case study questions of class 12 physics

• Electric Charges and Fields Class 12 Case Study Questions Physics Chapter 1

Last Updated on September 4, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 12 physics. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 12 physics. In this article, you will find case study questions for cbse class 12 physics chapter 1 Electric Charges and Fields.

Table of Contents

Case Study Questions on Electric Charges and Fields

Question 1:

Surface charge density is defined as charge per unit surface area of surface charge distribution. i.e., $\sigma=\frac{d q}{d s}$. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs having magnitude of $17.0 \times 10^{-22} \mathrm{Cm}^{-2}$ as shown. The intensity of electric field at a point is $E=\frac{\sigma}{\varepsilon_0}$, where $\varepsilon_0=$ permittivity of free space.

(i) $E$ in the outer region of the first plate is (a) $17 \times 10^{-22} \mathrm{~N} / \mathrm{C}$ (b) $1.5 \times 10^{-25} \mathrm{~N} / \mathrm{C}$ (c) $1.9 \times 10^{-10} \mathrm{~N} / \mathrm{C}$ (d) zero (ii) $E$ in the outer region of the second plate is (a) $17 \times 10^{-22} \mathrm{~N} / \mathrm{C}$ (b) $1.5 \times 10^{-15} \mathrm{~N} / \mathrm{C}$ (c) $1.9 \times 10^{-10} \mathrm{~N} / \mathrm{C}$ (d) zero (iii) $E$ between the plates is (a) $17 \times 10^{-22} \mathrm{~N} / \mathrm{C}$ (b) $1.5 \times 10^{-15} \mathrm{~N} / \mathrm{C}$ (c) $1.9 \times 10^{-10} \mathrm{~N} / \mathrm{C}$ (d) zero (iv) The ratio of $E$ from right side of $B$ at distances 2 cm and 4 cm , respectively is (a) $1: 2$ (b) $2: 1$ (c) $1: 1$ (d) $1: \sqrt{2}$ (v) In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is (a) spherical (b) cylindrical (c) straight line (d) none of these

(i) Option (d) is correct.

There are two plates $A$ and $B$ having surface charge densities,

$$ \sigma_A=17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2 $$

on $A$ and $\sigma_B=-17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2$ on $B$, respectively. According to Gauss’ theorem, if the plates have same surface charge density but having opposite signs, then the electric field in region I is zero.

$$ E_{\mathrm{I}}=E_A+E_B=\frac{\sigma}{2 \varepsilon_0}+\left(-\frac{\sigma}{2 \varepsilon_0}\right)=0 $$

(ii) (d): The electric field in region III is also zero.

$$ E_{\mathrm{III}}=E_A+E_B=\frac{\sigma}{2 \varepsilon_0}+\left(-\frac{\sigma}{2 \varepsilon_0}\right)=0 $$

(iii) (c): In region II or between the plates, the electric field

$$ \begin{aligned} & E_{\mathrm{II}}=E_A-E_B=\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0} \\ &=\frac{\sigma\left(\sigma_A \text { or } \sigma_B\right)}{\varepsilon_0}=\frac{17.0 \times 10^{-22}}{8.85 \times 10^{-12}} \\ & E=1.9 \times 10^{-10} \mathrm{NC}^{-1} \end{aligned} $$

(iv) (c): Since, electric field due to an infinite-plane sheet of charge does not depend on the distance of observation point from the plane sheet of charge. So, for the given distances, the ratio of $E$ will be $1: 1$. (v) (b): In order to estimate the electric field due to a thin finite plane metal plate, we take a cylindrical cross-sectional area $A$ and length $2 r$ as the gaussian surface.

Case study questions for other chapters of class 12 physics is given below.

Case Study Questions for Class 12 Physics

We hope the given case study questions for Electric Charges and Fields Class 12 helps you in your learning.

Topics from which case study questions may be asked

• Electric charges
• Conservation of charge
• Coulomb’s law – force between two-point charges
• Forces between multiple charges
• Superposition principle
• Continuous charge distribution
• Electric field
• Electric field due to a point charge
• Electric field lines
• Electric dipole
• Electric field due to a dipole
• Torque on a dipole in a uniform electric field
• Electric flux
• Statement of Gauss’s theorem
• Applications of Gauss’s theorem to find field due to: Infinitely long straight wire Uniformly charged infinite plane sheet Uniformly charged thin spherical shell (field inside and outside)

Two electric field lines never cross each other. If they intersect, then there will be two directions of electric field at the point of intersection which is not possible.

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Frequently Asked Questions (FAQs) on Electric Charges and Fields Case Study Questions

Q1: what are case study questions for cbse examinations.

A1: Case study questions in CBSE examinations typically involve scenarios or real-life examples, requiring students to apply their understanding of concepts to solve problems or analyze situations.

Q2: Why are case study questions important for understanding class 12 physics chapters?

A2: Case study questions provide a practical context for students to apply theoretical knowledge to real-world situations, fostering deeper understanding and critical thinking skills.

Q3: How should students approach answering case study questions for CBSE?

A3: Students should carefully read the case study, identify the key issues or problems presented, analyze the information provided, apply relevant concepts and principles of chemical reactions and equations, and formulate well-supported solutions or responses.

Q4: Are there any resources available online for students to practice case study questions on class 12 physics chapters for CBSE exams?

A4: Yes, several educational websites offer case study questions for CBSE students preparing for science examinations. We also offer a collection of case study questions for all classes and subject on our website. Visit our  website  to access these questions and enhance your learning experience. If you need more case study questions for your preparation, then you visit Physics Gurukul website.

Q5: How can students effectively prepare for case study questions on “Electric Charges and Fields” for CBSE exams?

A5: Effective preparation strategies include regular revision of concepts, solving practice questions, analyzing case studies from previous exams, seeking clarification on doubts, and consulting with teachers or peers for guidance and support.

Q6: How can teachers incorporate case study questions on “Electric Charges and Fields” class 12 physics into classroom teaching?

A6: Teachers can integrate case studies into lesson plans, group discussions, or interactive activities to engage students in active learning, promote problem-solving skills, and facilitate a deeper understanding of “Electric Charges and Fields”.

Q7: Electrostatic forces are much stronger than gravitational forces. Give one example.

A7: A charged glass rod can lift a piece of paper against the gravitational pull of the earth on this piece.

Q8: What is an electric line of force? What is its importance?

A8: An electric line of force is an imaginary straight or curved path along which a small positive test charge is supposed to move when free to do so. The tangent at a point on an electric line of force gives the direction of the resultant electric field at that point. The relative closeness of electric lines of force in a certain region provides us an estimate of the electric field strength in that region

Q9: Can a charged body attract another uncharged body? Explain.

A9: Yes, a charged body can attract another uncharged body. When the charged body is placed near the uncharged body, the induced charges of opposite kind are produced on the uncharged body and the uncharged body is attract by charged body.

Q10: In Coulomb’s law, on what factors the value of electrostatic force constant K depends?

A10: It depends on the nature of medium between the two charges and also on the system of units.

Q11: Is the mass of a body affected on charging?

A11: Yes, very slightly. The negatively charged body gains mass also along with electrons.

Q12: How does a free electron at rest move in an electric field?

A12: When the electron is released, it will move in a direction opposite to the direction of electric field.

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Case Study Questions Class 12 Physics

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CBSE will ask two Case Study Questions in the CBSE class 12 Physics questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them. You can download CBSE Class 12 Physics case study questions from the myCBSEguide App and our free student dashboard .

You can Score High

CBSE class 12 Physics question paper will carry questions for 70 marks. Certainly, the question paper is a bit easier this year. It is because the syllabus is already reduced and there are more internal choices. Besides this, the case study questions are a plus to winning the game with good marks.

In simple words, all circumstances are in favour of the sincere students who are working hard to score high this year. Although it has been a difficult time for students as they were not getting the personal attention of the teachers. We know that online classes have their own limits, but we still expect better scores, especially from students who are putting extra effort into their studies.

Class 12 Physics Case Study Questions

CBSE class 12 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, class 12 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line and then you should practice as many questions as possible.

Case Study Syllabus

We know that CBSE has reduced the syllabus. Hence, practice only relevant questions. don’t waste time on case study questions from deleted portion. It is of no use. You can download the latest class 12 Physics case study questions from the myCBSEguide App.

Physics Case Studies

Class 12 Physics has many chapters but all chapters are not important for case studies. As we know case studies are not exactly given from NCERT books but these may be extracted from some newspaper articles, magazines, journals or other books. So, it is very much important that you are studying only the most relevant case studies. Here, the myCBSEguide app helps you a lot. We have case study questions that are prepared by a team of expert teachers. These experts exactly know what types of questions can come in exams.

Case Study Questions

There are a number of study apps available over the internet. But if you are a CBSE student and willing to get an app for the CBSE curriculum, you have very limited options. And if you want an app that is specifically designed for CBSE students, your search will definitely end on finding myCBSEguide. Case study questions are the latest updates in CBSE syllabus. It is exclusively available in the myCBSEguide app.

Here are some example questions. For more questions, you can download the myCBSEguide App.

Physics Case Study -1

Read  the following source and answer any four out of the following questions: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charges positive and negative charges. Also, like charges repel each other whereas unlike charges attract each other.

• -3.2  × ×  10 -18  C
• 3.2  × ×  10  18  C
• -3.2  × ×  10 -17  C
• 3.2  × ×  10  -17   C
• -1.6  × ×  10 -18  C
• 1.6  × ×  10  -18  C
• 2.6  × ×  10 -18  C
• 1.6  × ×  10 -21  C
• 9.1  × ×  10 -31  kg
• 9.1  × ×  10 -31  g
• 1.6  × ×  10 -19  kg
• 1.6  × ×  10 -19  g
• there is only a positive charge in the body
• there is positive as well as negative charge in the body but the positive charge is more than the negative charge
• there is equally positive and negative charge in the body but the positive charge lies in the outer regions
• the negative charge is displaced from its position
• valence electrons only
• electrons of inner shells
• both valence electrons and electrons of the inner shell.
• none of the above

Physics Case Study -2

Read the following source and answer any four out of the following questions: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. Also, Resistivity is the electrical resistance of a conductor of unit cross-sectional area, and unit length. … A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents.

• nature of material
• temperature
• dimensions of material
• cross-sectional area
• length of wire
• wire’s nature
• all of the above
• more resistance
• less resistance
• same resistance

Physics Case Study -3

Read the source given below and answer any four out of the following questions: The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.

• The energy of the electrons in the orbit is quantized
• The electron in the orbit nearest the nucleus has the lowest energy
• Electrons revolve in different orbits around the nucleus
• The position and velocity of the electrons in the orbit cannot be determined simultaneously
• Single proton
• Multiple electrons
• emitted only
• absorbed only
• both (a) and (b)
• none of these
•  It must emit a continuous spectrum
•  It loses its energy
• Gaining its energy
• A discrete spectrum
• dequantized

Physics Case Study & myCBSEguide App

We at myCBSEguide provide the best case study questions for CBSE class 12 Physics. We have Physics case study questions for every chapter in 12th class Physics. Students can access the Physics case study questions with answers on the myCBSEguide App or on the student dashboard . Here are some features that make myCBSEguide the best learning app for CBSE students:

• Updated syllabus
• Up to date question bank
• Model papers and 10-year questions
• NCERT and Exemplar sulutions
• Best quality learning videos
• Detailed revision notes

12 Physics Question Paper Design

Here is the question paper design for CBSE class 12 Physics. It shows the typology of the questions and their weightage in CBSE board exams.

QUESTION PAPER DESIGN Theory (Class: 12)

Maximum Marks: 70 Duration: 3 hrs

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 typology of questions the same.

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Good question l will check answers sheet

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Case Study Question for Class 12 Physics Chapter 1 Electric Charges and Fields

• Last modified on: 2 years ago
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There is Case Study Questions in class 12 Physics in session 2020-21. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. Case Study Questions will have 5 MCQs out of which students will have to attempt any 4 questions. Here are the questions based on case study.

Case Study Question 1:

Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in given figure, the electric field at P is stronger than at Q.

(i) Electric lines of force about a positive point charge are (a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines

(ii) Which of the following is false for electric lines of force? (a) They always start from positive charge and terminate on negative charges. (b) They are always perpendicular to the surface of a charged conductor. (c) They always form closed loops. (d) They are parallel and equally spaced in a region of uniform electric field.

(iii) Which one of the following patterns of electric line of force is not possible in field due to stationary charges?

(iv) Electric field lines are curved (a) in the field of a single positive or negative charge (b) in the field of two equal and opposite charges. (c) in the field of two like charges. (d) both (b) and (c)

(v) The figure below shows the electric field lines due to two positive charges. The magnitudes E A , E B and E C of the electric fields at point A, B and C respectively are related as

(a) E A >E B >E C (b) E B >E A >E C (c) E A =E B >E C (d) E A >E B =E C

Case Study Question 2:

Smallest charge that can exist in nature is the charge of an electron. During friction it is only the transfer of electron which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron (1.6 x 10 -19 C) i.e., q=±ne where r= 1, 2, 3, 4 …. Hence no body can have a charge represented as 1.8e, 2.7e, 2e/5, etc. Recently, it has been discovered that elementary particles such as protons or neutrons are elemental units called quarks.

(i) Which of the following properties is not satisfied by an electric charge? (a) Total charge conservation. (b) Quantization of charge. (c) Two types of charge. (d) Circular line of force.

(ii) Which one of the following charges is possible? (a) 5.8 x 10 -18 C (b) 3.2 x 10 -18 C (c) 4.5 x 10 -19 C (d) 8.6 x 10 -19 C

(iii) If a charge on a body is 1 nC, then how many electrons are present on the body? (a) 6.25 x 10 27 (b) 1.6 x 10 19 (c) 6.25 X 10 28 (d) 6.25 X 10 9

(iv) If a body gives out 10 9 electrons every second, how much time is required to get a total charge of 1 from it? (a) 190.19 years (b) 150.12 years (c) 198.19 years (d) 188.21 years

(v) A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10 -7 C. Calculate the number of electrons transferred. (a) 2 x 10 12 (b) 3 x 10 12 (c) 2 x 10 14 (d) 3 x 10 14

Case Study Question 3:

When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However some work is done in rotating the dipole against the torque acting on it.

(i) The dipole moment of a dipole in a uniform external field Ē is B. Then the torque τ acting on the dipole is (a) τ=p x E (b) τ = P. Ē (c) τ = 2(p + Ē) (d) τ = (P + E)

(ii) An electric dipole consists of two opposite charges, each of magnitude 1.0 μC separated by a distance of 2.0 cm. The dipole is placed in an external field of 10 5 NC -1 . The maximum torque on the dipole is (a) 0.2 x 10 -3 Nm (b) 1x 10 -3 Nm (c) 2 x 10 -3 Nm (d) 4x 10 -3 Nm

(iii) Torque on a dipole in uniform electric field is minimum when θ is equal to (a) 0° (b) 90° (c) 180° (d) Both (a) and (c)

(iv) When an electric dipole is held at an angle in a uniform electric field, the net force F and torque τ on the dipole are (a) F= 0, τ = 0 (b) F≠0, τ≠0 (c) F=0, τ ≠ 0 (d) F≠0, τ=0

(v) An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle with the direction of the field. Assuming that potential energy of the dipole to be zero when 0 = 90°, the torque and the potential energy of the dipole will respectively be (a) pEsinθ, -pEcosθ (b) pEsinθ, -2pEcosθ (c) pEsinθ, 2pEcosθ (d) pEcosθ, – pEsinθ

Case Study Question 4:

A charge is a property associated with the matter due to which it experiences and produces an electric and magnetic field. Charges are scalar in nature and they add up like real number. Also, the total charge of an isolated system is always conserved. When the objects rub against each other charges acquired by them must be equal and opposite.

(i) The cause of a charging is: (a) the actual transfer of protons. (b) the actual transfer of electrons. (c) the actual transfer of neutrons. (d) none the above

(ii) Pick the correct statement. (a) The glass rod gives protons to silk when they are rubbed against each other. (b) The glass rod gives electrons to silk when they are rubbed against each other. (c) The glass rod gains protons from silk when they are rubbed against each other. (d) The glass rod gains electrons when they are rubbed against each other.

(iii) If two electrons are each 1.5 × 10 –10 m from a proton, the magnitude of the net electric force they will exert on the proton is (a) 1.97 × 10 –8 N (b) 2.73 × 10 –8 N (c) 3.83 × 10 –8 N (d) 4.63 × 10 –8 N

(iv) A charge is a property associated with the matter due to which it produces and experiences: (a) electric effects only (b) magnetic effects only (c) both electric and magnetic effects (d) none of these.

(v) The cause of quantization of electric charges is: (a) Transfer of an integral number of neutrons. (b) Transfer of an integral number of protons. (c) Transfer of an integral number of electrons. (d) None of the above.

Case Study Question 5:

Surface Charge Density. Surface charge density is defined as the charge per unit surface area the surface (Arial) charge symmetric distribution and follow Gauss law of electro statics mathematical term of surface charge density σ=ΔQ/ΔS

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign (± s). Having magnitude 8.8 × 10 –12 cm –2 as shown here. The intensity of electrified at a point is E =σ/ε 0   and flux is Φ=E.ΔS, where ΔS = 1 m 2 (unit arial plate)

(i) E in the outer region (I) of the first (A) plate is (a) 1.7 × 10 –22 N/C (b) 1.1 × 10 –12 V/m (c) Zero (d) Insufficient data

(ii) E in the outer region (III) of the second plate (B) is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) zero

(iii) E between (II) the plate is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) None of these

(iv) The ratio of E from left side of plate A at distance 1 cm and 2 m respectively is (a) 1 : 2 (b) 10 : 2 (c) 1 : 1 (d) 20 : 1

(v) In order to estimate the electric field due to a thin finite plane metal plate the Gaussian surface considered is (a) Spherical (b) Linear (c) Cylindrical (d) Cybic

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CBSE Class 12th - PHYSICS : Chapterwise Case Study Question & Solution

CBSE will ask two Case Study Questions in the CBSE class 12 Physics questions paper. Question numbers 15 and 16 are cased-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them.

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CBSE Class 12 Maths Exam 2025 : Chapter-Wise Important MCQs Questions with Answers; Download PDF

CBSE focuses on MCQs, which carry one mark and are now more important in exams. Class 12 Maths MCQs are especially tricky, involving calculations despite their short format. Proper practice across difficulty levels is crucial for scoring well. These questions, widely used in school and competitive exams, help assess learning effectively and improve students' grades with consistent preparation.

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Class 12 Physics Case Study Questions of Chapter 1 Electric Charges and Fields

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 1 Electric Charges and Fields Case Study and Passage-Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Case Study Questions   Electric Charges and Fields  to know their preparation level.

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In CBSE Class 12 Physics Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electric Charges and Fields Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Physics  Chapter 1 Electric Charges and Fields

Case Study/Passage-Based Questions

Answer: (a) 6.40 x 10-19 C

(ii) Extra electrons on this particular oil drop (given the presently known charge of the electron) are

Answer: (a) 4

(iii) A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m -1 .If the mass of the drop is 1.6 X 10 -3  g, the number of electrons carried by the drop is (g= 10 m s -2 )

Answer: (c) 1012 .

(iv) The important conclusion given by Millikan’s experiment about the charge is

Answer: (c) charge is quantized

(v) If in Millikan’s oil drop experiment, charges on drops are found to be 8μC,12μC,20μC8μC,12μC,20μC then quanta of charge is

Answer:  (d) 4μC

Case Study 2: Surface Charge Density. Surface charge density is defined as the charge per unit surface area the surface (Arial) charge symmetric distribution and follow Gauss law of electrostatics mathematical term of surface charge density σ=ΔQ/ΔS

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign (± s). Having magnitude 8.8 × 10 –12  cm –2  as shown here. The intensity of electrified at a point is E =σ/ε 0   and flux is Φ=E.ΔS, where ΔS = 1 m 2  (unit arial plate)

(i) E in the outer region (I) of the first (A) plate is (a) 1.7 × 10 –22  N/C (b) 1.1 × 10 –12  V/m (c) Zero (d) Insufficient data

Ans. (c) Zero C

(ii) E in the outer region (III) of the second plate (B) is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) zero

Ans. (d) Zero

(iii) E between (II) the plate is (a) 1 N/C (b) 0.1 V/m (c) 0.5 N/C (d) None of these

Ans. (d) None of these

(iv) The ratio of E from left side of plate A at distance 1 cm and 2 m respectively is (a) 1 : 2 (b) 10 : 2 (c) 1 : 1 (d) 20 : 1

Ans. (c) 1 : 1

(v) In order to estimate the electric field due to a thin finite plane metal plate the Gaussian surface considered is (a) Spherical (b) Linear (c) Cylindrical (d) Cybic

Ans. (c) Cylindrical

Case Study 3: Electric Charges and Fields, focuses on the basic properties of electric charges, the concept of electric fields, and the forces acting within these fields. The chapter introduces the idea that like charges repel, and unlike charges attract, a fundamental law of electrostatics. It also discusses the principle of superposition, which states that the net electrostatic force experienced by a charge due to multiple charges is simply the vector sum of the individual forces exerted by each charge independently. Further, the chapter introduces the concept of an electric field, defined as the electric force experienced by a unit positive charge at a point in space, and illustrates the way these fields are represented using field lines.

What is the basic property of electric charges?

A) Like charges repel, and unlike charges attract.

B) Like charges attract, and unlike charges repel.

C) All charges repel.

D) All charges attract.

Ans. A) Like charges repel, and unlike charges attract.

What does the principle of superposition state in terms of electrostatic forces?

A) The net electrostatic force is equal to the sum of the individual forces.

B) The net electrostatic force is equal to the difference of the individual forces.

C) The net electrostatic force is independent of the individual forces.

D) The net electrostatic force is inversely proportional to the individual forces.

Ans. A) The net electrostatic force is equal to the sum of the individual forces.

What is an electric field?

A) The region around a charge where magnetic force is experienced.

B) The region around a charge where gravitational force is experienced.

C) The region around a charge where electrostatic force is experienced.

D) The region around a charge where nuclear force is experienced.

Ans. C) The region around a charge where electrostatic force is experienced.

How is the electric field defined?

A) As the electric force experienced by a unit negative charge.

B) As the electric force experienced by a unit positive charge.

C) As the gravitational force experienced by a unit positive charge.

D) As the magnetic force experienced by a unit positive charge.

Ans. B) As the electric force experienced by a unit positive charge.

How are electric fields represented?

A) By field lines.

B) By electric charges.

C) By vectors.

D) By matrices.

Ans. A) By field lines.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Physics Chapter 1 Electric Charges and Fields with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Physics Electric Charges and Fields Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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CBSE 12th Standard Physics Subject Alternating Current Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

QB365 - Question Bank Software

Cbse 12th standard physics subject alternating current case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

(ii) The value of impedance is

(iii) What is the value of current in the circuit?

(iv) What is the value of the phase angle between current and voltage?

(v) From graph, which one is true from following?

A transformer is essentially an a.c. device. It cannot work on d.c. It changes alternating voltages or currents. It does not affect the frequency of a.c. It is based on the phenomenon of mutual induction. A transformer essentially consists of two coils of insulated copper wire having different number of turns and wound on the same soft iron core.  The number of turns in the primary and secondary coils of an ideal transformer are 2000 and 50 respectively. The primary coil is connected to a main supply of 120 V and secondary coil is connected to a bulb of resistance  \(0.6 \Omega\) (i) The value of voltage across the secondary coil is

(ii) The value of current in the bulb is

(iii) The value of current in primary coil is

(iv) Power in primary coil is

(v) Power in secondary coil is

(ii) The natural frequency of the circuit is

(iii) At what time is the energy stored completely electrical?

(iv) At what time is the energy stored completely magnetic?

(v) The value of X L is

(ii) The line power loss in the form of heat is

(iii) How much power must the plant supply, assuming there is negligible power loss due to leakage?

(iv) The voltage drop in the power line is

(v) The total value of voltage transmitted from the plant is

(ii) Transformer is used to

(iii) The number of turns in primary coil of a transformer is 20 and the number of turns in a secondary is 10. If the voltage across the primary is 220 ac V, what is the voltage across the secondary?

(iv) In a transformer the number of primary turns is four times that of the secondary turns. Its primary is connected to an a.c. source of voltage V. Then

(v) A transformer is used to light 100 W-110 V lamp from 220 V mains. If the main current is 0.5 A, the efficiency of the transformer is

*****************************************

Cbse 12th standard physics subject alternating current case study questions 2021 answer keys.

(i) (b): Given:  \(R=12 \Omega, X_{C}=14 \Omega, L=0.1 \mathrm{H}\) \(X_{L}=\omega L=2 \pi \cup L=2 \times 3.14 \times 50 \times 0.1=31.4 \Omega\) (ii) (d): Impedance  \(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\) \(=\sqrt{(12)^{2}+(31.4-14)^{2}}=21.13 \Omega\) (iii) (d):   \(I_{v}=\frac{E_{v}}{Z}=\frac{200 \mathrm{~V}}{21.13}=9.46 \mathrm{~A}\) (iv) (c):  \(\tan \phi=\frac{X_{L}-X_{C}}{R}=\frac{31.4-14}{12}=1.45\) \(\phi=\tan ^{-1}(1.45)=55^{\circ} 4^{\prime}\) (v) (c)

(i) (c) : As  \(\frac{E_{s}}{E_{p}}=\frac{n_{s}}{n_{p}} \Rightarrow E_{s}=E_{p} \cdot \frac{n_{s}}{n_{p}}\) \(=\frac{120 \times 50}{2000}=3 \mathrm{~V}\) (ii) (d) :  \(I_{s}=\frac{E_{s}}{R} \Rightarrow I_{s}=\frac{3}{0.6}=5 \mathrm{~A}\) (iii) (a) : As  \(\frac{I_{p}}{I_{s}}=\frac{E_{s}}{E_{p}}\) \(\Rightarrow I_{p}=\frac{E_{s}}{E_{p}} \times I_{s}=\frac{\ 3}{120} \times 5=0.125 \mathrm{~A}\) (iv) (d) : Power in primary  \(P_{p}=E_{p} \times I_{p}=120 \times 0.125\) = I5W (v) (a) : Power in secondary coil  \(P_{s}=E_{s} \times I_{s}=3 \times 5\) = 15W

(i) (d) :Energy,  \(E=\frac{1}{2} \frac{Q^{2}}{C}=\frac{\left(10 \times 10^{-3}\right)^{2}}{2 \times 50 \times 10^{-6}}=1 \mathrm{~J}\) (ii) (a): Frequency  \(v=\frac{1}{2 \pi \sqrt{L C}}\) \(=\frac{1}{2 \pi \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}=\frac{10^{3}}{2 \pi}=159.24 \mathrm{~Hz}\) (iii) (d) : Total time period  \(T=\frac{1}{v}=\frac{1}{159.24}=6.28 \mathrm{~ms}\) Total charge on capacitor at time t  \(Q^{\prime}=Q \cos \frac{2 \pi}{T} t\) For energy stored is electrical, we can write  \(Q^{\prime}=\pm Q\) Hence, energy stored in the capacitor is completely electrical at  \(t=0, \frac{T}{2}, T, \frac{3 T}{2}, \ldots .\) (iv) (d) : Magnetic energy is maximum when electrical energy is equal to zero Hence  \(t=\frac{T}{4}, \frac{3 T}{4}, \frac{5 T}{4}\) (v) (a):   \(X_{L}=\omega L=2 \pi \cup L=2 \times 3.14 \times 159.24 \times 20\)  x 10 -3 \(\Rightarrow \quad X_{L}=20 \Omega\)

(i) (d): Resistance of the two wire lines carrying power  \(=0.5 \Omega / \mathrm{km}\) Total resistance  \(=(15+15) 0.5=15 \Omega\) (ii) (c): Line power loss = I 2 R RMS current in the coil, \(I=\frac{P}{V_{1}}=\frac{800 \times 10^{3}}{4000}=200 \mathrm{~A}\) \(\therefore\)  Power loss  \(=(200)^{2} \times 15=600 \mathrm{~kW}\) (iii) (d): Assuming that the power loss is negligible due to the leakage of the current. The total power supplied by the plant = 800 kW + 600 kW = 1400 kW (iv) (b): Voltage drop in the power line = IR = 200 x 15 = 3000 V (v) (d): Total voltage transmitted from the plant = 3000 V + 4000 V = 7000 V

(i) (c) :In an ideal transformer, there is no power loss. The efficiency of an ideal transformer is  \(\eta=1(i . e\)  100%) i.e. input power = output power. (ii) (d): Transformer is used to obtain desired ac voltage and current. (iii) (c): For a transformer  \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\) where Ndenotes number of turns and V = voltage \(\therefore \frac{V_{s}}{220}=\frac{10}{20} \quad \therefore V_{s}=110 \mathrm{ac} \mathrm{V}\) (iv) (a) : In a transformer the primary and secondary currents are related by \(I_{s}=\left(\frac{N_{p}}{N_{s}}\right) I_{p}\) and the voltages are related by \(V_{s}=\left(\frac{N_{s}}{N_{p}}\right) V_{p}\) where subscripts p and s refer to the primary and secondary of the transformer Here,  \(V_{p}=V, \frac{N_{p}}{N_{s}}=4 \quad \therefore \quad I_{s}=4 I_{p}\) and  \(V_{s}=\left(\frac{1}{4}\right) V=\frac{V}{4}\) (v) (c) : The efficiency of the transformer is  \(\eta=\frac{\text { Output power }\left(P_{\text {out }}\right)}{\text { Input power }\left(P_{\text {in }}\right)} \times 100\) Here,  \(P_{\text {out }}=100 \mathrm{~W}, P_{\text {in }}=(220 \mathrm{~V})(0.5 \mathrm{~A})=110 \mathrm{~W}\) \(\therefore \quad \eta=\frac{100 \mathrm{~W}}{110 \mathrm{~W}} \times 100 \approx 90 \%\)

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