prepared for their final examination. However, the CBSE doesn’t conduct the class 11 board exam, but the CBSE Sample Paper Physics Class 11 will help the students in their school level final exam preparation.

Sample Paper of Physics Class 11 CBSE 2023-24 with Solution

Practising sample paper of Physics class 11 CBSE 2023-24 with solution aid students to know what should be the right method and word count to write an answer. The solutions ease the students to solve the difficult questions that are asked in the question paper of Physics. It allows them to revise the whole Physics topic very easily.

CBSE Sample Paper Physics Sample Paper 2024 Set 1

CBSE Sample Paper Physics Sample Paper 2024 Set 2

CBSE Sample Paper Physics Sample Paper 2024 Set 3

CBSE Sample Paper Physics Sample Paper 2024 Set 4

CBSE Sample Paper Physics Sample Paper 2024 Set 5

CBSE Sample Paper Physics Sample Paper 2024 Set 6

CBSE Sample Paper Physics Sample Paper 2024 Set 7

CBSE Sample Paper Physics Sample Paper 2024 Set 8

CBSE Sample Paper Physics Sample Paper 2024 Set 9

CBSE Sample Paper Physics Sample Paper 2024 Set 10

CBSE Sample Paper Physics Sample Paper 2023 Set 1

CBSE Sample Paper Physics Sample Paper 2023 Set 2

CBSE Sample Paper Physics Sample Paper 2023 Set 3

CBSE Sample Paper Physics Sample Paper 2023 Set 4

CBSE Sample Paper Physics Sample Paper 2023 Set 5

CBSE Sample Paper Physics Sample Paper 2023 Set 6

CBSE Sample Paper Physics Sample Paper 2023 Set 7

CBSE Sample Paper Physics Sample Paper 2023 Set 8

CBSE Sample Paper Physics Sample Paper 2023 Set 9

CBSE Sample Paper Physics Sample Paper 2023 Set 10

CBSE 11 Class Physics Sample Paper (Term-1) 2021-22 Set - 1

CBSE 11 Class Physics Sample Paper (Term-1) 2021-22 Set - 2

CBSE 11 Class Physics Sample Paper (Term-1) 2021-22 Set - 3

CBSE 11 Class Physics Sample Paper (Term-1) 2021-22 Set - 4

CBSE 11 Class Physics Sample Paper (Term-1) 2021-22 Set - 5

CBSE 11 Class Physics Sample Paper 2021 Set - 1

CBSE 11 Class Physics Sample Paper2021 Set - 2

CBSE 11 Class Physics Sample Paper 2021 Set - 3

CBSE 11 Class Physics Sample Paper 2021 Set - 4

CBSE 11 Class Physics Sample Paper 2021 Set - 5

CBSE 11 Class Physics Sample Paper 2021 Set - 6

CBSE 11 Class Physics Sample Paper 2021 Set - 7

CBSE 11 Class Physics Sample Paper 2021 Set - 8

CBSE 11 Class Physics Sample Paper 2021 Set - 9

Also, practicing the CBSE Sample Paper Physics Class 11 assists students to predict and solve the questions that have a little bit of a chance to appear in the actual examination. In other words, solving class 11 Physics sample paper 2023-24 CBSE is one of the final stages of the exam preparation.

Download Sample Paper of Physics Class 11 CBSE 2023-24 in PDF (Step-by-Step)

To download sample paper of Physics class 11 CBSE 2023-24 in PDF students should be thorough with these process:

• Visit Selfstudys website
• Click on CBSE and Choose Sample Papers
• Now, click on class 11
• A new page will open where students need to click on Physics
• Once clicked the PDF file of CBSE Class 9 Physics Sample paper will open.

Having access to the sample paper of Physics class 11 CBSE 2023-24 with solution in PDF, helps in being thorough with the questions that are prepared on the basis of the latest exam pattern. Students who download sample question paper for class 11 CBSE Physics PDF from Selfstudys can easily focus on their board Self-study and exam preparation.

Importance of CBSE Sample Paper Physics Class 11

Those who don’t understand the importance of CBSE Sample Paper Physics Class 11 won’t be able to take advantage of it. However, before thinking of taking leverage of CBSE Class 11 Physics sample papers, students must have completed their CBSE Class 11 Syllabus . Once the syllabus of class 11 Physics is covered, students can refer to the CBSE Class 11 Physics sample papers. The importance of sample papers are as follows:

• It assists in gathering details about the whole question papers for the upcoming final examination.
• Helps in practicing tons of questions based on the Class 11 Physics exam pattern.
• Since the sample papers are available free it helps candidates to check how much they are prepared for the upcoming Physics exam.

Advantages of Solving Class 11 Physics Sample Paper 2023-24 CBSE

Analyse level of exam preparation.

One of the best advantages of Solving Class 11 Physics Sample Paper 2023-24 CBSE is that one can be sure about their level of exam preparation. Solving the Physics sample papers enable the students to determine how much work they have to do more and what are their weakest points to focus on more.

Be Familiar with Exam Without Appearing for them

Everyone has a fear of appearing in the annual exams but those who solve the sample paper class 11 Physics 2023-24 CBSE during the preparation process, get familiar with the exam without appearing for them. It is the greatest advantage of Solving CBSE Sample Paper Physics Class 11.

Revise the whole Syllabus of Class 11 Physics

No matter what and how much students have studied in their class 11 Physics, they have to revise them time and again to keep their learning refresh. For the revision purpose CBSE sample paper Physics class 11 is the best apart from the handwritten CBSE Class 11 Revision Notes.

Is Class 11 Physics Sample Paper Very Difficult?

Those who have been thorough with their CBSE Class 11 Physics Syllabus, prescribed textbooks, and practice tons of questions from several resources wouldn’t find Class 11 Physics Sample Paper Very Difficult. It is because their basics will be already cleared and they have an in-depth knowledge of exam pattern and marking distributions. So, students can easily answer CBSE sample papers for class 11 Physics.

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CBSE Class 11 Physics Sample Papers with Solutions 2023-24

Cbse class 11 physics sample papers pdf.

Date: 19th Jul 2024

CBSE class 11 Physics sample papers with solutions are provided here for class 11 students to analyze their preparation and improve their performance accordingly in their CBSE class 11 Physics exam 2023-24.

We have collected cbse class 11 sample papers with solutions pdf to help you check the nature and understand the exam pattern & difficulty level of the questions asked in cbse class 11 Physics exam.

CBSE class 11 marking scheme, blueprint and previous year question papers are also provided along with the cbse sample papers. The best way to prepare for cbse exam is to understand the question pattern and practice them as given in CBSE sample papers for class 11.

CBSE class 12 Physics sample papers are based on the latest cbse syllabus and questions are based on most of the essential topics from the course. When students practice sample papers, they encounter repetitive topics and questions from the previous year’s exams.

CBSE class 11 students can solve these Physics sample papers to boost their confidence level & enhance their accuracy and preparation level for Physics exam 2023-24.

Students should remember to solve these Physics sample papers in a time-bound manner, as this will help them with the real-time pressure while taking the exam. Students can also develop time-management skills by solving these Physics sample papers.

These cbse class 11 Physics sample papers with solutions will help students tremendously in their preparation for class 11 Physics exam 2023-24.

To score high marks in your CBSE class 11 Physics exam, you have to regularly practice CBSE Class 11 Question Papers , Class 11 Sample Papers , Class 11 Notes , Class 11 NCERT Books , NCERT Solutions for Class 11 , Class 11 Exemplar and Class 11 Syllabus 2023-24 .

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Download CBSE class 11 Physics sample paper PDF

CBSE class 11 Physics students can easily make use of these CBSE class 11 Physics sample paper with solutions pdf for term 1 & term 2 by downloading them from below links:

How to Solve CBSE Class 11 Physics Sample Papers?

CBSE students can solve their class 11 Physics sample papers using the following suggestions:

• Students can start by reading the full paper from beginning to end & try to recall the answers to these questions quickly.
• Select any section & start writing all the answers to that section in a sequential manner.
• If you aren’t able to recall the answer to any question then leave a blank space for them & move to the next question. If you get some time after attempting all the questions then try to attempt the questions you left earlier.
• Keep an eye on the clock all the time and work on solving the Physics sample paper accordingly.
• In the last few minutes, double-check all your answers and make corrections to any mistakes.

Benefits of Solving CBSE class 11 Physics Sample Papers

Most of the students are already aware that solving CBSE class 11 Physics sample papers is very important for their exams 2023-24.

We have listed down some of the benefits of solving CBSE class 11 Physics sample papers:

• Solving CBSE class 11 Physics sample papers will help students in understanding the nature of questions that can be expected in the CBSE exam 2023-24.
• Students can attempt these CBSE class 11 Physics sample paper by tracking the time spent on solving these sample paper and checking their efficiency level.
• Regularly solving CBSE class 11 Physics sample papers will help students in scoring high marks in class 11 Physics exam by working on their mistakes.
• Solving CBSE class 11 Physics sample papers will also increase your speed and clear your basic concepts on each of the Physics class 11 topics.
• These sample papers will help students in self-analyzing their weakness and students can give more time and emphasis to those weak topics.
• Students can solve all the sample papers and make a note of all the important questions that were asked multiple times to be well-prepared for the Physics exam 2023-24.

Preparation Tips for CBSE Class 11 Physics Exam 2023-24

CBSE class 11 students preparing for Physics exam 2023-24 can go through the following preparation tips that can be helpful to score high marks:

• Give importance and emphasize clearing the basic concepts of NCERT class 11 Physics book.
• Solve as many questions as possible in order to have a clear understanding of the formulas.
• Students should understand the exam pattern and regularly practice class 11 Physics sample papers.
• Students should have a clear knowledge of the weightage of each chapter of NCERT class 11 Physics book.
• Write down all the important formulas and practice questions based on these formulas.

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CBSE Sample Papers for Class 11 Physics with Solution 2024-25

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CBSE Sample Papers for Class 11 Physics with Solution - Free PDF Download

Class 11 and 12 are the career-building years for every student. The subjects are no longer as easy as they had experienced in the earlier standards. Physics is one such subject that requires immense dedication, diligence, and practice. Most students prefer to solve class 11 Physics sample papers in their preparation for any examination. To help them in this matter, we have prepared class 11 Physics sample paper with all the essential inputs required for any examination. We have a series of sample papers that covers all the topics included in the syllabus.

Experts at Vedantu have followed the CBSE guidelines to prepare the CBSE class 11 Physics sample papers 2024-25 . This entire set of sample questions covers all the important points included in the Physics syllabus . The experts have taken notes from different examinations that the students face while preparing these sample questions. The students can download these sample papers for class 11 Physics from our website and use them in their exam preparations.

Download CBSE Class 11 Sample Papers 2024-25 PDF

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CBSE Sample Question Paper for Class 11 Physics with Solutions

Download CBSE Sample Papers Class 11 Physics

Class 11 Physics can be quite difficult if not studied properly. We have prepared the sample paper class 11 Physics 2024-25 and have made it available on our website. The students can download these sample papers for free and practice solving them whenever they want. Moreover, the students can open these files on any system, allowing them to use their laptops and android devices for the purpose. The students can now practice these problems, even before the examination, so that they can grow their confidence in this subject.

How Solving CBSE Sample Papers Class 11 Physics Can Help?

The CBSE sample papers for class 11 physics come with a lot of benefits for the students. Some of them are:

Growth in Confidence

Since Physics is a very tricky subject, most students fear this subject. Therefore, the Physics sample paper for class 11 can help the students to grow confidence. They can practice solving several papers and get an idea about how to score big in this examination.

Physics Have Numerical Questions That Require Practice

A large section of the Physics question paper will comprise several numerical questions related to various chapters included in the syllabus. The best way to prepare for these numerical questions is to practice solving them as much as possible. The students can practice such acts with the help of such sample papers in class 11 Physics.

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FAQs on CBSE Sample Papers for Class 11 Physics with Solution 2024-25

1. How can I Prepare for the Numerical Questions for the Class 11 Physics Examination?

You can prepare for the numerical questions by solving as much as you can from the class 11th Physics sample paper 2024-25.

2.  How can I be Confident About High Marks in the Class 11 Physics Examination?

If you wish to grow your confidence about the subject and its related examination, practice solving many sample questions on the topics.

3. How can I get Class 11 Physics Sample Papers for Free?

You can get the Physics class 11 sample paper 2024-25 for free on our website.

4. Why do students look for Class 11 Physics sample papers to solve?

Students need a lot of practice for a subject like Physics and sample papers are the right way to proceed. Physics has a lot of complex concepts that need utmost clarification for the purpose of understanding.  Solving these papers helps them evaluate their personal understanding of the chapters.  They also learn proper time management after having solved these papers. Completing the assessments on time becomes so much easier then.  These sample papers also familiarize them with the format of the question papers.

5. Where can I find sample papers for  CBSE Class 11 physics?

You can go to Vedantu and look for CBSE Class 11 Physics Sample Papers These papers have the solutions worked out and are the right guide before examinations. They have all the questions that might come for examinations and use the most strategic approach in solving them.  They will also help the students revise prior to sitting for competitive and engineering examinations. One must solve all the questions that are present so as to perform well in the tests.

6. How do I understand all concepts related to Class 11 physics for CBSE boards?

You can check out CBSE Class 11 Physics Sample Papers available on Vedantu. These papers are designed keeping the CBSE guidelines in mind and are great for reference. All the vital concepts that need proper comprehension are present here and so, one must go through each of these questions before sitting for tests. Making notes after going through these will also help as notes help one perform much better in all examinations. Make sure that you truly understand the concepts instead of just mugging them up.

7. How do I download CBSE Class 11 Physics Sample Papers from Vedantu?

You can go to Vedantu and then click on CBSE Class 11 Physics Sample Papers. The download PDF button will appear on the page if you scroll down a bit. Click on that to download the PDF format of the solved papers. This will ensure that you have access to these mock papers regardless of whether you have an Internet connection or not. The process is simple and convenient for all students. Vedantu has all the relevant study material to choose from.

8. How do I make notes for CBSE Class  Physics?

You can make proper notes by scanning the mock papers of CBSE Class 11 Physics. Making notes before studying will help you in deciphering the concepts better. You can refer to the papers and then understand which are the important portions and which portions are not so important. Make sure that you jot down all those questions that are repetitive.  This is an effective manner of preparing for your tests as it will help you cover all the parts of the chapters.

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CBSE Sample Papers Class 11 Physics 2024

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CBSE Sample Papers Class 11 Physics 2023-24

CBSE Sample Papers Class 11 Physics myCBSEguide provides CBSE Class 11 Sample Papers of Physics for the year 2023-24 with solutions in PDF format for free download. The CBSE Sample Papers for all – NCERT books and based on CBSE latest syllabus must be downloaded and practiced by students. Class 11 Physics New Sample Papers follow the blueprint of that year only. A student must check the latest syllabus and marking scheme. Sample papers for Class 11 Physics and other subjects are available for download as PDF in-app too. myCBSEguide provides sample papers with solutions for the year 2024.

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Class 11 – Physics Sample Paper (2023-24)

Maximum Marks: 70 Time Allowed: : 3 hours

General Instructions:

• There are 33 questions in all. All questions are compulsory.
• This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.
• Section A contains sixteen questions, twelve MCQ and four Assertion Reasoning based of 1 mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains two case study-based questions of four marks each and Section E contains three long answer questions of five marks each.
• There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.
• Use of calculators is not allowed.

Physics Sample Paper Section A

• Young’s modulus of steel is 1.9 {tex}\times{/tex}  10 11 N/m 2 . When expressed in CGS units of dynes/cm 2 , it will be equal to (1N = 10 5 dyne, 1m 2 = 10 4 cm 2 ) a) 1.9 {tex}\times{/tex}  10 10 b) 1.9 {tex}\times{/tex}  10 11 c) 1.9 {tex}\times{/tex}  10 12 d) 1.9 {tex}\times{/tex}  10 13
• The human ear is most sensitive to sound in the frequency range of: a) 1000 to 2000 Hz b) 200 to 400 Hz c) 10000 to 20000 Hz d) 20 to 20000 Hz
• Centre of gravity can be defined: a) as that point where the total gravitational torque on the body is greater than zero b) as the center of mass c) as that point where the total gravitational torque on the body is zero d) as that point where the total gravitational force on the body is zero
• What will be the height of a liquid column in a capillary tube on the surface of the moon? a) 1/6th of what was on earth’s surface b) Six times that on the earth’s surface c) It will remain unchanged d) 36 times that on earth’s surface
• A source of sound gives 5 beats per second when sounded with another source of frequency 100 second -1 . The second harmonic of the source, together with a source of frequency 205 sec -1 gives 5 beats per second. What is the frequency of the source? a) 100 second -1 b) 105 second -1 c) 205 second -1 d) 95 second -1
• Motion along a straight line is called __________. a) parabolic motion b) circular motion c) oscillatory motion d) rectilinear motion
• A transverse wave is represented by y = sin ( {tex}\omega{/tex} t – kx) For what value of the wavelength is the wave velocity equal to the maximum particle velocity? a) {tex}\frac{\pi A}{2}{/tex} b) {tex}\pi A{/tex} c) 2 {tex}\pi{/tex} A d) A
• Pressures inside two soap bubbles are 1.01 atm and 1.03 atm, ratio between their volumes is a) 3 : 1 b) none of these c) 27 : 1 d) 127 : 101
• it is closer to the earth and also it has the inactive inert gases in it.
• it is too far from the sun and has very low pressure in its outer surface.
• escape velocity of gas molecules is greater than their root mean square velocity.
• escape velocity of gas molecules is less than their root mean square velocity.
• Two rings of radii R and nR made from the same wire have the ratio of moments of inertia about an axis passing through their centre equal to 1 : 8. The value of n is a) 2 b) 4 c) {tex}2 \sqrt{2}{/tex} d) {tex}\frac{1}{2}{/tex}
• When water is heated from 0°C to 10°C its volume a) increases continuously b) first decreases and then increases c) decreases continuously d) first increases and then decreases
• Assertion: n small balls each of mass m colliding plastically each second on surface which velocity u. The force experienced by the surface is 2mnu. Reason: On elastic collision, the ball rebounds with the same velocity. a) Assertion and reason both are correct statements and reason is correct explanation for assertion. b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. c) Assertion is correct statement but reason is wrong statement. d) Assertion is wrong statement but reason is correct statement.
• Assertion: The value of AQ is always zero in adiabatic process. Reason: Adiabatic process is always a cyclic process. a) Assertion and reason both are correct statements and reason is correct explanation for assertion. b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. c) Assertion is correct statement but reason is wrong statement. d) Assertion is wrong statement but reason is correct statement.
• Assertion (A): If earth were a hollow sphere, gravitational field intensity at any point inside the earth would be zero. Reason (R): Net force on a body inside the sphere is zero. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true.
• Assertion (A): In projectile motion, the angle between the instantaneous velocity and acceleration at the highest point is 180°. Reason (R): At the highest point, velocity of projectile will be in horizontal direction only. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true.

Physics Sample Paper Section B

• At what temperature (in °C) will the speed of sound in air be 3 times its value at 0°C?
• In the relation p = {tex}( a / b ) e ^ { – ( a z / \theta )}{/tex} , p is the pressure, z is the distance, and {tex}\theta{/tex} is the temperature. What is the dimensional formula of b?
• A gas bubble, from an explosion underwater, oscillates with a period T proportional to p a d b E c , where p is the static pressure, d is the density of water and E is the total energy of the explosion. Find the values of a, b and c.

Physics Sample Paper Section C

• Eight rain drops of radius 1 mm each falling down with terminal velocity of 5 cms -1 coalesce to form a bigger drop. Find the terminal velocity of the bigger drop.
• Develop a relation between the co-efficient of linear expansion, co-efficient superficial expansion and coefficient of cubical expansion of a solid.
• A parachutist bails out from an aeroplane and after dropping through a distance of 40 m, he opens the parachute and decelerates at 2 ms -2 . If he reaches the ground with a speed of 2 ms -1 , how long is he in the air? At what height did he bailout from the plane?
• 500 J of work is done on a gas to reduce its volume by compression adiabatically. What is the change in internal energy of the gas?
• The coolant in a chemical or a nuclear plant, i.e. the liquid used to prevent the different parts of a plant from getting too hot should have high specific heat.
• The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Physics Sample Paper Section D

• If stretch in spring of force constant k is doubled, then the ratio of final to initial forces is: a) 4:1 b) 1:4 c) 2:1 d) 1:2
• A light body and a heavy body have the same kinetic energy. which one has greater linear momentum? a) light body b) both heavy and light body c) none of these d) heavy body
• A spring is cut into two equal halves. How is the spring constant of each half affected? a) becomes double b) none of these c) becomes 1/4th d) becomes half

When spring is compressed, its potential energy:

• What type of energy is stored in the spring of a watch? a) potential energy b) none of these c) mechanical energy d) kinetic energy
• If the temperature of the gas increases from 300 K to 600 K then the average kinetic energy becomes: a) same b) becomes double c) becomes half d) none of these
• What is the average velocity of the molecules of an ideal gas? a) Infinite b) Same c) None of these d) Zero
• Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will ________. a) decrease b) none of these c) increase d) remains same
• Find the ratio of average kinetic energy per molecule of Oxygen and Hydrogen: a) 1:1 b) 4:1 c) 1:2 d) 1:4

The velocities of the three molecules are 3v, 4v, and 5v. calculate their root mean square velocity?

• Will there be any change in weight of the body, during the oscillation? Figure In extensible string.
• If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?
• How long does the marble take to reach the floor?
• At what horizontal distance from the edge of the table does the marble land?
• What is its velocity as it strikes the floor?
• What is the angle between {tex}\overrightarrow A{/tex} and {tex}\overrightarrow B{/tex} if {tex}\overrightarrow A{/tex} and {tex}\overrightarrow B{/tex} denote the adjacent sides of a parallelogram drawn form a point and the area of the parallelogram is {tex}\frac { 1 } { 2 } A B{/tex} ?
• State and prove triangular law of vector addition.
• Find the torque t acting on the particle at a time t when it is at point Q with respect to O.
• Find the angular momentum L of the particle about O at this time t.
• Show that  {tex}\tau=\frac{d L}{d t}{/tex}  in this example.

Class 11 – Physics Sample Paper Solution (2023-24)

You are getting not only the question paper but also a complete solution with a step-by-step explanation. The solution of the question paper is based on the latest guidelines issued by CBSE for the session 2023-24.

To get more model papers with solutions, you can download the myCBSEguide app or visit https://mycbseguide.com/dashboard/ . CBSE has reduced the course content for class 11 Physics. This question paper and solution is based on the updated and reduced syllabus of CBSE for 2023-24.

Class 11 Physics Sample Paper Solution Section A

• (c) 1.9 {tex}\times{/tex}  10 12 Explanation: According to the problem, Young’s modulus, Y = 1.9  {tex}\times{/tex}  10 11 N/m 2 1N in SI system of units = 10 5 dyne in C.G.S system. Hence, Y = 1.9  {tex}\times{/tex}  10 11   {tex}\times{/tex}  10 5 dyne/m 2 In C.G.S length is measured in unit cm, so we should also convert m into cm. {tex}\therefore{/tex}  Y =  {tex}1.9 \times 10^{11}\left(\frac{10^{5} \mathrm{dyne}}{10^{4} \mathrm{cm}^{2}}\right){/tex}  [ {tex}\because{/tex}  1m = 100 cm] = 1.9  {tex}\times{/tex}  10 12 dyne/cm 2
• (d) 20 to 20000 Hz Explanation: Humans are most sensitive to (i.e. able to discern at lowest intensity) frequencies between 2,000 and 5,000 Hz. The human ear can respond to minute pressure variations in the air if they are in the audible frequency range, roughly 20 Hz – 20 kHz. This incredible sensitivity is enhanced by an effective amplification of the sound signal by the outer and middle ear structures.
• (c) as that point where the total gravitational torque on the body is zero Explanation: The point at which the entire weight of a body may be thought of as centered so that if supported at this point the body would balance perfectly, so it can also be defined as that point where the total gravitational torque on the body is zero.
• (b) Six times that on the earth’s surface Explanation: The height to which the liquid rises in a capillary tube is inversion all proportional to the acceleration due to gravity. {tex}h=\frac{2 S \cos \theta}{r \rho g}{/tex} thus if all other parameters are fixed {tex}h \propto\frac{1}{g}{/tex} On the surface of the moon, acceleration due to gravity is  {tex}1\over6{/tex} th as that on the surface of the earth. Therefore the water will rise to a height 6 times that on earth’s surface.
• (b) 105 second -1 Explanation: Frequency of the source = 100 {tex}\pm{/tex}  5 = 105 or 95 Hz The second harmonic of the source = 210 Hz or 190 Hz As the second harmonic gives 5 beats/sec with a sound of frequency 205 Hz, the second harmonic should be 210 Hz. {tex}\therefore{/tex}  Frequency of the source = 105 Hz
• (d) rectilinear motion Explanation: Rectilinear motion is another name for straight-line motion. This type of motion describes the movement of a particle or a body. A body is said to experience rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.
• (c) 2 {tex}\pi{/tex} A Explanation: Maximum particle velocity = Wave velocity {tex}\omega A=\frac{\omega}{k}{/tex} or  {tex}k=\frac{2 \pi}{\lambda}=\frac{1}{A}{/tex} {tex}\therefore \lambda{/tex} = 2 {tex}\pi{/tex} A
• (c) 27 : 1 Explanation: Excess pressures in the two bubbles will be (1.01 -1) or 0.01 atm and (1.03 – 1) or 0.03 atm. {tex}\frac{p_{1}}{p_{2}}=\frac{r_{2}}{r_{1}}{/tex} {tex}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{p_{2}}{p_{1}}=\frac{0.03}{0.01}{/tex}  = 3 {tex}\therefore \frac{V_{1}}{V_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{3}{/tex}  = 3 3 = 27 : 1
• (d) Option iv Explanation: The required escape velocity of gas molecules is less than their root mean square velocity.
• (a) 2 Explanation: As radius of second ring is n times, length and hence mass of wire used is also n times. {tex}\therefore \frac{I_{1}}{I_{2}}=\frac{M R^{2}}{n M(n R)^{2}}=\frac{1}{n^{3}}=\frac{1}{8}{/tex}   {tex}\therefore n=2{/tex}
• (b) first decreases and then increases Explanation: Density of water is maximum at 4 o C. So during heating from 0 o C to 10 o C, its density increases at first and then it decreases resulting in the decrease in volume at first and then increase in its volume.
• (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. Explanation: In elastic collision, kinetic energy remains conserved therefore the ball rebounds with the same velocity. According to Newton’s second law. F {tex}\times{/tex} t = change in linear momentum. {tex}\therefore{/tex}  F  {tex}\times{/tex}  1 = m  {tex}\times{/tex} n(u + u) or F = 2mnu
• (c) Assertion is correct statement but reason is wrong statement. Explanation: Assertion is correct statement but reason is wrong statement.
• (d) A is false but R is true. Explanation: At the highest point, the instantaneous velocity is acting horizontally and acceleration of projectile (acceleration due to gravity) is acting vertically downward. Therefore, angle between velocity and acceleration at the highest point is 90°.

Class 11 Physics Sample Paper Solution Section B

• We know that, speed of sound, v  {tex} \propto \sqrt{T} {/tex} , T being absolute temperature.​​​ where, T is in kelvin. {tex}\therefore \frac{V_{t}}{V_{0}}=\sqrt{\frac{273+t}{273+0}} {/tex}  = 3 (say, the required temperature is t℃) {tex} \Rightarrow \frac{273+t}{273} {/tex}  = 9 {tex}\Rightarrow{/tex}  t = (9  {tex}\times{/tex}  273) – 273 = 2184°C
• Since,  {tex}e ^ { – ( a z / \theta ) }{/tex}  is dimensionless (exponential function), we have  {tex}a z / \theta = 1{/tex} or,  {tex}a = \frac { \theta } { Z } = \frac { K } { L } = \left[ \mathrm { L } ^ { – 1 } \mathrm { K } \right]{/tex} We find that a/b = dimensions of pressure (because {tex}e ^ { – ( a z / \theta ) }{/tex}  is dimensionless) a/b = [ML -1 T -2 ] Therefore, {tex}b = \frac { a } { \left[ \mathrm { ML } ^ { – 1 } \mathrm { T } ^ { – 2 } \right] } = \frac { \left[ \mathrm { L } ^ { – 1 } \mathrm { K } \right] } { \left[ \mathrm { ML } ^ { – 1 } \mathrm { T } ^ { – 2 } \right] }{/tex} = [M -1 T 2 K]
• Let T = K p a d b E c , where K = a dimensionless constant. Putting the dimensions of various quantities, T = [ML -1 T -2 ] a [ML -3 ] b [ML 2 T -2 ] c or M 0 L 0 T = M a +b + c L ~a ~3b + 2c T -2a – 2c Equating the powers of M, L and T on both sides, we get a + b + c = 0, – a – 3b + 2c = 0, – 2a – 2c = 1 On solving, a =  {tex}-\frac{5}{6}{/tex} , b =  {tex}\frac{1}{2}{/tex} , c =  {tex}\frac{1}{3}{/tex}
• Here m 1 = 4 kg, m 2 , = 5 kg Force applied on block A = 12 N This force must atleast be equal to the kinetic friction applied on A by B. {tex}\therefore{/tex} 12 = f k = {tex}\mu_k{/tex} R = {tex}\mu_k{/tex} m 1 g or 12 = {tex}\mu_k{/tex} {tex}\times{/tex} 4 g or {tex}\mu_{k}=\frac{12}{4 g}=\frac{3}{g}{/tex} The block B is on a smooth surface. Hence to move A and B together, the (maximum) force F that can be applied on B is equal to the frictional forces applied on A by B and applied on B by A. F = {tex}\mu_k{/tex} m 1 g + {tex}\mu_k{/tex} m 2 g = {tex}\mu_k{/tex} (m 1 + m 2 )g {tex}=\frac{3}{g}{/tex} (4 + 5)g = 27 N As this force moves both the blocks together on a smooth table, so the acceleration produced is {tex}a=\frac{F}{m_{1}+m_{2}}=\frac{27}{4+5}{/tex} = 3 ms -2

Class 11 Physic Sample Paper Solution Section C

• Radius of each small drop, r = 1 mm = 0.1 cm Terminal velocity of each small drop, v = 5 cms -1 Volume of bigger drop = Volume of 8 small drops {tex}\frac{4}{3} \pi R^{3}=8 \times \frac{4}{3} \pi r^{3}{/tex} or R = 2r = 2  {tex}\times{/tex}  0.1 cm = 0.2 cm Terminal velocity of each small drop is given by v =  {tex}=\frac{2}{9} \frac{r^{2}}{\eta}\left(\rho-\rho^{\prime}\right) \mathrm{g}{/tex}  …(i) Terminal velocity of bigger drop is given by V =  {tex}\frac{2}{9} \frac{R^{2}}{\eta}\left(\rho-\rho^{\prime}\right) g{/tex}  …(ii) Dividing equation (ii) by (i), we get {tex}\frac{V}{v}=\frac{R^{2}}{r^{2}}{/tex} or V =  {tex}v \times \frac{R^{2}}{r^{2}}=5 \times \frac{(0.2)^{2}}{(0.1)^{2}}{/tex} = 5  {tex}\times{/tex}  4 = 20 cms -1
• Since, co-efficient of linear expansion,  {tex}\alpha{/tex} = {tex}\frac { \Delta L } { L \Delta T }{/tex} {tex}\Delta L{/tex} = change in length L = length {tex}\Delta T{/tex} = change in temperature, for an infinitesimally small change in temperature {tex}\alpha = \frac{dL}{LdT}{/tex} Similarly, co-efficient of superficial expansion,  {tex}\beta{/tex} = {tex}\frac {d S } { S dT }{/tex} dS = infinitesimal change in area S = original area dT = infintesimaly change in temperature S = L 2 ,   {tex}\beta= \frac{1}{L^2}\frac{dL^2}{dT} = 2L\frac{dL}{dT}{/tex} {tex}\beta = 2\alpha{/tex} Similarly, co-efficient of cubical expansion,  {tex}\gamma = \frac {dV}{VdT}{/tex} dV = infinitesimal change in volume V = original volume dT = infinitesimal change in temperature {tex}\gamma = \frac{1}{L^3}\frac{dL^3}{dT} = 3\frac{1}{L}\frac{dL}{dT} =3\alpha{/tex} {tex}\alpha = \frac { \beta } { 2 } = \frac { \gamma } { 3 }{/tex}
• When the parachutist falls freely : u = 0, v = 9.8 ms -2 , s = 40m, t = ?, u = ? As s = ut +  {tex}\frac{1}{2} g t^{2}{/tex} {tex}\therefore 40=0+\frac{1}{2} \times 9.8 \times t^{2}{/tex} or t =  {tex}\sqrt{\frac{80}{9.8}}=\frac{20}{7} \mathrm{~s}{/tex}  = 2.86 s Also, v = u + gt = 0 + 9.8  {tex}\times \frac{20}{7}{/tex}  = 28 ms -1 When the parachutist decelerates uniformly: u = 28 ms -1 , a = -2 ms -2 , 5 = 2 ms -1 Time taken, t =  {tex}\frac{v \ – \ u}{a}=\frac{2 \ – \ 28}{-2}{/tex}  = 13 s Distance, s = ut +  {tex}\frac{1}{2} at^{2}{/tex}  = 28  {tex}\times{/tex}  13  {tex}-\frac{1}{2} \times{/tex}  2  {tex}\times{/tex}  (13) 2 = 364 – 169 = 195 m Total time of parachutist in air = 2.86 + 13 = 15.86 s Height at which parachutist bails out = 40 + 195 = 235 m
• {tex}\because{/tex}  process is adiabatic {tex}\therefore{/tex}   {tex}\Delta{/tex}  = 0 Work done on the gas, {tex}\Delta{/tex} W = -500J According to the first law of thermodynamics. {tex}\Delta{/tex} Q = {tex}\Delta{/tex} U + {tex}\Delta{/tex} W {tex}\Rightarrow{/tex}   {tex}\Delta{/tex} U = – {tex}\Delta{/tex} W = 500J
• This is because heat absorbed by a substance (coolant) is directly proportional to the specific heat of the substance.
•  This is because in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of a harbour town is without extremes of hot and cold.
• Conveyor belt is accelerating with 1 m/s². A man is stationary with respect to horizontal conveyor belt Hence, acceleration of man = acceleration of belt = 1m/s² Hence, Net force on the man = ma = 65  {tex}\times{/tex}  1 = 65 N direction of this force is opposite to direction of conveyor belt. Coefficient of friction between the man’s shoes and belt = 0.2. Let a’ is the acceleration of the belt can the man continue to be stationary relative to the belt. In equilibrium condition, Friction force = ma’ 0.2 {tex}\times{/tex}  m {tex}\times{/tex}  g = ma’ a’ = 0.2 {tex}\times{/tex}  g = 0.2 {tex}\times{/tex}  10 = 2m/s²
• Force P 1 ( {tex}\Delta A{/tex} ) acting vertically downward at the top face A,
• Force P 2  ( {tex}\Delta A{/tex} ) acting vertically upward at the bottom face B, and

Weight mg of the liquid in the volume element acting vertically downwards. Here,  {tex}\Delta A{/tex}  is the cross-section area of the given volume element. For volume element to be in equilibrium net force acting on it should be zero. Therefore, we have {tex}P_1\triangle A + mg – P_2\triangle A = 0\ or\ P_1\triangle A + mg = P_2\triangle A{/tex} But m = mass of volume element =  {tex}\Delta A{/tex} .h. {tex}\rho{/tex} , where  {tex}\rho{/tex}  = density of the liquid {tex}\therefore{/tex}   {tex}P_1\triangle A + \triangle A h\rho g = P_2\triangle A{/tex} {tex}\Rightarrow{/tex}   {tex}P_2 = P_1 + h\rho g{/tex} According to the above equation, if the pressure P 1 is increased in any way, the pressure P 2  must increase by exactly the same amount. The pressure applied to any part of an enclosed liquid at rest is transmitted undiminished to every portion of the liquid as well as the walls of the container.

Class 11 Physics Sample Paper Section D Solution

• (c) 2:1 Explanation: 2:1
• (d) heavy body Explanation: heavy body
• (a) becomes double Explanation: becomes double
• (a) potential energy Explanation: potential energy
• (b) becomes double Explanation: becomes double
• (d) Zero Explanation: Zero
• (d) remains same Explanation: remains same
• (a) 1:1 Explanation: 1:1

Class 11 Physics Sample Paper Section E Solution

• The maximum weight is 885 N, when platform moves from lowest to upward direction. And the minimum weight is 95.5 N, when platform moves from the highest point to downward direction.
• t = ? and y = -1.00 m {tex}\therefore{/tex}  y =  {tex}\frac{-1}{2}{/tex} gt 2 {tex}\Rightarrow t = \sqrt { \frac { – 2 y } { g } } = \sqrt { \frac { ( – 2 ) ( – 1.00 ) } { 9.8 } }{/tex} = 0.452 s
• x = ?, when t = 0.452 s {tex}\therefore{/tex}  x = v x0 t = 1.00  {tex}\times{/tex}  0.452 s = 0.452 m
• v = ?,  {tex}\theta {/tex}  = ? at t = 0.452 s The x-component of velocity is constant throughout the motion, v x  = v x0  = 1.00 m/s The y-component of velocity is given by v y  = v y0  – gt = 0 – 9.8  {tex}\times{/tex}  0.452 = – 4.43 m/s {tex}\therefore v = \sqrt { v _ { x } ^ { 2 } + v _ { y } ^ { 2 } }{/tex}   {tex}= \sqrt { ( 1.00 ) ^ { 2 } + ( – 4.43 ) ^ { 2 } }{/tex}  = 4.54 m/s, the magnitude of the resultant velocity of the motion. {tex}\theta = \tan ^ { – 1 } \left| \frac { v _ { y } } { v _ { x } } \right| = \frac { 4.43 } { 1.00 } = 77.3 ^ { \circ }{/tex} As the marble hits the floor, its velocity is 4.54 m/s directed 77.3° downward with respect to the horizontal.
• Area of a parallelogram  {tex}= | \vec { A } \times \vec { B } |{/tex}  =  {tex}A B \sin \theta{/tex}  ( {tex}\therefore{/tex}  Applying cross product) Given, area of parallelogram  {tex}= \frac { 1 } { 2 } A B{/tex} So, {tex}\frac { 1 } { 2 } A B = A B \sin \theta{/tex} {tex}\frac { 1 } { 2 } = \sin \theta{/tex} {tex}\theta = \sin ^ { – 1 } \left( \frac { 1 } { 2 } \right){/tex} {tex}\theta = 30 ^ { \circ }{/tex}

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• The force of gravity, F= mg produces the torque {tex}\tau{/tex} . Let {tex}\vec{r}{/tex} be the position vector of Q. Then the magnitude of the torque is given by {tex}\tau=r F \sin \theta{/tex} {tex}=r \times m g \times \frac{x_{0}}{r}=m g x_{0}{/tex}   {tex}\left[\because \sin \theta=\frac{x_{0}}{r}\right]{/tex} The direction of the torque is directed into the plane of paper and perpendicular to it, as shown by  {tex}\otimes{/tex} .
• The magnitude of the angular momentum is  {tex}L=r p \sin \theta=r m v \sin \theta{/tex} But the velocity v at point Q is given by v = u + at = 0 + gt = gt {tex}\therefore \quad L=r m g t . \frac{x_{0}}{r}=m g x_{0} t{/tex} The direction of angular momentum is the same as that of torque.
• Now L = mgx 0 t Differentiating both sides with respect to f, we get {tex}\frac{d L}{d t}=\frac{d}{d t}\left(m g x_{0} t\right)=m g x_{0}=\tau{/tex} Hence the relation  {tex}\tau=\frac{d L}{d t}{/tex}  holds in this example.

CBSE Sample Papers for Class 11

• Mathematics
• Accountancy
• Business Studies
• Computer Science
• Informatics Practices
• English Core
• Hindi Elective
• Political Science
• Physical Education
• Other Subjects

To download sample papers for class 11 Physics, Chemistry, Biology, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solutions, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Papers all are made available through the best app for CBSE students  and myCBSEguide website.

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CBSE Class 11 Physics Term 2 Study Materials

In This Post I Have Provided  CBSE Class 11   Physics   Term 2  Chapter-Wise Revision Notes, Case-Based Questions and Short/Long Type, Practice Question With Solution.

This Post Contains  CBSE Class 11 Physics Term 2  Chapter-Wise Notes, Case-Based and Short/Long Type Questions, Practice Questions With Solution for Session 2021-22. 

Every Student Know These types of Questions Comes With High-Scoring Opportunities in any exams. The article of  CBSE Class 11 Physics Term 2  is an important resource and is very helpful for the students who are preparing for term 2 examinations.

CBSE Class 11 Physics Term 2 Syllabus 2021-22

CBSE Class 11 Physics Syllabus is a must watch for the students. Here we are providing information on Class 11th Physics Syllabus Physics subjects, sub subjects as well as marks distribution of all chapters. Class 11 Physics Syllabus is prepared after proper study and research by the subject experts.

It contains all the necessary information that a student should know before starting the preparation for the exam. Syllabus for Class 11 Physics for the academic year 2021-2022 with marking scheme and time duration covering all the important subjects and sub-topics.

Physics Syllabus Class 11 2022

Physics Syllabus all Students of CBSE class 11 Physics Term 2 can now read the new Physics Syllabus here. NCERT has prepared and revised the syllabus of this subject. Also, the NCERT Book for Class 11 Physics Syllabus has a list of chapters which are very useful for the preparation of competitive exams.

CBSE Class 11 Physics Term 2 Syllabus in Hindi and English Medium

It is also advised to refer CBSE Class 11 Physics Term 2 Syllabus 2022 while preparing for Class 11 Physics paper. Also, you can download free CBSE Class 11 Physics Syllabus 2022 PDF for Class 11 Physics Chapters by NCERT.

CBSE Class 11 Physics Term 2 Syllabus in Hindi and English Medium Updated Class 11 Physics NCERT Syllabus PDF in Hindi and English as per the latest syllabus after NCERT (National Council of Education Research and Training). Students studying in Class 11 CBSE Hindi and English Medium can download NCERT Books for Class 11 Physics Hindi and English Medium.

CBSE Class 11 Physics Syllabus 2022

CBSE Official Link For Syllabus

CBSE has prepared and revised the syllabus of this subject. Also, the NCERT Book for Class 11 Physics Syllabus has a list of chapters which are very useful for the preparation of competitive exams.

It is also advised to go through the syllabus while preparing for class 11 physics exams. Apart from this, one can also download Free CBSE Class 11 Physics Syllabus 2022 PDF for Chapters of Class 11 Physics Syllabus by NCERT.

CBSE Class 11 Others Subjects Best Study Materials

Mathematics | Physics | Chemistry | Biology | English

Given Below Are The CBSE Class 11 Physics Term 2 Chapters Name With Their Respective Download Links.

NCERT Class 11 Physics Revision Notes and Study Material

As we know that in this year  CBSE  decided the examinations in two phases  Term-1  and  Term-2  due to Covid-19. The term-1 exam successfully has been completed and now the student are prepared for the term-2 examination. 

On the basis of and performance of term-1 and term-2 examinations the final result will be prepared by the  CBSE  for the students. That is why it is very important that now the students has to be prepared for the  term-2  examination in well manner.

Since term-2 examination is very important for the students for good result in their examinations 2021-22. That is why here I am going to provide systematic study materials for  Class 11 Physics  chapter wise notes and also important questions on the basis of latest  CBSE  exam pattern 2021-22.

These questions are very helpful to the students for the good preparation for their  CBSE Class 11 Physics Term 2 examinations. So, every student must try to study in very good manner with the help of these study materials and also try to solve given questions with the help of given Solutions and Answers.

These  CBSE Class 11 Physics Term 2  study materials and revision notes are prepared by our expert which are on the basis of latest  CBSE  exam pattern. These notes are very helpful for the preparing  term-2 examinations 2022.

Benefit Of CBSE Class 11 Physics Term 2 Good Study Material and Quick Revision Notes

There are many important benefits of good study materials and quick revision notes for  Class 11 Physics  which are listed below

* Student can easily understand the pattern of  CBSE  question papers for Physics subject.

* Student can prepared for their term 2 examination in very short time.

* Good study materials helps to the student for the better preparation for their term 2 examination 2022.

* Study material helps to the student to enhance their knowledge for term 2 Exam.

* With the help of a variety of questions and important quick revision notes student can clear their doubt and enhance their knowledge for term 2 Exam 2022.

* Quick Revision notes are very helpful in the last time revision for their final examination in very short time.

Best Study Materials Resource For Class 11

If you are preparing for any other Board examinations or any competitive examinations like  JEE ,  NEET ,  GATE  and  NTSE  etc, then this website is very helpful to the student. 

Maths And Physics With Pandey Sir  is a very dedicated website for the students which always provide a good class notes, revision notes, important questions with solution/answer, case based questions, sample papers etc for their students, so that they can prepared for their Board examinations or any other competitive examination.

CBSE Previous Year Question Paper for Class 11 Physics with Solutions

NCERT Solutions Class 11 All Subjects Notes Sample Papers Past Years Papers

CBSE Past Years Papers for Class 11 Physics

• CBSE Class 11 Physics Previous Year Question Paper 2020 
• CBSE Class 11 Physics Previous Year Question Paper 2019 Set 1
• CBSE Class 11 Physics Previous Year Question Paper 2019 Set 2
• CBSE Class 11 Physics Previous Year Question Paper 2017 Set 1
• CBSE Class 11 Physics Previous Year Question Paper 2016 set 1
• CBSE Class 11 Physics Previous Year Question Paper 2016 set 2
• CBSE Class 11 Physics Previous Year Question Paper 2016 set 3
• CBSE Class 11 Physics Previous Year Question Paper 2016 set 4
• CBSE Class 11 Physics Previous Year Question Paper 2016 set 5
• CBSE Class 11 Physics Previous Year Question Paper 2015 set 1
• CBSE Class 11 Physics Previous Year Question Paper 2015 set 2
• CBSE Class 11 Physics Previous Year Question Paper 2014
• CBSE Class 11 Physics Previous Year Question Paper 2013
• CBSE Class 11 Physics Previous Year Question Paper 2012
• CBSE Class 11 Physics Previous Year Question Paper 2011
• CBSE Class 11 Physics Previous Year Question Paper 2010

Question papers of class 11 Physics

CBSE Past Years Paper for class 11 Physics is made available by subjects experts so students can practice well and get prepared for their exam. Physics is a subject that requires constant practice along with an idea as to what kind of questions are going to be asked. Theory and numerical go hand in hand in Physics that require a mix of memory and comprehension skills. By solving the CBSE Past Years Papers for Class 11 Physics, students get enough practice that they can face the real exam with full confidence. Practising the CBSE Past Years Papers also makes them comfortable with the question paper format.

Here we have provided the CBSE Class 11 Physics Past Years Paper for students practice.  By practicing them, students will get to know how much they are actually prepared to face the exam. They will also get to know the areas where they are lagging behind. Working on them they can improve their performance in the exam.

Class 11 Physics List of Chapters NCERT Book

• Chapter 1 – Physical World
• Chapter 2 – Units and measurement
• 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 of rotational motion
• Chapter 8 – Gravitation
• Chapter 9 – Mechanical Properties of solids
• Chapter 10 – Mechanical Properties of fluids
• Chapter 11 – Thermal Properties of Matters
• Chapter11 – Thermodynamics
• Chapter 13 – Kinetic theory
• Chapter 14 – Oscillation
• Chapter 15 – Waves

JEE Main PYQ Solutions

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Cbse test series | class xi physics.

UNIT 1 : Physical World & Measurement

• Question Paper

UNIT 2: Kinematics (Motion In A Straight Line)

UNIT 3: Laws Of Motion

UNIT 4: Work , Power & Energy 

UNIT 5: System Of Particles & Rigid Body

UNIT 6: Gravitation

UNIT 7: Properties Of Bulk Matter

UNIT 8: Thermodynamics

UNIT 9: Behaviour Of Perfect Gas & Kinetic Theory Of Gases

UNIT 10: Oscillations & Waves

CBSE Sample Papers for Class 11 Physics Term 2 Set 1 with Solutions

Students can access the CBSE Sample Papers for Class 11 Physics with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 11 Physics Standard Term 2 Set 1 with Solutions

Time Allowed: 2 Hours Maximum Marks: 40

General Instructions:

• There are 12 questions in all. All questions are compulsory.
• This question paper has three sections: Section A, Section B and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks. :
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the choices in such questions.
• You may use log tables if necessary but use of calculator is not allowed.

Section – A (2 Marks Each)

Question 1. Distinguish clearly between temperature and heat. Answer: The degree of hotness of a body is called temperature. All bodies in thermal contact with each other ultimately attain the same temperature. The flow of heat energy from one body to another due to temperature difference between the two is called heat. Two bodies in thermal equilibrium with each other may not have the same amount of heat, but they must have the same temperature.

Question 2. A silica glass rod has a diameter of 1 cm and is 10 cm long. Estimate the largest mass that can be hung from it without breaking it. Breaking strength of glass is 50 × 10 6 Nm -2 . OR What are the factors affecting Elasticity? Answer: Using Stress = \(\frac{F}{A}\), we get F = stress × A = (50 × 10 6 ) × π × (\(\frac{10^{-2}}{2}\)) 2 N Largest mass = \(\frac{F}{g}\) = \(\frac{\left(50 \times 10^{6}\right)\left(\frac{\pi \times 10^{-4}}{4}\right)}{10}\)kg = 392.7 kg OR Factors affecting Elasticity are – (1) Hammering and rolling: This result is in an increase in the elastidty of material. (2) Annealing: Annealing results in decrease in the elasticity of material. (3) Temperature: Elastidty decreases with rise in temperature (Exception: the elasticity of invar steel (alloy) does not change with change in temperature). (4) Impurities: This- type of effect depends upon the nature of impurities. If the added impurity is more elastic than the substance then its elasticity is increased. If the added impurity is less elastic than the substance, then its elasticity is decreased.

Question 3. By applying the first law of thermodynamics to isobaric process, obtain relation between two specific heats of a gas. Answer: In an isobaric process, pressure remains constant. If an amount of heat dQ is supplied to one mole of a gas at constant pressure and its temperature increases by dT, then dQ = C P dT Here C P is molar specific heat of the gas at constant pressure. Therefore, for an isobaric process, the first law of thermodynamics becomes: C P dt = d\J + FcN …………… (i) From perfect gas equation it follows that PdV = RdT In the eqn. (i), substituting PdV and d\J, we have C P dT – C V dT C P = C V + R.

Section – B (3 Marks Each)

Question 4. Estimate the temperature of the surface of the Sun from the following data: (i) Average radius of the earth’s orbit = 1.5 × 10 5 km (ii) Average radius of the Sun = 7.0 × 10 5 km (iii) Solar radiant power on the Earth = 1400 Wm -2 at Noon time

Assume the Sun to be a perfect black body will your estimate be greater or less than the actual surface temperature of the Sun ? Explain. Answer: Here, Radius of Sun R = 7 × 10 5 km = 7 × 10 8 m ∴ Surface area of Sun = 4πR 2 = 4π(7 × 10 8 ) 2 = 6.16 × 10 18 m 2 σ = 5.67 × 10 -8 Wm -2 K -4 Using E = AσT 4 , E = (6.16 × 10 18 ) (5.67 × 10 -8 )T 4 = 3.49 × 10 11 × T 4 W …………….. (i) E = 4π 2 × Solar radiant power on Earth Here, Radius of Earth = r = 1.5 × 10 5 km 1.5 × 10 8 m E = 4π(1.5 × 10 8 ) 2 × 1400 = 3.96 × 10 20 watt …………….. (ii) Equating eqn (i) and (ii), we get 3.96 × 10 11 × T 4 = 3.96 × 10 20 T 4 = \(\frac{3.96 \times 10^{20}}{3.49 \times 10^{11}}\) T = 1.83 × 10 2 K.

Question 5. In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from -3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine. Answer: Given: T 1 = 27°C = 27 + 273 = 300 K T 2 = -3°C = -3 + 273 = 270 K Efficiency, η = 1 – \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\) = 1 – \(\frac{270}{300}\) = 1 – 0.9 = 0.1 = \(\frac{1}{10}\) Efficiency of refrigerator is 50% of perfect engine. η = 50% of η = 0.5 × \(\frac{1}{10}=\frac{1}{20}\) ∴ Coefficient of performance, β = \(\frac{Q_{2}}{W}=\frac{I-\eta^{\prime}}{\eta^{\prime}}\) β = \(\frac{1-\frac{1}{20}}{\frac{1}{20}}\) = \(\frac{1-0.05}{0.05}\) = \(\frac{0.95}{0.05}\) = 19 Q 2 = 19% of work done by motor on refrigerator = 19 × 1 kW Q 2 = 19 kJ/s

Question 6. To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8 × 10 8 Nm -2 , and the density of sea water is 10 3 kgm -3 .) Answer: Bulk Modulus, B = 9.8 × 10 8 N/m 2 Density of sea water (ρ) = 10 3 kg/m 3 Volume decrease (Percentage), (\(\frac{\Delta V}{V}\) × 100) = 0.1% \(\frac{\Delta V}{V}=\frac{0.1}{100}\) = \(\frac{1}{1000}\) = 1 × 10 -3 Let rubber ball be taken up to depth h ∴ change in pressure, P = hρg Bulk Modulus, K = \(\left|\frac{P}{\Delta V / V}\right|=\frac{h \rho g}{\Delta V / V}\) or, h = \(\frac{B \times(\Delta V / V)}{\rho g}\) = \(\frac{9.8 \times 10^{8} \times 1 \times 10^{-3}}{10^{3} \times 9.8}\) h = 100 m

Question 7. Calculate (i) r.m.s velocity, and (ii) mean kinetic energy of one gram molecule of hydrogen at S.T.P (Given, density of hydrogen at S.T.E is 0.09 kg/m 3 .) Answer: Here, ρ = 0.09 kg/m 3 At S.T.E P = 1.01 × 10 5 Pa

(i) According to kinetic theory of gases P = \(\frac{1}{3}\)ρc 2 or c = \(\sqrt{\frac{3 P}{\rho}}\) = \(\sqrt{\frac{3 \times 1.01 \times 10^{5}}{0.09}}\) = 1834.8 m/s

(ii) Volume occupied by one mole of hydrogen at S.T.P. = 22.4 litres = 22.4 × 10 -3 m 3 Mass of hydrogen, M = Volume × Density = 22.4 × 10 -3 × 0.09 kg = 2.016 × 10 -3 kg Mean K.E. of one gram molecule of hydrogen at S.T.P. = 1/2 Mc 2 = 1/2 × 2.016 × 10 -3 × (1834.8) 2 = 3.4 × 10 3 J

(a) Time taken to pass through mean position t = \(\frac{T}{4}\) = \(\frac{3.9 \times 10^{-3}}{4}\) = 9.8 × 10 -4

(b) Nodes- A, B, C, D, E (zero displacement) Anti-nodes-A’,C’ (maximum displacement)

(c) At A’, C’, there are consecutive anti-nodes, so distance between A’ and C’, λ \(\frac{v}{v}=\frac{360}{256}\) λ = 1.41m

(b) As, υ ∝ √T (T = Temperature) \(\frac{v_{20}}{v_{0}}\) = \(\sqrt{\frac{273+20}{273+0}}\) = \(\sqrt{\frac{293}{273}}\) \(\frac{v_{20}}{v_{0}}\) = 1.04 or υ 0 = \(\frac{v_{20}}{1.04}\) = \(\frac{348.16}{1.04}\) = 334.8 m/s

Water and mercury in tube reflects the sound into air column to form stationary wave and reflection is more in mercury than water as mercury is more denser than water. So, intensity of sound heard will be larger but reading does not change as medium in tube (air) and running fork are same.

OR Length for each loop = \(\frac{\lambda}{2}\) Now, L = \(\frac{n \lambda}{2}\) λ = \(\frac{2 L}{n}\) …………… (1) But υ = vλ or λ = \(\frac{v}{\mathrm{v}}\)

Putting in eqn. (1), \(\frac{v}{v}=\frac{2 L}{n}\) v = \(\frac{n}{2 L}\) v = \(\frac{n}{2 L} \sqrt{\frac{T}{\mu}}\) [∵ υ = \(\sqrt{\frac{T}{\mu}}\)] for n = 1,v 1 = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}=\) v 0 for n = 2, v 2 = \(\frac{2}{2 L} \sqrt{\frac{T}{m}}=\) = 2v 0 Therefore, v 1 : v 2 : v 3 : v 4 = n 1 : n 2 : n 3 : n 4 v 1 : v 2 : v 3 : v 4 =1 : 2 : 3 : 4

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Home » 11th Class » CBSE Class 11 Physics Sample Paper 2024 PDF

CBSE Class 11 Physics Sample Paper 2024 PDF

CBSE Class Class 11 Physics Sample Paper 2024 are now available for download in PDF format. CBSE Sample Paper as per the guidelines of CBSE board marking scheme is provided below for Physics subject. This is the model question paper as per latest Central Board of Secondary Education syllabus published for the 11th class exams 2024 at cbse.nic.in,.

CBSE Class 11 Physics Sample Paper 2024

The CBSE Sample Paper for class 11 Physics has model questions of the subject Physics, in the same exam pattern that you will get in your annual exam. Model questions are type of questions that you can expect in the CBSE class 11 Physics question paper 2024.

CBSE Class 11 Physics Sample Paper 2024 PDF Download Link – Click Here to Download Sample Question Paper

You can practice more CBSE class 11 study material here – CBSE class 11

Central Board of Secondary Education Board Exam Sample Question Paper for Class 11 is given here. CBSE Class 11 Physics Sample Question Paper gives an idea of question paper pattern and marking scheme.

More Sample Papers for CBSE Class 11

Here on this page you got the Physics model question paper. However there are more subjects to study in 11th standard. Therefore here are the subject-wise specimen paper of CBSE for class 11 students.

• Class 11 Sample Paper Accountancy
• Class 11 Sample Paper Biology
• Class 11 Sample Paper Business Studies
• Class 11 Sample Paper Chemistry
• Class 11 Sample Paper Economics
• Class 11 Sample Paper English
• Class 11 Sample Paper Hindi
• Class 11 Sample Paper History
• Class 11 Sample Paper Maths
• Class 11 Sample Paper Physical Education
• Class 11 Sample Paper Physics
• Class 11 Sample Paper Political Science
• Class 11 Sample Paper Sociology

Class 11 Physics

Arabic is an interesting subject for 11th standard pupils. Here are some key points.

• You should study all topics given in CBSE Class 11 Physics Syllabus 2023-24 .
• Moreover refer Class 11 Physics NCERT Books to study the chapters.
• You can also use NCERT Solutions for Class 11 Physics  to solve exercises.
• Solve the CBSE Previous Year Question Papers Class 11 Physics  to know what type of questions were asked last year.
• Finally study from CBSE Class 11 Physics Notes and summary to revise the chapters quickly before exam.
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The class-wise model question papers for CBSE board are as follows.

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Ncert exemplar class 11 physics chapter 4 laws of motion, ncert exemplar class 11 physics, leave a reply cancel reply, cbse board quick links.

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CBSE Class 11 Physics 2022 For Term 1 & Term 2 – Syllabus, Notes, NCERT Solutions, Question Papers

Physics deals with the studies of matter, its motion, and behavior through space and time. Physics is one of the most important scientific disciplines to understand how the universe behaves. By studying Class 11 Physics and developing depth concept, you will improve your chance to score better marks in  11th CBSE Exams .

Apart from that, with a thorough concept on the subject can help you to crack various exams like,

• JIPMER etc.

Here you will find a complete guide on CBSE Class 11 Physics 2022 For Term 1 & Term 2.

Table of Contents

CBSE Class 11 Physics 2022 For Term 1 & Term 2

Physics plays an important role in CBSE Class 11 Science Stream. You can easily score better marks in the exam with the depth concept. To build depth concept you should study NCERT 11th physics. It has used Simple and clear language is used. Multiple examples are available to build the student’s better base. The experts have organised the chapter into sections comprising one or more concepts or sub-concepts. All answers to the exercises and or hints are given where required. So a student can become more proficient in the subject.

CBSE Class 11 Physics Books 2022 For Term 1 & Term 2

CBSE Class 11 Physics books are created by NCERT following latest pattern. CBSE student, you should have the book to build the depth concept on the subject before the board exam.

Here you will find CBSE 11th standard Physics exercise-wise NCERT all chapters as in the textbook Part 1 and Part 2.

You will find the following chapters in 11th CBSE Physics

• 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

Crack The Exam Easily and Chase Your Dream

CBSE Class 11 Physics Syllabus 2022 For Term 1 & Term 2

Here you will get the detailed CBSE Class 11 Syllabus For Physics along with section-wise marks distribution. You can take a look at the units included in CBSE Mathematics Syllabus along with the marks distribution:

CBSE Physics Class 11 Syllabus and Exam Structure

• 11th Physics Syllabus 2022 For Term 1 & Term 2  

CBSE Class 11 Physics Solutions 2022 For Term 1 & Term 2

After the completion of the Physics textbook, you should have Class 11 Physics Solutions to prepare correctly 11th Exam.

You will easily find the exercise wise explanation of all the problems. You will also find various Miscellaneous Exercise questions from every exercise.

CBSE Class 11 Physics NCERT Solutions 2022 For Term 1 & Term 2

NCERT Experts has created Class 11 Physics Solutions to provide students depth concept on every chapter. They have covered all questions from all exercises based on the latest NCERT pattern.

You should not worry about the content because the NCERT experts have created such a way that any student can easily understand without having depth concept in the section.

• 11th Physics NCERT Solutions   

CBSE Class 11 Physics Question Papers

After completion of the Class 11 Physics syllabus and chapter-wise exercise, you should solve the class 11 exam question papers. You will get an idea about the exam and type of questions coming in the exam.

By solving more question papers on CBSE Class 11, you will understand the exam pattern more accurate way. You will also know the time management and analysis of your preparation level. You should have the following materials to find the complete question papers related to CBSE Class 11th Exam,

• Sample Question Papers
• Previous Year Question Papers
• Question Bank

Class 11 Physics Sample Question Papers 2022 For Term 1 & Term 2

CBSE every year, releases sample question papers and marking schemes for all subjects of class 11. By solving these sample papers, you will have a preview of the board question paper.

Analysing the CBSE 11th Physics sample papers, you can know about the pattern of the question paper and the format of questions to be asked in the board exam. 

• 11th Physics Sample Papers 2022 For Term 1 & Term 2 

Class 11 Physics Previous year Papers

You must solve the 11th CBSE Maths previous year papers before the board exams. You will get an idea about various important topics which are generally asked in the board exam. It will also help you to build concept on various types of questions.

You can analysis your preparation level and identify the conceptual weaknesses. You can easily cover those before the board exam.

• 11th Physics Previous year Papers

Class 11 Physics Question Bank 2022 For Term 1 & Term 2

The CBSE class 11 Physics question bank is designed by experts by analysing the chapter wise exercise and the previous exams. You will find multiple questions for every single topic. By solving many questions, you can develop the depth concept in every single section of Class 11 Physics.

CBSE Class 11 Physics Revision Notes 2022 For Term 1 & Term 2

The CBSE 11th Physics revision notes are a special piece of study material that students can utilise properly. The notes are prepared by some of the Experts. All the CBSE class 11 Physics revision notes are prepared by analysing previous years of question papers and syllabus in mind. 

CBSE Class 11 Physics Formulas 2022 For Term 1 & Term 2

Many of the class students struggle to learn basic formulae. We are providing you with the complete set of class 11 Physics formulas in the PDF format. Make sure that you should have Class 11 Physics Formulas handy. It will be helpful for you to revise and memorise various chapter wise formulas available in class 11 Physics  book.

We have covered a detailed guide on  CBSE Class 11 Physics 2022 For Term 1 & Term 2. Feel Free to ask any questions in the comment section below. 

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Class 11 Physics: CBSE Sample Question Papers- Term II (2021-22)- 2 | Sample Papers for Class 11 Medical and Non-Medical - NEET PDF Download

Time : 120 Minutes

Maximum marks : 35.

General Instructions: 

• There are 12 questions in all. All questions are compulsory. 
• This question paper has three sections: Section A, Section B and Section C. 
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks. 
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the choices in such questions. 
• You may use log tables if necessary but use of calculator is not allowed..

Section - A

Q.1. Distinguish clearly between temperature and heat.

The degree of hotness of a body is called temperature. All bodies in thermal contact with each other ultimately attain the same temperature. The flow of heat energy from one body to another due to temperature difference between the two is called heat. Two bodies in thermal equilibrium with each other may not have the same amount of heat, but they must have the same temperature.

Q.2. Why the does the temperature of a gas decrease, when it is allowed to expand adiabatically?

For an adiabatic change, the first law of thermodynamics may be expressed as dU + PdV = 0   or    PdV = – dU      ...(i) During expansion, dVis + ve. Therefore the equation (i) will hold, if dU is – ve, i.e., temperature decreases.

State second law of thermodynamics.

Kelvin’s statement:  No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work Clausius statement: It is not possible to transfer heat from a body at lower temperature to another at higher temperature without the help of some external energy.

Q.3. By applying the first law of thermodynamics to isobaric process, obtain relation between two specific heats of a gas.

In an isobaric process, pressure remains constant. If an amount of heat dQ is supplied to one mole of a gas at constant pressure and its temperature increases by dT, then dQ = C P dT Here C P is molar specific heat of the gas at constant pressure. Therefore, for an isobaric process, the first law of thermodynamics becomes : C P dt = dU + PdV     ...(i) From perfect gas equation it follows that PdV = RdT In the eqn. (i), substituting PdV and dU, we have C P dT = C V dT + RdT  C P = CV + R.

Section - B

Q.4. If the coefficient performance of a refrigerator is 5 and operates at the room temperature (27°C), find the temperature inside the refrigerator.

Given: β = 5, T1 = 27ºC = 27º + 273 = 300 K Coefficient performance of refrigerator, T 2 + 5T 2 = 1500 6T 2 = 1500 T 2 = 1500/6 = 250 K T 2 = 250 - 273 T 2 = -23ºC

Q.5. In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

Given:   T 1 = 27°C = 27 + 273 = 300 K   T 2 = −3°C = −3 + 273 = 270 K Efficiency of refrigerator is 50% of perfect engine. ∴ Coefficient of performance, Q 2 = 19% of work done by motor on refrigerator =19 × 1 kW Q 2 = 19 k J/s

Q.6. To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8 × 10 8 Nm −2 , and the density of sea water is 10 3 kgm −3 .)

Bulk Modulus, B = 9.8 × 10 8 N/m 2   Density of sea water (ρ) = 10 3 kg/m 3 Volume decrease (Percentage), Let rubber ball be taken up to depth h ∴ change in pressure, P = hρg Bulk Modulus,

Q.7. Discuss strings stretched between fixed points.

Let a spring be stretched between two points as shown in figure. Vibrations will set up in this string when it is disturbed, say from middle. The figure given below shows that vibration set up in this string appears to be similar to a half wave having wavelength λ/2. Length of the spring l = λ/2 or    λ = 2l   ......(i) If T is tension in the string and m is mass per unit length of the string, then velocity of disturbance is given by        ...(ii) also = Wavelength × Frequency =   λf or v = 2l × f     using equ (i) or f = v/2l    ...(iii) From equation (ii) and (iii)

Q.8. State Pascal’s law of transmission of fluid pressure. Explain how is Pascal’s law applied in  a hydraulic lift. (with suitable diagram).

Pascal law: If gravity effect is neglected the pressure at every point of liquid in equilibrium at rest is same. Hydraulic lift: It is used to lift the heavy loads. Its working is based on Pascal’s law. A simple hydraulic lift is shown in Figure. Here C and D are two cylinders of different areas of cross section. They are connected to each other with a pipe E. Each cylinder is provided with airtight frictionless piston. Let a, and A be the areas of cross-sections of the pistons in C and D respectively, where a < < A. The cylinders are filled with an incompressible liquid. Let a downward force f be applied on the piston of C. Then the pressure exerted on the liquid, P = f/a. According to Pascal’s law, this pressure is transmitted equally to piston of cylinder D. ∴ Upward force acting on the piston of cylinder D will be

A razor blade can be made to float on water. What forces act on this blade ?

When a razor blade is made to float on water, three forces act on the blade : (i) Weight of the blade acting vertically downwards. (ii)  Reaction on blade exerted by the liquid surface acting vertically upwards. (iii)  Force of the surface tension on circumference of the blade acting tangentially to the liquid surface.

Q.9. The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. The velocity of two waves superimposing to form stationary waves is 360 ms -1 and their frequencies are 256 Hz.

Given:  frequency of wave, ν = 256 Hz Time period, T = 1/v = 1/256 = 3.9 x 10 -3 s (i) Time taken to pass through mean position t = T/4 (ii) Nodes-  A, B, C, D, E ( zero displacement) Anti-nodes-A’, C’ (maximum displacement) (iii) At  A’, C’, there are consecutive anti-nodes, so distance between A’ and C’,

Q.10. Show that the motion of a particle represented by y = sin ωt – cos ωt is simple harmonic with a period of 2π/ω

∴ (sin ωt-cos ωt) represents S.H.M. ∴ Timeperiod = 2π/ω

Q.11. What is standing wave ? Give its characteristics.

Standing or stationary wave results when two progressive waves of same wavelength and amplitude travelling with same speed in opposite directions superimpose. Characteristics :  Energy is not transferred from particle to particle. Amplitude of vibration of particles is maximum at antinode and minimum (say zero) at nodes. Time period of all particles is same except at nodes. Wavelength and frequency of standing wave is same as that of the component wave.  Pressure variation is minimum at antinode and maximum at nodes.

Obtain an equation of a progressive wave.

For a simple harmonic motion, displacement y = r sin (ωt + ϕ) where r is amplitude, ω is angular velocity and ϕ is phase difference  But  ω = 2π/T = 2πf and where λ is wavelength and x is path difference. ∴

Section - C

Q.12. Case Study : Molecular Perspective In water, there are two types of molecules. Some molecules are at the surface, called exterior molecules, and some molecules are inside, called interior molecules. The interior molecules are attracted to all the molecules around them. The exterior molecules are attracted to only the other surface molecules and to those below the surface. So that the energy state of the molecules on the interior is much lower than that of the molecules on the exterior. The molecules always try to maintain a lower energy state and hence the exterior molecules experience a downward force. This force is known as cohesive force. As a result they try to maintain a minimum surface area, thus allowing more molecules to have a lower energy state. Thus surface tension is created. The water molecules attract one another due to the water molecule’s polar property. The hydrogen ends, which are positive in comparison to the negative ends of the oxygen cause water to “stick “ together. This is why there is surface tension. Water has very high surface tension. It is 72.8 milli newton per meter at 20°C. (i) The energy state of the interior molecules of a fluid is (a) Higher than the exterior molecules of the fluid (b) Lower than the exterior molecules of the fluid (c) Equal to the exterior molecules of the fluid (d) Higher than or Equal to the exterior molecules of the fluid

Correct Options is  (b). In water, there are two types of molecules. Some molecules are at the surface, called exterior molecules, and some molecules are inside, called interior molecules. The interior molecules are attracted to all the molecules around them. The exterior molecules are attracted to only the other surface molecules and to those below the surface. So that the energy state of the molecules on the interior is much lower than that of the molecules on the exterior.

(ii) The tendency of water to maintain .............. surface area is known as surface tension. (a) Maximum (b) Minimum  (c) Fixed (d) Energetic

Correct Option is  (b). The energy state of the molecules on the interior is much lower than that of the molecules on the exterior. The molecules always try to maintain a lower energy state. As a result, they try to maintain a minimum surface area, thus allowing more molecules to have a lower energy state. Thus, surface tension is created.

(iii) The water molecules attract one another due to the water molecule’s (a) Polar covalent bond (b) Non-polar covalent bond (c) Ionic bond (d) Metallic bond

Correct Option is  (a). The water molecules attract one another due to the water molecule’s polar property. The hydrogen ends, which are positive in comparison to the negative ends of the oxygen cause water to “stick” together.

(iv) Which one of the following is the correct molecular structure of water molecule? (a) 

Correct Option is  (a). Water is a polar molecule having positively charged sides where the two hydrogen atoms are found and a negatively charged end where the oxygen atom is located.

(v) Surface tension of water is (a) 72.8 mili N (b) 72.8 N/ m (c) 72.8 mili N /m (d) 72.8 mili N /m 2    

Correct Option is  (c). Water has very high surface tension. It is 72.8 milli newton per metre at 20°C.

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• NCERT Solutions
• NCERT Class 11
• NCERT 11 Physics

NCERT Solutions for Class 11 Physics

Download chapter-wise ncert solutions for class 11 physics.

NCERT Solutions for Class 11 Physics for all chapters can be accessed here by following the links provided below. The highly experienced subject matter experts have explained each and every minute concept in the best way possible to help students face the exam without fear. Physics is a subject which contains a lot of concepts and numerical problems. For this purpose, a thorough understanding of the concepts is necessary to solve the problems effortlessly. NCERT Solutions  is prepared in order to provide the students with authentic information completely based on the latest syllabus prescribed by the CBSE board.

From Class 11 Physics, students will study the various concepts which are important both from the exam and competitive exam perspective. The solutions provided here can be accessed by the students in online mode or can be downloaded as a PDF. Each chapter of NCERT Solutions for Class 11 Physics can be downloaded in a PDF format by clicking on the “Download PDF” button at the top of each chapter page given below.

NCERT Solutions for Class 11 Physics (Chapter-wise)

• 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

The following chapters have been removed from the NCERT Class 11 Physics textbook 2023-24.

Physical World

Class 11 is an important stage in the life of a student because they are exposed to learning various new concepts. The solutions at BYJU’S not only helps them to understand it but also provides them with a basic knowledge of their applications in our daily life. So learning the chapters present in the NCERT textbook will help them to obtain a strong hold on the basic concepts. There are numerical problems and examples present in each chapter, which are explained in a step-wise manner to improve the problem-solving abilities of the students.

Download NCERT Solutions for Class 11 Physics to speed up the exam preparation.

NCERT Solutions Class 11 Physics

NCERT Solutions Class 11 Physics is an important study material which helps students to master the subject. Physics is a crucial subject from the exam perspective, which is difficult to master as it deals with various properties of nature and matter. It forms a strong base for students who wish to pursue higher levels of education in the field of Science. The solutions are planned out to aid the Class 11 students in achieving good marks in the exam. The expert teachers answer each and every question in an elaborate manner so that students do not miss any concept which is important for the exam. By regular practice using the NCERT Solutions, students will be able to know the right way to approach the questions which would appear in the exam.

NCERT Solutions Class 11 Physics Chapter Details and Exercises

Chapter 1: physical world.

In this chapter, we will be studying how things work and the reason behind it. Students will become more aware of the things surrounding them, the origin and the history of Science. They will also come across various aphorisms and their justification using Physics concepts in real life in this chapter. The concepts covered here are gravitational force , electromagnetic force, strong and weak nuclear force. The theories and observations done by various scientists and their conclusion is provided briefly here. You will gain more idea about the factors which are the reasons for the development of Science and Technology in our country. The existence of an electron is explained in this chapter in an elaborate manner to acquaint students with the basic conceptual knowledge.

Topics Covered in Class 11 Physics Chapter 1 Physical World

Physics: Scope and excitement; nature of physical laws; Physics, technology and society

Also, access the following resources for Class 11 Chapter 1 Physical World, at BYJU’S:

• Physical World Class 11 Notes Chapter 1
• Physics Revision Notes for Class 11 Chapter 1 Physical World
• Physical World and Measurement MCQs
• NCERT Exemplar Class 11 Physics Solutions for Chapter 1 – Introduction

Chapter 2: Units and Measurements

With the senses like ears, eyes etc., scientists gather information and make observations. A few of them are like figuring out the colour and texture, while others would be complex where measurement is necessary. It is a basic concept in Science without which scientists will not be able to conduct any experiments. Students will learn about the units of physical quantities and the method to be followed to evaluate them. It will also provide a good idea of the errors which would occur while measuring things and significant figures.

Topics Covered in Class 11 Physics Chapter 2 Units and Measurements

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. Length, mass and time measurements; accuracy and precision of measuring instruments; errors in measurement; significant figures. Dimensions of physical quantities, dimensional analysis and its applications.

Also, access the following resources for Class 11 Chapter 2 Units and Measurements, at BYJU’S:

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

Chapter 3: Motion in a Straight Line

This chapter will cover essential topics like comparing objects as point object and the method of plotting graphs and finding the values based on them. The method plotting x-t motion graph can be learnt easily by using the NCERT Solutions from BYJU’S. The time taken by a biker, the speed of a car and the time taken by a bus to travel can be found effortlessly by referring to these solutions. Students will learn the difference between magnitude of displacement and the total length of the path covered. They will also study about the average speed, average velocity, instantaneous speed and velocity as per the CBSE guidelines.

Topics Covered in Class 11 Physics Chapter 3 Motion in a Straight Line

Frame of reference. Motion in a straight line: Position-time graph, speed and velocity. Uniform and non-uniform motion, average speed and instantaneous velocity. Uniformly accelerated motion, velocity-time and position-time graphs, relations for uniformly accelerated motion (graphical treatment). Elementary concepts of differentiation and integration for describing motion.

Also, access the following resources for Class 11 Chapter 3 Motion in a Straight Line, at BYJU’S:

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

Chapter 4: Motion in a Plane

Students will learn about the scalar and vector quantities and their difference. The method of performing algebraic expressions on various scalar quantities are also explained here. Besides, true or false questions based on these concepts and their justification are elaborated under each answer. We will also learn the concept of magnitude of vectors and their co-linearity, magnitude of displacement vector and the various types of displacement of a moving vehicle. It will help us to understand the time taken by a person to cross a river, experience of riding a bicycle in rain and a ball thrown by a cricketer in the ground.

Topics Covered in Class 11 Physics Chapter 4 Motion in a Plane

Scalar and vector quantities: Position and displacement vectors, general vectors and notation, equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors. Relative velocity. Unit vectors. Resolution of a vector in a plane – rectangular components. Motion in a plane. Cases of uniform velocity and uniform acceleration – projectile motion. Uniform circular motion.

Also, access the following resources for Class 11 Chapter 4 Motion in a Plane, at BYJU’S:

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

Chapter 5: Laws of Motion

This chapter is very important in mechanics. Problems based on the conservation of momentum are frequently asked in the exam. By going through this chapter, students will become well-versed with the laws of motion and their applications in our daily activities. Various numerical problems based on mechanics are solved clearly in this chapter to boost the exam preparation of students. Having a good knowledge of these concepts, students will be able to pursue well in their higher education levels.

Topics Covered in Class 11 Physics Chapter 5 Laws of Motion

Intuitive concept of force. Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces. Static and kinetic friction, laws of friction, rolling friction, lubrication. Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on level circular road, vehicle on banked road).

Also, access the following resources for Class 11 Chapter 5 Laws of Motion, at BYJU’S:

• NCERT Exemplar Class 11 Physics Solutions for Chapter 5 – Laws Of Motion
• Laws Of Motion Class 11 Notes

Chapter 6: Work, Energy and Power

The Work Energy Theorem is an important concept in this chapter. Students are highly recommended to learn this chapter effectively as it is continued in future classes also. The work done by a person and the energy required to do that work is explained briefly with suitable examples to make the students well-versed with the concepts. Problems on determining the energy and power are solved in a step-wise manner based on the mark weightage allotted by the CBSE board. Practising these concepts regularly will make them master the chapter effortlessly.

Topics Covered in Class 11 Physics Chapter 6 Work, Energy and Power

Scalar product of vectors. Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power. Notion of potential energy, potential energy of a spring, conservative forces; conservation of mechanical energy (kinetic and potential energies); non-conservative forces; elastic and inelastic collisions in one and two dimensions.

Also, access the following resources for Class 11 Chapter 6 Work, Energy and Power, at BYJU’S:

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

Chapter 7: System of Particles and Rotational Motion

This chapter will help students understand the motion of extended bodies, such as the system of particles. The centre of mass of a system of particles is the main concept in this chapter. The term motion and its usefulness are briefly explained as per the syllabus of the CBSE board. As many concepts are covered under this chapter, students are highly recommended to make use of a perfect study material while answering the exercise questions from the textbook.

Topics Covered in Class 11 Physics Chapter 7 System of Particles and Rotational Motion

Centre of mass of a two-particle system, momentum conservation and centre of mass motion. Centre of mass of a rigid body; centre of mass of circular ring, disc, rod and sphere. Vector product of vectors; moment of a force, torque, angular momentum, conservation of angular momentum with some examples.

Equilibrium of rigid bodies, rigid body rotation and equation of rotational motion, comparison of linear and rotational motions; moment of inertia, radius of gyration. Values of M.I. for simple geometrical objects (no derivation). Statement of parallel and perpendicular axes theorems and their applications.

Also, access the following resources for Class 11 Chapter 7 System of Particles and Rotational Motion, at BYJU’S:

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

Chapter 8: Gravitation

We know that all material objects are attracted towards the Earth. The things which we throw up fall down. Climbing a hill is difficult when compared to going downhill. Gravitation is a very popular concept as most of them which we study in the future are in accordance with this phenomenon. To understand the more complex concepts in this chapter, students should know the difference between gravity and gravitation. The other concepts discussed in this chapter are potential energy difference between two points, acceleration due to gravity and interplanetary motion.

Topics Covered in Class 11 Physics Chapter 8 Gravitation

Kepler’s laws of planetary motion. The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Gravitational potential energy; gravitational potential. Escape speed, orbital velocity of a satellite. Geostationary satellites.

Also, access the following resources for Class 11 Chapter 8 Gravitation, at BYJU’S:

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

Chapter 9: Mechanical Properties of Solids

A rigid body has a definite shape and size, but in reality, it is bent, stretched and compressed. A force is necessary in order to know the shape of a solid. The concepts of importance in this chapter are plasticity and elasticity. Students will learn about the mechanical properties of a solid, Young’s Modulus of copper and steel and the yield strength. Problems based on these concepts appear in the exams for more marks, so it is necessary to practise them on a daily basis. Students will also get to know about the stress-strain relationship graph, which is very important for the exam.

Topics Covered in Class 11 Physics Chapter 9 Mechanical Properties of Solids

Elastic behaviour, stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear, modulus of rigidity.

Also, access the following resources for Class 11 Chapter 9 Mechanical Properties of Solids, at BYJU’S:

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

Chapter 10: Mechanical Properties of Fluids

Here, we will learn about the physical properties of liquids and gases. They are called as fluids as they have the capacity to flow. This is an important property which separates solids from liquids and gases. The other important concepts covered here are Bernoulli’s principle , Reynold’s number, streamline flow, viscosity and surface tension. The fundamental property of a fluid is that it can flow. It has no resistance to change its shape. Students will get a clear picture about the mechanical properties of fluids which are of more marks in the final exam.

Topics Covered in Class 11 Physics Chapter 10 Mechanical Properties of Fluids

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes). Effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, Reynold’s number, streamline and turbulent flow.

Bernoulli’s theorem and its applications. Surface energy and surface tension, angle of contact, application of surface tension ideas to drops, bubbles and capillary rise.

Also, access the following resources for Class 11 Chapter 10 Mechanical Properties of Fluids, at BYJU’S:

• Mechanical Properties of Fluids Class 11 Notes
• Physics Revision Notes for Class 11 Chapter 10 Mechanical Properties of Fluids
• NCERT Exemplar Class 11 Physics Solutions for Chapter 10 – Mechanical Properties of Fluids

Chapter 11: Thermal Properties of Matter

We all are familiar with heat and temperature. The method used to determine the temperature is covered in this chapter. Students will also get a clear idea about Newton’s Law of Cooling, which is an important concept for various competitive exams. Temperature is the hotness possessed by a body that can be measured using a thermometer. The various thermal properties of matter are listed in a step-wise manner in this chapter to make students more confident about the concepts.

Topics Covered in Class 11 Physics Chapter 11 Thermal Properties of Matter

Heat, temperature, thermal expansion; specific heat capacity – calorimetry; change of state – latent heat. Heat transfer – conduction, convection and radiation, thermal conductivity, Newton’s law of cooling.

Also, access the following resources for Class 11 Chapter 11 Thermal Properties of Matter, at BYJU’S:

• CBSE Class 11 Physics Notes Chapter 11 Thermal Properties of Matter
• Physics Revision Notes for Class 11 Chapter 11 Thermal Properties of Matter
• NCERT Exemplar Class 11 Physics Solutions for Chapter 11 – Thermal Properties of Matter

Chapter 12: Thermodynamics

Thermodynamics is a concept of heat and temperature and its conversion to other energy forms. It is one of the main topics which the question paper setters focus on, and most of the questions in the exams can appear from this chapter. In order to attain remarkable grades, it is crucial for students to learn all these concepts effectively. The laws of thermodynamics, specific heat capacity, various thermodynamic processes and the Carnot engine are some of the important concepts covered in this chapter.

Topics Covered in Class 11 Physics Chapter 12 Thermodynamics

Thermal equilibrium and definition of temperature (zeroth law of thermodynamics). Heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics: Reversible and irreversible processes. Heat engines and refrigerators.

Also, access the following resources for Class 11 Chapter 12 Thermodynamics, at BYJU’S:

• CBSE Class 11 Physics Notes Thermodynamics
• Physics Revision Notes for Class 11 Chapter 12 Thermodynamics
• NCERT Exemplar Class 11 Physics Solutions for Chapter 12 – Thermodynamics

Chapter 13: Kinetic Theory

This chapter is a more scoring section when it comes to exam preparation. Apart from the properties of solids and liquids, the properties of gases are easy to understand. Students will learn concepts like molecular nature of matter, behaviour of gases, kinetic theory of  an ideal gas, law of equipartition of energy , specific heat capacity and mean free path in this chapter. The problems are also solved in a systematic manner following the CBSE guidelines so that scoring good marks would be an easy task for the Class 11 students.

Topics Covered in Class 11 Physics Chapter 13 Kinetic Theory

Equation of state of a perfect gas, work done on compressing a gas. Kinetic theory of gases: Assumptions, concept of pressure. Kinetic energy and temperature; rms speed of gas molecules; degrees of freedom, law of equipartition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number.

Also, access the following resources for Class 11 Chapter 13 Kinetic Theory, at BYJU’S:

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

Chapter 14: Oscillations

Oscillations are one of the basic concepts in Physics. It is very important for the students to learn the concepts from this chapter, as it will be continued in further classes as well. The concepts explained in this chapter are periodic and oscillatory motions , simple harmonic motion, uniform circular motion etc. To learn these concepts effectively, it is necessary for the students to follow suitable study material for this purpose. By learning the chapter on a daily basis, students will be able to face the more complex questions that would arise in the exam.

Topics Covered in Class 11 Physics Chapter 14 Oscillations

Periodic motion – period, frequency, displacement as a function of time. Periodic functions. Simple harmonic motion (SHM) and its equation; phase; oscillations of a spring – restoring force and force constant; energy in SHM – kinetic and potential energies; simple pendulum – derivation of expression for its time period; free, forced and damped oscillations (qualitative ideas only), resonance.

Also, access the following resources for Class 11 Chapter 14 Oscillations, at BYJU’S:

• CBSE Class 11 Physics Notes Chapter 14 Oscillations
• Physics Revision Notes for Class 11 Chapter 14 Oscillations
• NCERT Exemplar Class 11 Physics Solutions for Chapter 14 – Oscillations

Chapter 15: Waves

This chapter contains comprehensive questions on important concepts like wave dynamics. Questions present in this chapter will assist students through concepts like the types of waves , speed of sound in air, the tension on strings and the dependence of the speed of sound in air. The important formulas and concepts are highlighted in the solutions to make a revision of concepts easier for the students. The various wave characteristics, such as amplitude, frequency, phase, wavelength, displacement of waves and resonance, are covered in this chapter.

Topics Covered in Class 11 Physics Chapter 15 Waves

Wave motion. Longitudinal and transverse waves, speed of wave motion. 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 Chapter 15 Waves, at BYJU’S:

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

As there are many concepts present in each chapter, students have to revise the concepts on a regular basis to get a grip on them. It will help them to understand the right way of approaching the difficult questions in the final exam. NCERT Class 11 Books can be accessed by the students to get a clear understanding of the syllabus.

Features of BYJU’S NCERT Solutions for Class 11 Physics

The Class 11 Physics NCERT Solutions provided by BYJU’S have the following features:

• Elaborate explanations for all the textbook questions.
• A systematic process for answering numerical problems.
• Clear-cut answers for all the questions based on theory.
• Authentic information from highly experienced subject matter experts.
• Solutions are present in both online and offline modes.

Note: Class 11 Physics students (who are studying under the CBSE syllabus) can bookmark this page in their browser to easily access NCERT Solutions for Class 11 Physics in the future.

CBSE Marking Scheme 2023-24

CBSE board has divided the academic sessions covering the entire syllabus to help students learn the concepts effectively. A systematic approach is followed based on the interconnectivity of topics and concepts by the experts. The exams will be conducted by the CBSE board at the end of the academic year as per the bifurcated syllabus. This basically improves the condition of having board-conducted classes at the end of the year.

Why Should One Opt for BYJU’S NCERT Solutions?

The NCERT Solutions at BYJU’S are the best study material for students under the CBSE board. The content provided is accurate, and students can rely on these solutions to perform better in the annual exam. The solutions strictly follow the NCERT Class 11 Physics Syllabus , which is prescribed by the CBSE board. Some of the advantages of using the NCERT Solutions provided by us are given below.

• Top-notch experts prepare the solutions

The NCERT Solutions have been prepared by a set of highly experienced faculty at BYJU’S with the main aim of making quality educational content for each student. The precise and informative solutions are considered to be the best among all the other materials available online.

• Students will be able to answer complex questions in the exam

The NCERT Solutions at BYJU’S are famous for being student-friendly and easy to understand. Answers to difficult questions and problems are explained in steps to help students understand the method of solving them easily. The solutions provided are concept-focused and not question focused, which enables the students to face the different questions that would arise in the exam.

• The best study material for revision

After learning all the chapters, students are advised to revise all the concepts a month prior to the exam. This will help them to remember all the problems and concepts which are important from the exam perspective. It will also help them to memorise all the formulas and problems which are of high marks as per the CBSE exam pattern.

After Going through Our Class 11 Physics NCERT Solutions, Also Explore,

• NCERT Exemplar Problems for Class 11 Physics
• CBSE Revision Notes for Class 11 Physics

Have any questions/issues regarding our NCERT Solutions for Class 11 Physics? Our support team is always available to resolve your queries. Register with BYJU’S and get in touch with them NOW!

Frequently Asked Questions on NCERT Solutions for Class 11 Physics

Why should i choose the ncert solutions for class 11 physics from byju’s, which chapter is the easiest in the ncert solutions for class 11 physics, which are the chapters present in the ncert solutions for class 11 physics, are the ncert solutions for class 11 physics the best reference materials for the students to rely on.

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What Is Project 2025, and Why Is Trump Disavowing It?

The Biden campaign has attacked Donald J. Trump’s ties to the conservative policy plan that would amass power in the executive branch, though it is not his official platform.

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By Simon J. Levien

Donald J. Trump has gone to great lengths to distance himself from Project 2025, a set of conservative policy proposals for a future Republican administration that has outraged Democrats. He has claimed he knows nothing about it or the people involved in creating it.

Mr. Trump himself was not behind the project. But some of his allies were.

The document, its origins and the interplay between it and the Trump campaign have made for one of the most hotly debated questions of the 2024 race.

Here is what to know about Project 2025, and who is behind it.

What is Project 2025?

Project 2025 was spearheaded by the Heritage Foundation and like-minded conservative groups before Mr. Trump officially entered the 2024 race. The Heritage Foundation is a think tank that has shaped the personnel and policies of Republican administrations since the Reagan presidency.

The project was intended as a buffet of options for the Trump administration or any other Republican presidency. It’s the latest installment in the Heritage Foundation’s Mandate for Leadership series, which has compiled conservative policy proposals every few years since 1981. But no previous study has been as sweeping in its recommendations — or as widely discussed.

Kevin Roberts, the head of the Heritage Foundation, which began putting together the latest document in 2022, said he thought the American government would embrace a more conservative era, one that he hoped Republicans would usher in.

“We are in the process of the second American Revolution,” Mr. Roberts said on Real America’s Voice, a right-wing cable channel, in early July, adding pointedly that the revolt “will remain bloodless if the left allows it to be.”

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