CBSE Important Questions for Class 12 Physics - 2025-26

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.

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

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• 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. 

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