CUET Physics Question Paper 2024 PDF, Download Previous Year Papers

CUET 2025 Physics Previous Year Question Paper- Every year, the National Testing Agency (NTA) conducts the Common University Entrance Test (CUET) to offer admissions in UG and PG courses in premier institutions nationwide. It is performed once a year to facilitate admissions in central universities. Since the exam is about to be conducted, candidates might be stressed about their preparation. But no worries, since we bring you the best material to boost your preparation. Candidates preparing for the CUET 2025 for the physics domain will find this article helpful as a reference for their preparation for the CUET 2025 physics exam.

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1. CUET Physics Previous Year Questions
2. CUET 2025 Physics Important Topics
3. CUET MCQ For All Popular Science Subjects
4. CUET 2025 Physics Preparation Tips

In this article, we will deepen our knowledge base and learn about the CUET 2025 Physics exam, the types of questions which will be asked, and the essential topics covered in the exam. Candidates preparing for the CUET UG exam can score well in the exam concerning the CUET physics question paper. As it is said that practice makes a man perfect, here we present you with the previous year's questions in the most precise manner for your understanding and reference. Candidates can also download the CUET Previous Year Question Papers of the physics question paper from the official website in PDF form.

CUET Physics Previous Year Questions

The following section discusses some of the essential questions from previous years and the options for better insight into the nature of questions and the paper pattern. Previous Year Question Papers will strategise your preparation to ensure successful results within no time.

1- Two charged particles, placed at a distance d apart in 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.

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 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 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 2025 Physics Important Topics

For the CUET 2025 Physics exam, students should focus on the following essential topics:

• 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 2025 candidates who want to score well in Physics, solving previous year question papers is a good idea. Not only do they familiarize you with the exam pattern and common topics asked, but 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 in an effective manner.

CUET Previous Year Question Papers

We also have previous year papers on English, Political Science, Economics, Chemistry, Accounts, Sociology, History, Geography, Legal Studies to give you a holistic view of the exam pattern of other vital subjects of CUET UG 2025.

Download Physics Previous Year Question Paper PDF

Official Physics question papers from previous years have been provided to help students become familiar with the exam. Understanding the exam pattern is necessary, as it evolves with time. Be sure to follow the CUET 2025.Physics guidelines.

A comprehensive analysis of previous years' official papers, focusing on question types and repetition trends, can give students a valuable advantage in the CUET 2025 examination.

CUET Physics 2025 Study Material

Packed with an extensive collection of CUET Previous Year Questions (PYQS), this e-book helps students easily grasp the latest exam pattern and focus on the most critical topics for CUET 2025.

CUET MCQ For All Popular Science Subjects

Apart from physics, we also bring you Previous Year MCQS for other typical science subjects like biology, chemistry, computers, and mathematics, as well as languages like English and other significant general ability subjects. If we cover all subjects, our chances of scoring better significantly increase. Please visit the links below.

CUET 2025 Physics Preparation Tips

Candidates preparing for the Common University Entrance Test should be fluent in physics to score better on the CUET 2025 physics paper. Physics is one of the toughest domains in the exam. Aspirants can ace the exam by practising the CUET 2025 physics previous years' questions. Candidates can prepare for the CUET exam with the help of the CUET question paper. Candidates can prepare for the exam with the help of CUET 2025 physics preparation tips:

• Candidates should be aware of the types of questions asked in the exams while preparing for the CUET exam 2025.

• Candidates preparing for the exam should know about the CUET physics syllabus 2024.

• Candidates can also prepare for the exam with the help of mock tests and the CUET 2025 previous year question paper in physics.

• Candidates should prepare all the essential topics frequently asked in the CUET 2025 previous year's question papers.

• Aspirants should attempt the CUET 2025 physics 2025 mock tests to understand their CUET Exam preparation.

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• CUET 2024 Admit Card

• CUET City Slip