CUET Physics Question Paper PDF, Download Previous Year Papers

CUET 2026 Physics Previous Year Question Paper: The Common University Entrance Test (CUET), conducted yearly by the National Testing Agency (NTA), opens doors to undergraduate programs at India’s leading universities. With CUET 2026 approaching, effective preparation is the key. This Careers360 article offers vital support for CUET 2026 Physics aspirants. Here, you’ll find solved CUET Previous Year Questions, CUET Previous Year Question Papers for Physics, important topics, and essential CUET preparation tips to maximise your chances of success in CUET 2026.

This Story also Contains

1. CUET Physics Previous Year Questions
2. CUET 2026 Physics Important Topics
3. CUET Physics Previous Year Question Paper PDF Download
4. CUET MCQs For All Science Subjects
5. Chapter-wise MCQs for CUET Physics
6. CUET 2026 Physics Preparation Tips
7. CUET 2026 Physics Score Improvement Techniques

CUET Physics Previous Year Questions

Given below is a selection of important questions from previous years' CUET papers. Practising and solving these questions will help you become familiar with the nature of the question paper and the exam pattern. Going through these questions thoroughly is crucial in your CUET preparation for acing the CUET 2026 exam.

1- Two charged particles, placed at a distance d apart in a vacuum, exert a force F on each other. Now, each of the charges is doubled. To keep the force unchanged, the distance between the charges should be changed to_____.

Fill in the blank with the correct answer from the options given below.

(1) $4 d$

(2) $2 d$

(3) $d$

(4) $\frac{d}{2}$

Correct Answer: 2

Solution: To solve this, we apply Coulomb's Law:

$F=\frac{k q_1 q_2}{d^2}$

Now, both charges are doubled, so:

$q_1^{\prime}=2 q_1, \quad q_2^{\prime}=2 q_2$

The new force becomes:

$F^{\prime}=\frac{k\left(2 q_1\right)\left(2 q_2\right)}{d^{\prime 2}}=\frac{4 k q_1 q_2}{d^{\prime 2}}$

To keep the force unchanged ( $F^{\prime}=F^{\prime}$ ):

$\frac{4 k q_1 q_2}{d^{\prime 2}}=\frac{k q_1 q_2}{d^2}$

Cancel $k q_1 q_2$ from both sides:

$\frac{4}{d^{\prime 2}}=\frac{1}{d^2}$

$d^{\prime 2}=4 d^2 \Rightarrow d^{\prime}=2 d$

2- Two parallel plate capacitors of capacitances $2 \mu \mathrm{~F}$ and $3 \mu \mathrm{~F}$ are joined in series and the combination is connected to a battery of V volts. The values of potential across the two capacitors $\mathrm{V}_1$ and $\mathrm{V}_2$ and energy stored in the two capacitors $\mathrm{U}_1$ and $\mathrm{U}_2$ respectively are related as ______ .

Fill in the blank with the correct answer from the options given below.

(1) $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{3}{2}$

(2) $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{2}{3}$

(3) $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{3}{2}$ and $\frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{2}{3}$

(4) $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{2}{3}$ and $\frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{3}{2}$

Correct Answer: 3

Solution: To solve this, we note that the capacitors are connected in series. In a series combination:

Charge on each capacitor is the same: $Q$

Voltage across each capacitor: $V_i = \frac{Q}{C_i}$

Energy stored in each capacitor: $U_i = \frac{1}{2} C_i V_i^2$

Given: $C_1 = 2~\mu\mathrm{F}, \quad C_2 = 3~\mu\mathrm{F}$

Step 1: Find the ratio of voltages.

Since $V_1 = \frac{Q}{C_1}$ and $V_2 = \frac{Q}{C_2}$:

$\frac{V_1}{V_2}=\frac{\frac{Q}{C_1}}{\frac{Q}{C_2}}=\frac{C_2}{C_1}=\frac{3}{2}$

Step 2: Find the ratio of energies.

We use:

$U_1=\frac{1}{2} C_1 V_1^2, \quad U_2=\frac{1}{2} C_2 V_2^2$

So,

$\frac{U_1}{U_2}=\frac{C_1 V_1^2}{C_2 V_2^2}=\frac{C_1}{C_2} \cdot\left(\frac{V_1}{V_2}\right)^2=\frac{2}{3} \cdot\left(\frac{3}{2}\right)^2=\frac{2}{3} \cdot \frac{9}{4}=\frac{18}{12}=\frac{3}{2}$

Therefore: $\frac{V_1}{V_2}=\frac{3}{2}, \quad \frac{U_1}{U_2}=\frac{2}{3}$

3- Two charged metallic spheres with radii $R_1$ and $R_2$ are brought in contact and then separated. The ratio of final charges $Q_1$ and $Q_2$ on the two spheres respectively will be _____.

Fill in the blank with the correct answer from the options given below.

(1) $\frac{\mathrm{Q}_1}{\mathrm{Q}_2}=\frac{\mathrm{R}_2}{\mathrm{R}_1}$

(2) $\frac{Q_1}{Q_2}<\frac{R_1}{R_2}$

(3) $\frac{Q_1}{Q_2}>\frac{R_1}{R_2}$

(4) $\frac{\mathrm{Q}_1}{\mathrm{Q}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}$

Correct Answer: 4

Solution: To solve this, we use the fact that when two metallic spheres are brought into contact, charge redistributes such that their potentials become equal.

The electric potential of a sphere is given by:

$V=\frac{Q}{4 \pi \varepsilon_0 R}$

Let the final charges after contact be $Q_1$ and $Q_2$, and radii be $R_1$ and $R_2$. Since potentials are equal:

$\frac{Q_1}{R_1}=\frac{Q_2}{R_2}$

Rearranging: $\frac{Q_1}{Q_1}=\frac{R_1}{R_2}$

4- The current through a $4 / 3 \Omega$ external resistance connected to a parallel combination of two cells of 2 V and $1 \mathrm{~V}$ emf and internal resistances of $1 \Omega$ and $2 \Omega$ respectively is. Fill in the blank with the correct answer from the options given below.

(1) $1 \mathrm{~A}$

(2) $2 / 3 \mathrm{~A}$

(3) $3 / 4 \mathrm{~A}$

(4) $5 / 6 \mathrm{~A}$

Correct Answer: 4

Solution: To solve this, we use the concept of parallel combination of cells with different emfs and internal resistances.

Let the two cells have:

- Cell 1: $\mathcal{E}_1=2 \mathrm{~V}, r_1=1 \Omega$

- Cell 2: $\mathcal{E}_2=1 \mathrm{~V}, r_2=2 \Omega$

Let the external resistance be:

$R=\frac{4}{3} \Omega$

Step 1: Find the net emf and net internal resistance of the parallel combination.

Use the formula for equivalent emf $\left(\mathcal{E}_{\text {eq }}\right)$ and equivalent internal resistance ( $r_{\text {eq }}$ ) for two cells in parallel:

$\mathcal{E}_{\mathrm{eq}}=\frac{\mathcal{E}_1 / r_1+\mathcal{E}_2 / r_2}{1 / r_1+1 / r_2}=\frac{2 / 1+1 / 2}{1 / 1+1 / 2}=\frac{2+0.5}{1+0.5}=\frac{2.5}{1.5}=\frac{5}{3} \mathrm{~V}$

$r_{\mathrm{eq}}=\frac{r_1 r_2}{r_1+r_2}=\frac{1 \cdot 2}{1+2}=\frac{2}{3} \Omega$

Step 2: Total resistance in the circuit

$R_{\text {total }}=R+r_{\mathrm{eq}}=\frac{4}{3}+\frac{2}{3}=2 \Omega$

Step 3: Current using Ohm's Law

$I=\frac{\mathcal{E}_{\text {eq }}}{R_{\text {total }}}=\frac{5 / 3}{2}=\frac{5}{6} \mathrm{~A}$

5- A metallic wire of uniform area of cross section has a resistance R , resistivity $\rho$ and power rating P at V volts. The wire is uniformly stretched to reduce the radius to half the original radius. The values of resistance, resistivity and power rating at $V$ volts are now denoted by $\mathrm{R}^{\prime}, \rho^{\prime}$ and $\mathrm{P}^{\prime}$ respectively. The corresponding values are correctly related as $\qquad$

Fill in the blank with the correct answer from the options given below.

(1) $\rho^{\prime}=2 \rho, \mathrm{R}^{\prime}=2 \mathrm{R}, \mathrm{P}^{\prime}=2 \mathrm{P}$

(2) $\rho^{\prime}=(1 / 2) \rho, \mathrm{R}^{\prime}=(1 / 2) \mathrm{R}, \mathrm{P}^{\prime}=(1 / 2) \mathrm{P}$

(3) $\rho^{\prime}=\rho, R^{\prime}=16 \mathrm{R}, \mathrm{P}^{\prime}=(1 / 16) \mathrm{P}$

(4) $\rho^{\prime}=\rho, R^{\prime}=(1 / 16) R, P^{\prime}=16 \mathrm{P}$

Correct Answer: 3

Solution: To solve this, we use the following facts:

Original resistance:

$R=\rho \frac{L}{A}$

Resistivity $\rho$ is a material property, so it remains unchanged:

$\rho^{\prime}=\rho$

When the wire is stretched uniformly and the radius becomes half, the new area becomes:

$A^{\prime}=\pi(r / 2)^2=\frac{1}{4} A$

Volume remains constant:

$A \cdot L=A^{\prime} \cdot L^{\prime} \Rightarrow L^{\prime}=4 L$

New resistance becomes:

$R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \frac{4 L}{A / 4}=16 \cdot \rho \frac{L}{A}=16 R$

Power rating at constant voltage $V$ is:

$P=\frac{V^2}{R} \Rightarrow P^{\prime}=\frac{V^2}{R^{\prime}}=\frac{V^2}{16 R}=\frac{P}{16}$

6- Three magnetic materials are listed below

(A) paramagnetics

(B) diamagnetics

(C) ferromagnetics

Choose the correct order of the materials in increasing order of magnetic susceptibility.

(1) (A), (B), (C)

(2) (C), (A), (B)

(3) (B), (A), (C)

(4) (B), (C), (A)

Correct Answer: 3

Solution: Let us understand the relative magnetic susceptibility:

• Diamagnetic materials: Have negative susceptibility (weakly repelled by magnetic fields)

• Paramagnetic materials: Have small positive susceptibility

• Ferromagnetic materials: Have very large positive susceptibility

Hence, the increasing order of susceptibility is:

Diamagnetic<Paramagnetic<Ferromagnetic

7- Two infinitely long straight parallel conductors carrying currents $I_1$ and $I_2$ are held at a distance d apart in a vacuum. The force $F$ on a length $L$ of one of the conductors due to the other is___________.

Fill in the blank with the correct answer from the options given below.

(1) proportional to L but independent of $\mathrm{I}_1 \times \mathrm{I}_2$

(2) proportional to $I_1 \times I_2$ but independent of length $L$

(3) proportional to $\mathrm{I}_1 \times \mathrm{I}_2 \times \mathrm{L}$

(4) proportional to $\frac{L}{I_1 \times I_2}$

Correct Answer: 3

Solution: To solve this, we apply the formula for magnetic force per unit length between two parallel conductors:

$F=\frac{\mu_0}{2 \pi} \cdot \frac{I_1 I_2 L}{d}$

Where:

$\mu_0$ is the permeability of free space,

$I_1$ and $I_2$ are the currents,

$L$ is the length of wire under consideration,

$d$ is the distance between the wires.

So, the force is directly proportional to $I_1 \cdot I_2 \cdot L$.

8- A square loop with each side 1 cm, carrying a current of 10 A, is placed in a magnetic field of 0.2 T. The direction of the magnetic field is parallel to the plane of the loop. The torque experienced by the loop is ______. Fill in the blank with the correct answer from the options given below.

(1) zero

(2) $2 \times 10^{-4} \mathrm{Nm}$

(3) $2 \times 10^{-2} \mathrm{Nm}$

(4) 2 Nm

Correct Answer: 2

Solution: Given:

- Side of square loop $=1 \mathrm{~cm}=0.01 \mathrm{~m}$

- Current $I=10 \mathrm{~A}$

- Magnetic field $B=0.2 \mathrm{~T}$

- Magnetic field is parallel to the plane of the loop.

We use the formula for torque on a current loop:

$\tau=n I A B \sin \theta$

Where:

- $n=1$ (single loop),

- $A=$ area $=(0.01)^2=10^{-4} \mathrm{~m}^2$,

- $\theta=0^{\circ}$ (since magnetic field is in the plane of loop, angle with normal $=90^{\circ}$ so $\left.\sin \theta=\sin \left(90^{\circ}\right)=1\right)$.

So, $\tau=1 \cdot 10 \cdot 10^{-4} \cdot 0.2 \cdot 1=2 \times 10^{-4} \mathrm{Nm}$

9- In an ac circuit, the current leads the voltage by $\pi / 2$. The circuit is ______. Fill in the blank with the correct answer from the options given below.

(1) purely resistive

(2) should have circuit elements with resistance equal to reactance.

(3) purely inductive

(4) purely capacitive

Correct Answer: 4

Solution: In an AC circuit, the phase relationship between current and voltage depends on the nature of the circuit elements:

Purely resistive: Voltage and current are in phase.

Purely inductive: Current lags voltage by $\frac{\pi}{2}$.

Purely capacitive: Current leads voltage by $\frac{\pi}{2}$.

Since the current is given to lead the voltage by $\frac{\pi}{2}$, this implies the circuit is purely capacitive.

Phase difference $\phi=\frac{\pi}{2} \Rightarrow$ Purely capacitive

10- In a pair of adjacent coils, for a change of current in one of the coils from 0 A to 10 A in 0.25 s , the magnetic flux in the adjacent coil changes by 15 Wb. The mutual indutance of the coils is ______ . Fill in the blank with the correct answer from the options given below.

(1) 120 H

(2) 12 H

(3) 1.5 H

(4) 0.75 H

Correct Answer: 3

Solution: The formula for mutual inductance is given by:

$M = \frac{\Delta \Phi}{\Delta I}$

Given:

$\Delta \Phi = 15\, \text{Wb}, \quad \Delta I = 10\, \text{A} - 0\, \text{A} = 10\, \text{A}$

Substituting the values:

$M = \frac{15}{10} = 1.5\, \text{H}$

Therefore, the mutual inductance of the coils is:$\boxed{1.5\, \text{H}}$

CUET 2026 Physics Important Topics

For the CUET 2026 Physics exam, students are advised to focus on the following essential topics as these hold a significantly higher weightage and have been asked frequently in the past years:

• Electrostatics

• Current Electricity

• Magnetic Effects of Current and Magnetism

• Electromagnetic Induction

• Optics

• Dual Nature of Matter and Radiation

• Atoms and Nuclei

• Electromagnetic Waves

• Electronic Devices and Communication Systems

CUET Physics Previous Year Question Paper PDF Download

For CUET 2026 candidates who want to score well in Physics, solving previous year CUET question papers is recommended by experts and toppers alike. Not only do they familiarise you with the exam pattern and common topics asked, but they also enhance your time management and speed of solving problems. Download the CUET Physics Previous Year Question Paper PDFs to have a clear idea about the kind of questions you will be asked and enhance your preparation effectively.

Download Official CUET Physics Previous Year Question Paper PDF

Official Physics question papers from previous years have been provided by the NTA to help students become familiar with the exam. Understanding the exam pattern is necessary, as it evolves with time.

A comprehensive analysis of CUET previous years' official papers, focusing on question types and repetition trends, can give students a valuable advantage in the CUET 2026 examination.

CUET Previous Year Question Papers

We also have previous year papers in English, Political Science, Economics, Chemistry, Accounts, Sociology, History, Geography, and Legal Studies to provide you with a holistic view of the exam pattern of other vital subjects in CUET 2026.

CUET Physics 2026 Study Material

Packed with an extensive collection of CUET Previous Year Questions (PYQS), this e-book will help students easily grasp the latest exam pattern and focus on the most critical topics for CUET 2026.

CUET MCQs For All Science Subjects

Apart from physics, we also provide you with previous year MCQs for other typical science subjects, such as biology, chemistry, computer science, and mathematics, as well as languages like English and other significant general ability subjects. Please visit the links below to practise more efficiently during your CUET 2026 preparation.

Chapter-wise MCQs for CUET Physics

CUET Physics is one of the most challenging parts of the paper. The topics are complex, and the questions are more application-based. Here are some important topic MCQs that are expected in the CUET 2026 exam for you to practice.

CUET 2026 Physics Preparation Tips

Candidates preparing for the Common University Entrance Test should be fluent in physics to score better on the CUET 2026 physics paper. Physics is one of the toughest domains in the exam. Aspirants can ace the exam by practising the CUET 2026 physics questions from previous years. Given below are some important preparation tips that all students are advised to follow:

• Candidates should be aware of the types of questions asked in the CUET exam 2026 while preparing for the exam.

• Candidates preparing for the exam should be familiar with the CUET Physics syllabus.

• Candidates can also prepare for the exam with the help of mock tests and the CUET 2026 previous year question paper in physics.

• Candidates should prepare all the essential topics frequently asked in the CUET 2026 previous year's question papers.

• Aspirants should attempt the CUET 2026 mock tests to evaluate their CUET Exam preparation.

CUET 2026 Physics Score Improvement Techniques

These techniques help students avoid common pitfalls, optimise time management during the actual CUET exam day, and strategically approach the MCQs to achieve their target scores.

• Focus on High-Weightage Topics for maximum score impact.

• Write key formulas with units, derivations, and memory tricks. Review daily for instant recall during the exam.

• Look for obviously wrong answers first, use dimensional analysis, and apply basic physics principles to eliminate options.

• Practice maximum CUET 2026 pattern MCQs to understand question types and common trap answers.

• Avoid random guessing when completely unsure about the topic.