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CBSE Class 12 Physics Important Questions

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Important Questions and Answers for CBSE Physics Class 12 - FREE PDF Download

CBSE Class 12 Physics Important Questions are a great way for students to prepare for their exams. These questions highlight the most important topics from each chapter according to the latest CBSE Class 12 Physics Syllabus, making it easier to focus on what matters. By working through these questions, students can strengthen their understanding and get comfortable with different types of problems.

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These important questions help boost confidence as students practice. They can see how different concepts are connected and improve their problem-solving skills. With this resource, students can study effectively and be well-prepared to do their best in the exams.


CBSE Class 12 Physics Chapter-wise Important Questions

CBSE Class 12 Physics Chapter-wise Important Questions and Answers cover topics from all 14 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.


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10 Most Important Questions From Class 12 Physics Chapters

1. What is the force between two small charged spheres of charges 

$2\times {{10}^{-7}}C$ and $3\times {{10}^{-7}}C$ placed $30cm$ apart in air?

Ans: We are given the following information:

Repulsive force of magnitude,$F=6\times {{10}^{-3}}N$

Charge on the first sphere, ${{q}_{1}}=2\times {{10}^{-7}}C$

Charge on the second sphere, ${{q}_{2}}=3\times {{10}^{-7}}C$

Distance between the two spheres, $r=30cm=0.3m$

The electrostatic force between the two spheres is given by Coulomb’s law as,

$F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Where, ${{\varepsilon }_{0}}$is the permittivity of free space and, 

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

Now on substituting the given values, Coulomb’s law becomes, 

\[F = \frac{9\times  10^{9}\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^{2}}\]

Therefore, we found the electrostatic force between the given charged spheres to be $F=6\times {{10}^{-3}}N$. Since the charges are of the same nature, we could say that the force is repulsive.

2. Three capacitors of capacitance \[2pF,\text{ }3\text{ }pF\text{ }and\text{ }4pF\] are connected in parallel. 

a) What is the total capacitance of the combination?

Ans: Provided that,

Capacitances of the given capacitors are, 

\[{{C}_{1}}=2pF\text{ };\text{ }{{C}_{2}}=3pF\text{ };\text{ }{{C}_{3}}=4pF\]

The formula for equivalent capacitance $(~C')$ of the capacitors’ parallel combination is given by

\[C'={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]

\[\Rightarrow C'=2+3+4=9pF\]

Therefore, the total capacitance of the combination is \[9pF.\]

b) Determine the charge on each capacitor if the combination is connected to a \[100\text{ }V\] supply.

Ans: We have,

Supply voltage, \[V=100\text{ }V\]

Charge on a capacitor with capacitance C and potential difference V is given by,

$q=CV$……(i)

For $C=2pF$,

Charge $=VC=100\times 2=200pF$

For $C=3pF$,

Charge $=VC=100\times 3=300pF$

For $C=4pF$,

Charge $=VC=100\times 4=400pF$


3. At room temperature ${{27.0}^{\circ }}C$, the resistance of a heating element is $100\Omega $. What is the temperature of the element if the resistance is found to be $117\Omega $, given that the temperature coefficient of the material of the resistor is $1.70\times {{10}^{-4}}^{\circ }{{C}^{-1}}$?

Ans: In the above question it is given that at room temperature $(T={{27.0}^{\circ }}C)$, the resistance of the heating element is $100\Omega $ (say R).

Also, the heating element’s temperature coefficient is given to be $\alpha =1.70\times {{10}^{-4}}{}^\circ {{C}^{-1}}$.

Now, it is said that the resistance of the heating element at an increased temperature (say ${{T}_{1}}$) is $117\Omega $ (say ${{R}_{1}}$). To compute this unknown increased temperature ${{T}_{1}}$, the formula for the temperature coefficient of a material can be used. It is known that the temperature co-efficient of a material provides information on the nature of that material with respect to its change in resistance with temperature. Mathematically,

$\alpha =\frac{{{R}_{1}}-R}{R\left( {{T}_{1}}-T \right)}$

$\Rightarrow {{T}_{1}}-T=\frac{{{R}_{1}}-R}{R\alpha }$

Substituting the given values,

$\Rightarrow {{T}_{1}}-27=\frac{117-100}{100\times 1.70\times {{10}^{-4}}}$

$\Rightarrow {{T}_{1}}-27=1000$

$\Rightarrow {{T}_{1}}={{1027}^{\circ }}C$

Clearly, it is at ${{1027}^{\circ }}C$ when the resistance of the element is $117\Omega $.


4. A $3.0cm$ Wire Carrying a Current of $10A$ is Placed Inside a Solenoid Perpendicular to Its Axis. The Magnetic Field Inside the Solenoid Is Given to Be $0.27T$. What is the Magnetic Force on the Wire?

Ans: We are given the following, 

Length of the wire, $l=3cm=0.03m$

Current flowing in the wire, $I=10A$

Magnetic field, $B=0.27T$

Angle between the current and magnetic field, $\theta =90{}^\circ $

(Since the magnetic field produced by a solenoid is along its axis and the current carrying wire is kept perpendicular to the axis) 

The magnetic force exerted on the wire is given as, 

$F=BIl\sin \theta $

Substituting the given values, 

$F=0.27\times 10\times 0.03\sin 90{}^\circ $

$\Rightarrow F=8.1\times {{10}^{-2}}N$

Clearly, the magnetic force on the wire is found to be $8.1\times {{10}^{-2}}N$. The direction of the force can be obtained from Fleming’s left-hand rule. 


5. A long solenoid with 15 turns per cm has a small loop of area $2.0c{{m}^{2}}$placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from \[2.0A\] to \[4.0A\]in\[0.1s\], what is the induced emf in the loop while the current is changing?

Ans: We are given the following information:

Number of turns on the solenoid \[=15turns/cm=1500turns/m\]

Number of turns per unit length, \[n=1500\text{ }turns\]

The solenoid has a small loop of area,$A=2.0c{{m}^{2}}=2\times {{10}^{-4}}{{m}^{2}}$

Current carried by the solenoid changes from\[2A\text{ }to\text{ }4A\].

Now, the change in current in the solenoid, \[di=4-2=2A\]

Change in time, \[dt=0.1\text{ }s\]

Induced emf in the solenoid could be given by Faraday’s law as:

$\varepsilon =\frac{d\phi }{dt}$………………………… (1)

Where, induced flux through the small loop, $\phi =BA$…………… (2) 

Equation (1) would now reduce to:

$\varepsilon =\frac{d}{dt}\left( BA \right)=A{{\mu }_{0}}n\times \left( \frac{di}{dt} \right)$

Substituting the given values into this equation, we get, 

$\varepsilon =2\times {{10}^{-4}}\times 4\pi \times {{10}^{-7}}\times 1500\times \frac{2}{0.1}$

$\therefore \varepsilon =7.54\times {{10}^{-6}}V$

Therefore, the induced voltage in the loop is found to be, $\varepsilon =7.54\times {{10}^{-6}}V$. 


6. Obtain the resonant frequency ${{\omega }_{r}}$ of a series LCR circuit with $L=2.0H$, $C=32\mu F$ and $R=10\Omega $ . What is the Q-value of this current?

Ans: It is given that,

Inductance, $L=2H$ 

Capacitance, \[C=32\mu F=32\times {{10}^{-6}}F\] 

$R=10\Omega $ 

It is known that,

Resonant frequency, ${{\omega }_{r}}=\frac{1}{\sqrt{LC}}$ 

$\Rightarrow {{\omega }_{r}}=\frac{1}{\sqrt{2\times 32\times {{10}^{-6}}}}$ 

$\Rightarrow {{\omega }_{r}}=\frac{1}{8\times {{10}^{-3}}}$

$\Rightarrow {{\omega }_{r}}=125{rad}/{s}\;$

$Q-value=\frac{{{\omega }_{r}}L}{R}$ 

$\Rightarrow Q=\frac{1}{R}\sqrt{\frac{L}{C}}$ 

$\Rightarrow Q=\frac{1}{10}\sqrt{\frac{2}{32\times {{10}^{-6}}}}$

$\Rightarrow Q=\frac{1}{10\times 4\times {{10}^{-3}}}$

$\Rightarrow Q=25$ 

Therefore, the resonant frequency is $125rad/s$ and Q-value is $25$.


7. The Amplitude of the Magnetic Field Part of a Harmonic Electromagnetic Wave in Vacuum is B0 = 510nT. What is the Amplitude of the Electric Field Part of the Wave?

Ans: The amplitude of the magnetic field of an electromagnetic wave in a vacuum, is:

${B_0} = 510nT = 510 \times {10^{ - 9}}T$

Speed of light in a vacuum, $c = 3 \times {10^8}m/s$

The amplitude of the electric field of the electromagnetic wave can be given as:

$E = cB = 3 \times {10^8} \times 510 \times {10^{ - 9}} = 153N/C$.


8. In Young's double-slit experiment using monochromatic light of wavelength \[\lambda \]. The intensity of light at a point on the screen where path difference is \[\lambda \], is \[K\] units. What is the intensity of light at a point where the path difference is \[\frac{\lambda }{3}\]?

Ans: The intensity of the two light waves be \[{{I}_{1}}\] and \[{{I}_{2}}\]. Their resultant intensities can be evaluated as: \[{I}'={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \]

Where,

\[\phi =\]The phase difference between two waves for monochromatic light waves,

Since \[{{I}_{1}}={{I}_{2}}\]

So \[{I}'=2{{I}_{1}}+2{{I}_{1}}\cos \phi \]

The formula for phase difference can be given as:

\[Phase\,difference=\frac{2\pi }{\lambda }\times \,Path\,difference\]

Since, path difference is \[\lambda \],

Phase difference is \[\phi =2\pi \]

\[{I}'=2{{I}_{1}}+2{{I}_{1}}=4{{I}_{1}}\]

Given,

\[{{I}_{1}}=\frac{{{K}'}}{4}\] …… (1)

When path difference \[=\frac{\lambda }{3}\]

phase difference, \[\phi =\frac{2\pi }{3}\]

Hence, the resultant intensity,

\[{{I}_{R}}={{I}_{1}}+{{I}_{1}}+2\sqrt{{{I}_{1}}{{I}_{1}}}\cos \frac{2\pi }{3}\]

\[\Rightarrow {{I}_{R}}=2{{I}_{1}}+2{{I}_{2}}\left( -\frac{1}{2} \right)={{I}_{1}}\]

Using equation (1), we can state that

\[{{I}_{R}}={{I}_{1}}=\frac{K}{4}\]

Hence, for monochromatic light waves, the intensity of light at a point where the path difference is \[\frac{\lambda }{3}\] is \[\frac{K}{4}\]units.


9. Obtain the binding energy (in MeV) of a nitrogen nucleus $\left( {}_{7}^{14}N \right)$, given $m\left( {}_{7}^{14}N \right)=14.00307u$

Ans: We are given:

Atomic mass of nitrogen $\left( {}_{7}{{N}^{14}} \right)$, $m=14.00307u$

A nucleus of ${}_{7}{{N}^{14}}$nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus would be, $\Delta m=7{{m}_{H}}+7{{m}_{n}}-m$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$

Substituting these values into the above equation, we get, 

$\Delta m=7\times 1.007825+7\times 1.008665-14.00307$

$\Rightarrow \Delta m=7.054775+7.06055-14.00307$

$\therefore \Delta m=0.11236u$

But we know that, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow \Delta m=0.11236\times 931.5MeV/{{c}^{2}}$

Now, we could give the binding energy as, 

${{E}_{b}}=\Delta m{{c}^{2}}$

Where, $c=\text{speed of light =3}\times \text{1}{{\text{0}}^{8}}m{{s}^{-2}}$

Now, ${{E}_{b}}=0.11236\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b}}=104.66334MeV$

Therefore, we found the binding energy of a Nitrogen nucleus to be $104.66334MeV$. 


10. Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of $10$ and the second has a voltage gain of $20$. If the input signal is $0.01$ volt, calculate the output AC signal.

Ans: It is given that,

Voltage gain of the first amplifier, ${{V}_{1}}=10$

Voltage gain of the second amplifier, ${{V}_{2}}=20$

Input signal voltage, ${{V}_{i}}=0.01V$

Output signal voltage, ${{V}_{o}}=?$

When two amplifiers are connected one after another in series (cascaded) the total voltage gain is the product of the voltage gains of both amplifiers.

$\Rightarrow V={{V}_{1}}\times {{V}_{2}}$

$\Rightarrow V=10\times 20$

$\Rightarrow V=200$

Voltage amplification, $V=\frac{{{V}_{o}}}{{{V}_{i}}}$

$\Rightarrow 200=\frac{{{V}_{o}}}{0.01}$

$\Rightarrow {{V}_{o}}=200\times 0.01$

$\Rightarrow {{V}_{o}}=2V$

Thus, the output AC signal for the given amplifier is $2V$.

Here are 10 important questions from all chapters. For more detailed, chapter-wise questions, refer to the table above and download the PDF. This will help you gain a better understanding of each chapter and enhance your exam preparation.


How do Physics Important Questions Class 12 Help you with Exams?

  • Physics important questions help you focus on key topics that are likely to appear in exams. This makes your study time more effective.

  • Practising these questions reinforces your understanding of essential concepts. It helps you see how different topics are related.

  • Regular practice with important questions boosts your confidence. You become more familiar with the exam format and types of questions.

  • These questions help you identify areas where you need more practice. Knowing your weak points allows you to focus on them.

  • Working through important questions improves your problem-solving skills. You learn different methods for tackling various physics problems.

  • Using these questions for revision allows for quick and efficient study sessions. You can review key concepts without getting overwhelmed.

  • Important questions prepare you for different formats, including multiple-choice and long-answer questions. This variety ensures you're ready for anything on the exam.

  • Discussing important questions with classmates can enhance your understanding. Sharing ideas helps clarify doubts and improves learning.


Class 12 Physics Important Questions are a valuable resource for students preparing for their exams. These questions focus on key concepts and topics, making it easier to study and understand the material. By practising these important questions, students can build their confidence, improve their problem-solving skills, and get familiar with the exam format. Using these resources effectively can lead to better results and a stronger grasp of Physics concepts. 


Additional Study Materials for Class 12 Physics 

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FAQs on CBSE Class 12 Physics Important Questions

1. What are "important questions" for the CBSE Class 12 Physics exam 2025-26?

Important questions for the CBSE Class 12 Physics exam 2025-26 are a curated set of questions that have a high probability of appearing in the board exams. They are selected based on an analysis of previous year papers, topic weightage as per the latest syllabus, and recurring conceptual themes. These questions cover key derivations, numericals, and conceptual problems from the entire NCERT syllabus.

2. Which chapters carry the most weight in the Class 12 Physics board exam?

As per the CBSE pattern, certain units hold higher weightage. For Class 12 Physics, Optics (covering Ray Optics and Wave Optics) and Electrostatics (including Electric Charges and Fields, and Potential and Capacitance) are typically the most scoring units. Current Electricity and Magnetic Effects of Current & Magnetism also contribute significantly to the total marks.

3. How are these important questions different from the questions in the NCERT textbook?

While NCERT textbook questions form the foundation, "important questions" are a more focused subset. They differ in a few ways:

  • Frequency: Important questions are often those that have been asked repeatedly in past CBSE board exams.
  • Exam-Orientation: They are framed to match the board's question style, including HOTS (High Order Thinking Skills) and application-based problems.
  • Prioritisation: They help you prioritise topics that carry more marks, unlike the uniform chapter-end exercises in the textbook.

4. What is the best way to use a list of important questions for effective revision?

For effective revision, first, ensure you have a strong understanding of the concepts from the NCERT textbook. Then, use the important questions as a self-assessment tool. Try to solve them without looking at the solution, mimicking exam conditions. This helps you identify your weak areas, practise time management, and understand how concepts are tested in the board exam.

5. How do these important questions align with the latest CBSE 2025-26 exam pattern?

Important questions are specifically designed to align with the CBSE Class 12 exam pattern for 2025-26. This means they include a mix of question types as per the official blueprint, such as MCQs, assertion-reasoning questions, short-answer (SA), long-answer (LA), and case-study based questions. They reflect the current marking scheme and the emphasis on conceptual understanding over rote memorisation.

6. Should I focus only on numerical problems or also on derivations and conceptual questions from the important questions list?

A balanced approach is crucial for scoring well in the Physics board exam. You must focus on all three types:

  • Derivations: Key derivations from topics like electromagnetism and optics are frequently asked and carry high marks.
  • Numerical Problems: These test your application of formulas and concepts, especially from chapters like Current Electricity and EMI.
  • Conceptual Questions: These "why" and "how" questions test the depth of your understanding and are essential for a high score.
Important questions provide practice across all these categories.

7. What is the connection between Previous Year Questions (PYQs) and the important questions for the upcoming board exam?

Previous Year Questions (PYQs) are the primary source for creating a list of important questions. Analysing PYQs helps identify recurring patterns, high-frequency topics, and the specific style of questions the CBSE prefers. Therefore, practising important questions is essentially a structured way of practising the most relevant PYQs, making your preparation highly targeted for the 2025-26 board exam.

8. How do important questions help in improving time management during the Physics exam?

Practising a comprehensive list of important questions significantly improves time management. By repeatedly solving questions of varying difficulty and type (e.g., 3-mark numericals, 5-mark derivations), you become familiar with the optimal time to allocate for each section. This practice reduces hesitation and calculation time during the actual exam, allowing you to complete the paper confidently within the stipulated time.

9. Do important questions for CBSE boards also help in preparing for competitive exams like NEET or JEE?

Yes, to a significant extent. The important questions for CBSE boards build a strong conceptual foundation, which is the first step for any competitive exam. While NEET and JEE have a different question pattern and a higher difficulty level, mastering the core concepts through board-level important questions is essential. It helps you clear your fundamentals before moving on to advanced problem-solving.

10. Are there certain "High Order Thinking Skills" (HOTS) questions that are consistently considered important for Class 12 Physics?

Yes, HOTS questions are a key part of the CBSE paper and are often considered important. These questions typically involve applying concepts to unfamiliar scenarios. For Physics, important HOTS questions often come from topics like LCR circuits in AC, applications of Gauss's Law, lens combinations in Optics, and logic gates in Semiconductor Electronics. They test your analytical ability rather than direct recall.