| Updated On - Nov 11, 2024
CUET Physics Question Paper 2024 (Set A) is available here. NTA conducted CUET 2024 Physics paper on 16 May in Shift 2A from 3 PM to 4 PM. CUET Physics Question Paper 2024 is based on objective-type questions (MCQs). Candidates get 60 minutes to solve 40 MCQs out of 50 in CUET 2024 question paper for Physics.
CUET Physics Question Paper 2024 (Set A) PDF Download
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CUET Physics Questions with Solution
| Serial No. | Question | Answer | Solution |
|---|---|---|---|
| 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: 1. 4d 2. 2d 3. d 4. d/2 |
Option (2). 2d | According to Coulomb's law, doubling each charge requires doubling the distance to keep the force constant. |
| 2 | Two parallel plate capacitors of capacitances 2 µF and 3 µF are joined in series and connected to a battery of V volts. The values of potential across the two capacitors V1 and V2 and energy stored in them U1 and U2 are related as: 1. V2/V1 = U2/U1 = 2/3 2. V2/V1 = U2/U1 = 3/2 3. V2/V1 = 2/3, U2/U1 = 3/2 4. V2/V1 = 3/2, U2/U1 = 2/3 |
Option (4). V2/V1 = 3/2, U2/U1 = 2/3 | In series, potential is inversely proportional to capacitance, and energy is proportional to C × V². |
| 3 | Two large plane parallel sheets with equal but opposite surface charge densities +σ and -σ. A point charge q placed at points P1, P2, and P3 experiences forces F1, F2, and F3 respectively. Then: 1. F1 = 0, F2 = 0, F3 = 0 2. F1 = 0, F2 ≠ 0, F3 = 0 3. F1 ≠ 0, F2 ≠ 0, F3 ≠ 0 4. F1 = 0, F3 ≠ 0, F2 = 0 |
Option (2). F1 = 0, F2 ≠ 0, F3 = 0 | Charges outside the sheets (P1 and P3) experience no net force due to cancellation, but within (P2) they do. |
| 4 | Two charged metallic spheres with radii R1 and R2 are brought into contact and then separated. The ratio of final charges Q1 and Q2 on the two spheres is: 1. Q2/Q1 = R1/R2 2. Q2/Q1 < R1/R2 3. Q2/Q1 > R1/R2 4. Q2/Q1 = R2/R1 |
Option (4). Q2/Q1 = R2/R1 | On contact, charge distribution is proportional to the ratio of their radii. |
| 5 | Two resistances of 100 Ω and 200 Ω are connected in series across a 20 V battery. The reading in a 200 Ω voltmeter across the 200 Ω resistance is: 1. 4 V 2. 20/3 V 3. 10 V 4. 16 V |
Option (2). 20/3 V | In series, current is uniform; voltage drop across 200 Ω resistor is 20/3 V. |
| 6 | The current through a 4/3 Ω external resistance connected to a parallel combination of two cells with emfs of 2 V and 1 V, internal resistances 1 Ω and 2 Ω respectively, is: 1. 1 A 2. 2/3 A 3. 3/4 A 4. 5/6 A |
Option (4). 5/6 A | Using equivalent emf and resistance for cells in parallel, current is calculated as 5/6 A. |
| 7 | A metallic wire of resistance R is stretched uniformly to reduce the radius to half. The new resistance, resistivity, and power rating at V volts are R', ρ', and P' respectively, such that: 1. ρ' = 2ρ, R' = 2R, P' = 2P 2. ρ' = 1/2ρ, R' = 1/2R, P' = 1/2P 3. ρ' = ρ, R' = 16R, P' = 1/16P 4. ρ' = ρ, R' = 1/16R, P' = 16P |
Option (3). ρ' = ρ, R' = 16R, P' = 1/16P | When stretched, resistance increases by square of length ratio; resistivity remains unchanged. |
| 8 | Three magnetic materials: (A) Paramagnetic, (B) Diamagnetic, (C) Ferromagnetic. The correct order of increasing magnetic susceptibility is: 1. (A), (B), (C) 2. (C), (A), (B) 3. (B), (A), (C) 4. (B), (C), (A) |
Option (3). (B), (A), (C) | Diamagnetic < Paramagnetic < Ferromagnetic in terms of magnetic susceptibility. |
| 9 | Two infinitely long parallel conductors carrying currents I1 and I2 are at distance d apart. The force F on length L of one conductor due to the other is: 1. proportional to L but independent of I1 × I2 2. proportional to I1 × I2 but independent of L 3. proportional to I1 × I2 × L 4. proportional to L/I1 × I2 |
Option (3). proportional to I1 × I2 × L | Force between two parallel currents is proportional to product of currents and length. |
| 10 | In a circuit where current 3I enters at A, with semicircular sections ABC and ADC of equal radii r, resistances 2R and R respectively, the magnetic field at the center is: 1. μ0I/4r out of the plane 2. μ0I/4r into the plane 3. μ03I/4r out of the plane 4. μ03I/4r into the plane |
Option (4). μ03I/4r into the plane | The magnetic field at the center of the loop is determined using Ampere's Law and is directed into the plane. |
| 11 | A square loop with each side 1 cm, carrying a current of 10 A, is placed in a magnetic field of 0.2 T, parallel to the plane of the loop. The torque experienced by the loop is: 1. zero 2. 2×10⁻⁴ Nm 3. 2×10⁻² Nm 4. 2 Nm |
Option (1). zero | When the magnetic field is parallel to the plane of the loop, the torque experienced by the loop is zero. |
| 12 | In an AC circuit, the current leads the voltage by π/2. The circuit is: 1. purely resistive 2. should have resistance equal to reactance 3. purely inductive 4. purely capacitive |
Option (4). purely capacitive | In a purely capacitive AC circuit, the current leads the voltage by 90° (π/2 radians). |
| 13 | In a pair of adjacent coils, if current in one coil changes from 0 A to 10 A in 0.25 s, causing a magnetic flux change of 15 Wb in the adjacent coil, the mutual inductance of the coils is: 1. 120 H 2. 12 H 3. 1.5 H 4. 0.75 H |
Option (3). 1.5 H | Mutual inductance M is the ratio of flux change to current change, calculated as M = 15/10 = 1.5 H. |
| 14 | A wire of irregular shape (figure a) and a circular loop (figure b) are placed in uniform magnetic fields as shown. The induced current direction will be: 1. clockwise in both 2. anticlockwise in both 3. clockwise in (a) and anticlockwise in (b) 4. anticlockwise in (a) and clockwise in (b) |
Option (3). clockwise in (a) and anticlockwise in (b) | According to Lenz’s law, the induced current direction opposes the change in magnetic flux. |
| 15 | Match the variation of opposition to AC flow versus frequency in List-I with circuit characteristics in List-II: 1. (A) - (I), (B) - (II), (C) - (III), (D) - (IV) 2. (A) - (IV), (B) - (III), (C) - (II), (D) - (I) 3. (A) - (I), (B) - (II), (C) - (IV), (D) - (III) 4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II) |
Option (4). (A) - (III), (B) - (IV), (C) - (I), (D) - (II) | Matching the graphs and circuit characteristics as per opposition to AC flow with frequency. |
| 16 | In an electromagnetic wave, the ratio of energy densities of electric and magnetic fields is: 1. 1 : 1 2. 1 : c 3. c : 1 4. 1 : c² |
Option (1). 1 : 1 | In electromagnetic waves, electric and magnetic fields have equal energy densities. |
| 17 | The correct arrangement of electromagnetic spectrum in decreasing order of wavelength is: 1. Radio waves, X-rays, Infrared waves, microwaves, visible waves 2. Infrared waves, microwaves, Radio waves, X-rays, visible waves 3. Radio waves, microwaves, Infrared waves, visible waves, X-rays 4. X-rays, visible waves, Infrared waves, microwaves, Radio waves |
Option (3). Radio waves, microwaves, Infrared waves, visible waves, X-rays | Electromagnetic spectrum is arranged from longest wavelength (Radio waves) to shortest (X-rays). |
| 18 | Match the electromagnetic waves in Column-I with production methods in Column-II: 1. (A) - (I), (B) - (II), (C) - (III), (D) - (IV) 2. (A) - (II), (B) - (III), (C) - (IV), (D) - (I) 3. (A) - (II), (B) - (I), (C) - (IV), (D) - (III) 4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II) |
Option (2). (A) - (II), (B) - (III), (C) - (IV), (D) - (I) | Correctly matching wave types with devices: magnetron for microwaves, etc. |
| 19 | For a point object placed 20 cm from point P, distance of its image from P (given refractive index µ = 4/3) is: 1. 16 cm left of P in air 2. 16 cm right of P in water 3. 20 cm right of P in water 4. 20 cm left of P in air |
Option (1). 16 cm left of P in air | Using refraction formula, the image is calculated to be 16 cm to the left of P in air. |
| 20 | For fixed radii of curvature of a lens, power is proportional to: 1. P ∝ (µ - 1) 2. P ∝ µ² 3. P ∝ 1/µ 4. P ∝ µ⁻² |
Option (1). P ∝ (µ - 1) | For given curvature, power of a lens depends on (µ - 1), per Lensmaker's formula. |
| 21 | The graph correctly representing the variation of image distance ‘v’ for a convex lens of focal length ‘f’ versus object distance ‘u’ is: 1. Line approaching focal length asymptotically 2. Straight line 3. Curve approaching but not touching focal points 4. Exponential curve |
Option (3). Curve approaching but not touching focal points | For a convex lens, the relationship between u, v, and f creates a hyperbolic curve. |
| 22 | Using monochromatic light for diffraction in a single slit of width 0.1 mm, with a central maximum of 5 mm width on a screen 50 cm away, the wavelength of light used is: 1. 2.5 × 10⁻⁷ m 2. 4 × 10⁻⁷ m 3. 5 × 10⁻⁷ m 4. 7.5 × 10⁻⁷ m |
Option (3). 5 × 10⁻⁷ m | Using the formula for single-slit diffraction width, the wavelength is calculated as 5 × 10⁻⁷ m. |
| 23 | Radiation of frequency 2ν₀ incident on a metal with threshold frequency ν₀ results in maximum kinetic energy of photoelectrons as: 1. No photoelectrons emitted 2. All have kinetic energy equal to hν₀ 3. Maximum kinetic energy can be hν₀ 4. Maximum kinetic energy will be 2hν₀ |
Option (3). Maximum kinetic energy can be hν₀ | Using Einstein's photoelectric equation, maximum kinetic energy is hν₀ when frequency is 2ν₀. |
| 24 | A point source causing photoelectric emission from a metallic plate is moved away. The variation of photoelectric current with distance from the source is best represented by: 1. Constant 2. Decreasing curve leveling off 3. Exponential decay 4. Straight decreasing line |
Option (2). Decreasing curve leveling off | Photoelectric current decreases with distance due to the inverse square law of light intensity. |
| 25 | A proton accelerated through potential difference V has de Broglie wavelength λ. On doubling the potential, the de Broglie wavelength of the proton: 1. remains unchanged 2. becomes double 3. becomes four times 4. decreases |
Option (4). decreases | The de Broglie wavelength is inversely proportional to the square root of the accelerating potential, so doubling the potential decreases λ. |
| 26 | The kinetic energy of an electron in the ground state of a hydrogen atom is K. The values of its potential energy and total energy respectively are: 1. -2K, -K 2. +2K, -K 3. -K, +2K 4. +K, +2K |
Option (1). -2K, -K | In a hydrogen atom, potential energy U = -2K, and total energy E = -K. |
| 27 | Two nuclei with mass numbers A and B have density ratios of: 1. A : B 2. √A : √B 3. A² : B² 4. 1 : 1 |
Option (4). 1 : 1 | Nuclear density is approximately constant and independent of mass number. |
| 28 | The shortest wavelengths in the hydrogen spectrum, in decreasing order, for spectral series Pfund, Balmer, Brackett, and Lyman are: 1. (A), (B), (C), (D) 2. (A), (C), (B), (D) 3. (B), (A), (D), (C) 4. (A), (C), (D), (B) |
Option (2). (A), (C), (B), (D) | The correct wavelength order for decreasing values in the hydrogen spectrum is Pfund, Brackett, Balmer, Lyman. |
| 29 | Silicon can be doped to get an n-type semiconductor by using: 1. (A) and (C) only 2. (B) and (C) only 3. (A), (B), (C), and (D) 4. (C) and (D) only |
Option (1). (A) and (C) only | Elements like arsenic and phosphorus (Group V) create n-type semiconductors by donating electrons. |
| 30 | The correct sequence of graphs for forward biased p-n junction, Zener diode, Photo diode, and Solar cell is: 1. (D), (C), (A), (B) 2. (A), (C), (D), (B) 3. (B), (A), (D), (C) 4. (C), (B), (D), (A) |
Option (2). (A), (C), (D), (B) | Each graph type is matched to the diode characteristic curves of p-n junction, Zener, Photo diode, and Solar cell. |
| 31 | A wire carrying current I, bent as shown, is placed in a uniform magnetic field B, emerging normally out of the plane. The force on the wire is: 1. 4BIR, downward 2. 3BIR, upward 3. BI(2R + πR), downward 4. 2πBIR, from P to Q |
Option (1). 4BIR, downward | Forces on the straight sections cancel out; the net force from semicircular arcs is 4BIR. |
| 32 | The minimum deviation angle in an equilateral prism with refractive index √2 is: 1. 60° 2. 75° 3. 30° 4. 90° |
Option (1). 60° | Using the prism formula and given refractive index, the minimum deviation is 60°. |
| 33 | The quantization of electric charge is evidenced by transfer of integral number of: 1. photons 2. nuclei 3. electrons 4. neutrons |
Option (3). electrons | Electric charge is quantized, as it transfers in integer multiples of the elementary charge on electrons. |
| 34 | Introducing a 4 mm thick insulating slab between capacitor plates requires increasing the plate distance by 3.2 mm to restore original capacitance. The dielectric constant is: 1. 2 2. 5 3. 3 4. 7 |
Option (2). 5 | Dielectric constant K is calculated using deff = d/K, yielding K = 5. |
| 35 | A copper ball of density 8 g/cc and 1 cm diameter, immersed in oil of density 0.8 g/cc, remains suspended in an electric field of 600π V/m. The charge on the ball is: 1. 2×10⁻⁶ C 2. 2×10⁻⁵ C 3. 1×10⁻⁵ C 4. 1×10⁻⁶ C |
Option (3). 1×10⁻⁵ C | Equating electric force to gravitational force, charge Q is found to be 1×10⁻⁵ C. |
| 36 | A metal wire under constant potential difference, when heated, shows that the drift velocity of the electron: 1. increases, thermal velocity decreases 2. decreases, thermal velocity decreases 3. increases, thermal velocity increases 4. decreases, thermal velocity increases |
Option (4). decreases, thermal velocity increases | Higher temperature increases collisions, reducing drift velocity, while thermal velocity rises. |
| 37 | For the given mixed resistor combination, calculate the total resistance between A and B: 1. 9 Ω 2. 18 Ω 3. 4 Ω 4. 14 Ω |
Option (2). 18 Ω | Simplifying step-by-step, the total resistance is found to be 18 Ω. |
| 38 | A cell of emf 1.1 V and internal resistance 0.5 Ω is connected to a 0.5 Ω wire. Another cell of same emf is added in series, but current remains unchanged. The second cell's internal resistance is: 1. 1 Ω 2. 2.5 Ω 3. 1.5 Ω 4. 2 Ω |
Option (3). 1.5 Ω | Applying Kirchhoff’s law and solving, the internal resistance of the second cell is 1.5 Ω. |
| 39 | Wires P, Q, R, S of resistances 3, 3, 3, and 4 Ω form a Wheatstone bridge. The resistance to shunt S and balance the bridge is: 1. 14 Ω 2. 12 Ω 3. 15 Ω 4. 7 Ω |
Option (4). 7 Ω | For balance, the required shunt resistance is calculated as 7 Ω. |
| 40 | A thin bar magnet with magnetic moment M is bent into a semicircle. Its new magnetic moment is: 1. M/π 2. M/2 3. M 4. 2M/π |
Option (3). M | Bending doesn’t affect magnetic moment, which depends only on pole strength and effective length. |
| 41 | Ferromagnetic material in transformers should have: 1. Low permeability, High Hysteresis loss 2. High permeability, Low Hysteresis loss 3. High permeability, High Hysteresis loss 4. Low permeability, Low Hysteresis loss |
Option (2). High permeability, Low Hysteresis loss | High permeability and low hysteresis loss reduce energy dissipation in transformers. |
| 42 | A conducting ring of radius r in a magnetic field varying at rate x has electric field intensity at any ring point as: 1. rx 2. rx/2 3. 2rx 4. 4rx |
Option (1). rx | Using Faraday’s law, induced electric field E = rx in a conducting loop. |
| 43 | A 50 Hz AC current of crest 1 A flows through a transformer primary. With mutual inductance 0.5 H, the crest voltage induced in secondary is: 1. 75 V 2. 150 V 3. 100 V 4. 200 V |
Option (2). 150 V | Induced voltage is calculated using V = M(dI/dt), yielding 150 V. |
| 44 | A solenoid with 2×10⁴ turns per meter, diameter 0.1 m, has a coil of 100 turns and 0.01 m radius placed inside. If solenoid current decreases from 4 A to 0 in 0.05 s, total charge through coil is: 1. 16 µC 2. 32 µC 3. 16π µC 4. 32π µC |
Option (4). 32π µC | Using Faraday’s law and induced emf, the total charge through coil is 32π µC. |
| 45 | Lower half of a convex lens is opaque. The image of an object placed in front will: 1. No change in image 2. Show only half of the object 3. Intensity reduced 4. Show half of object and reduced intensity |
Option (4). Show half of object and reduced intensity | Blocking half reduces intensity and visibility but does not affect image completeness. |
| 46 | Two slits 0.1 mm apart with screen 2 m away produce fringe separation with 500 nm light. The separation is: 1. 1 cm 2. 0.15 cm 3. 1.5 cm 4. 0.1 cm |
Option (3). 1.5 cm | Using fringe formula, separation ∆y = λD/d is calculated as 1.5 cm. |
| 47 | For an astronomical telescope with 10 m focal length objective and 10 cm eyepiece, tube length and magnification are: 1. 20 cm, 1 2. 1000 cm, 1 3. 1010 cm, 1 4. 1010 cm, 100 |
Option (4). 1010 cm, 100 | Tube length is sum of focal lengths, and magnification is ratio of focal lengths. |
| 48 | According to Bohr's Model: (A) Radius ∝ n (B) Speed ∝ 1/n (C) Total energy ∝ 1/n² (D) Radius ∝ n² Correct options: 1. (A), (B), (C) 2. (A), (B), (D) 3. (A), (B), (C), (D) 4. (B), (C), (D) |
Option (4). (B), (C), (D) | Bohr's model: radius ∝ n², speed ∝ 1/n, total energy ∝ 1/n². |
| 49 | In a full-wave rectifier, if input frequency is 50 Hz, the output frequency is: 1. 50 Hz 2. 100 Hz 3. 25 Hz 4. 0 Hz |
Option (2). 100 Hz | A full-wave rectifier doubles the input frequency to 100 Hz. |
| 50 | For an electric dipole in a non-uniform electric field with dipole moment parallel to the field, force F and torque τ are: 1. F = 0, τ = 0 2. F ≠ 0, τ = 0 3. F = 0, τ ≠ 0 4. F ≠ 0, τ ≠ 0 |
Option (2). F ≠ 0, τ = 0 | In a non-uniform field, dipole has force but no torque when aligned with the field. |
CUET Questions
1. If \( A = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 1 & 10 \\ -2 & -1 \end{bmatrix} \), then the matrix \( AB \) is:
If \( A = \begin{bmatrix} 5 & 1 \\ -2 & 0 \end{bmatrix} \) and \( B^T = \begin{bmatrix} 1 & 10 \\ -2 & -1 \end{bmatrix} \), then the matrix \( AB \) is:
- \( \begin{bmatrix} 1 & 10 \\ -1 & 0 \end{bmatrix} \)
- \( \begin{bmatrix} 15 & -11 \\ -2 & 4 \end{bmatrix} \)
- \( \begin{bmatrix} 3 & 49 \\ -2 & -20 \end{bmatrix} \)
- \( \begin{bmatrix} 1 & 9 \\ -2 & -20 \end{bmatrix} \)
2.
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[
Z = 400x + 250y,
\]
subject to constraints:
\[
x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0.
\]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below:
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[ Z = 400x + 250y, \]
subject to constraints:
\[ x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below:
- (A), (B), and (C) only
- (A), (B), and (D) only
- (B) and (C) only
- (B), (C), and (D) only
3. The correct solution of \(-22 < 8x - 6 \leq 26\) is the interval:
The correct solution of \(-22 < 8x - 6 \leq 26\) is the interval:
- \([-2, 4]\)
- \((-2, 4]\)
- \((-2, 4)\)
- \([-2, 4)\)
4. In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
- \(\frac{609}{625}\)
- \(\frac{16}{625}\)
- \(\frac{513}{625}\)
- \(\frac{112}{625}\)
5. If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
- \( \frac{1}{4t\sqrt{t}} \)
- \({2t^2} \)
- \( \frac{\log_e 3}{4t\sqrt{t}} \)
- \( \frac{2t^2}{e^{\frac{1}{2}\log_e t}} \)
6. The cost of a machinery is ₹8,00,000. Its scrap value will be one-tenth of its original cost in 15 years. Using the linear method of depreciation, the book value of the machine at the end of the 10th year will be:
The cost of a machinery is ₹8,00,000. Its scrap value will be one-tenth of its original cost in 15 years. Using the linear method of depreciation, the book value of the machine at the end of the 10th year will be:
- ₹4,80,000
- ₹3,20,000
- ₹3,68,000
- ₹4,32,000
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