CUET 2024 Question Paper with Answer Key PDF for Physics (May 16 - Shift 2)

Content Curator | Updated On - Dec 5, 2024

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CUET 2024 Physics Question Paper with Answer Key PDF

CUET 2024 Physics Question Paper with Answer Key Set A	Download	Check Solutions
CUET 2024 Physics Question Paper with Answer Key Set B	Download	Check Solutions
CUET 2024 Physics Question Paper with Answer Key Set c	Download	Check Solutions
CUET 2024 Physics Question Paper with Answer Key Set D	Download	Check Solutions
CUET 2024 Physics Question Paper with Answer Key	Download	Check Solutions

CUET Physics Questions with Solution Set A

Question	Answer	Detailed Solution
Question 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	(2) 2d	Using Coulomb's Law: F=k(q1q2)/r². When both charges are doubled, the force becomes 4F. To maintain the same force, the distance must increase by a factor of 2. Therefore, r' = 2d.
Question 2: Two parallel plate capacitors of capacitances 2µF and 3µF 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 U₁ and U₂, respectively, are related as: (1) V₁/V₂ = 3/2, U₁/U₂ = 3/2 (2) V₁/V₂ = 2/3, U₁/U₂ = 2/3 (3) V₁/V₂ = 3/2, U₁/U₂ = 2/3 (4) V₁/V₂ = 2/3, U₁/U₂ = 3/2	(1) V₁/V₂ = 3/2, U₁/U₂ = 3/2	In a series combination, the same charge Q flows through both capacitors. The potential difference divides inversely proportional to capacitance: V₁/V₂ = C₂/C₁ = 3/2. For capacitors in series, since Q is the same, U₁/U₂ = C₂/C₁ = 3/2.
Question 3: Two large plane parallel sheets with equal but opposite surface charge densities +σ and −σ have a point charge q placed at points P1, P2, and P3. The forces F1, F2, and F3 experienced by q are: (See Diagram in PDF) (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	(2) F₁ = 0, F₂ ≠ 0, F₃ = 0	The electric field is zero outside the parallel plates (at P₁ and P₃) because the fields cancel. Inside the plates (at P₂), the electric field is uniform, and q experiences a force F₂ ≠ 0.
Question 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) Q₁/Q₂ = R₂/R₁ (2) Q₁/Q₂ = R₁/R₂ (3) Q₁/Q₂ > R₁/R₂ (4) Q₁/Q₂ = R₁/R₂	(2) Q₁/Q₂ = R₁/R₂	When the spheres are brought into contact, charge redistributes in proportion to their capacitances. Capacitance of a sphere is proportional to its radius: Q₁/Q₂ = R₁/R₂
Question 5: Two resistances of 100Ω and 200Ω are connected in series across a 20 V battery. The reading in a 200Ω voltmeter connected across the 200Ω resistance is: (1) 4 V (2) 2 V (3) 10 V (4) 16 V	(3) 10 V	Total resistance = 300Ω. Current = 20V/300Ω = 1/15 A. Voltage across the 200Ω resistor = (1/15 A) * 200Ω = 10 V. The voltmeter reading is equal to the voltage drop across the 200Ω resistor.
Question 6: The current through a 4/3Ω external resistance connected to a parallel combination of two cells of 2 V and 1 V emf and internal resistances of 1Ω and 2Ω respectively is: (1) 1 A (2) 2/3 A (3) 3/4 A (4) 5/6 A	(4) 5/6 A	Equivalent emf = (22 + 11)/(1+2) = 5/3 V. Equivalent resistance = (1*2)/(1+2) = 2/3 Ω. Total resistance = 2/3Ω + 4/3Ω = 2Ω. Current = (5/3 V) / (2Ω) = 5/6 A.
Question 7: A metallic wire of uniform cross-sectional area has a resistance R, resistivity ρ, 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 are now denoted as R', ρ', and P'. The corresponding relations are: (1) ρ' = 2ρ, R' = 2R, P' = 2P (2) ρ' = ρ/2, R' = R/2, P' = P/2 (3) ρ' = ρ, R' = 16R, P' = P/16 (4) ρ' = ρ, R' = R/16, P' = 16P	(3) ρ' = ρ, R' = 16R, P' = P/16	Stretching the wire reduces its radius to half, increasing its length to four times. R' = ρ(4L)/[π(r/2)²] = 16R. Resistivity does not change (ρ' = ρ). Power rating at constant voltage: P' = V²/R' = (V²/R)/16 = P/16.
Question 8: Three magnetic materials are listed: (A) Paramagnetics, (B) Diamagnetics, and (C) Ferromagnetics. Choose the correct order of the materials in increasing magnetic susceptibility: (1) (A), (B), (C) (2) (C), (A), (B) (3) (B), (A), (C) (4) (B), (C), (A)	(3) (B), (A), (C)	Diamagnetic materials have the lowest susceptibility, followed by paramagnetic, and ferromagnetic with the highest.
Question 9: Two infinitely long straight parallel conductors carrying currents I₁ and I₂ 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: (1) Proportional to L but independent of I₁ × I₂ (2) Proportional to I₁ × I₂ but independent of length L (3) Proportional to I₁ × I₂ × L (4) Proportional to L/I₁ × I₂	(3) Proportional to I₁ × I₂ × L	The force per unit length between two parallel current-carrying conductors is given by: F/L = (μ₀I₁I₂)/(2πd). For a length L, the force is: F = (μ₀I₁I₂L)/(2πd). The force is directly proportional to I₁, I₂, and L.
Question 10: In the circuit shown below, a current 3I enters at A. The semicircular parts ABC and ADC have equal radii r but resistances 2R and R, respectively. The magnetic field at the center of the circular loop ABCD is: (See Diagram in PDF) (1) (μ₀I)/(4r) out of the plane (2) (μ₀I)/(4r) into the plane (3) (3μ₀I)/(4r) out of the plane (4) (3μ₀I)/(4r) into the plane	(1) (μ₀I)/(4r) out of the plane	Current splits inversely proportional to resistances: I_ABC = I, I_ADC = 2I. Magnetic field at the center for a semicircular wire is (μ₀I)/(4r). Fields add vectorially; since currents are in opposite directions, they add up: B_net = (μ₀I)/(4r) + (2μ₀I)/(4r) = (3μ₀I)/(4r). Direction is out of the plane using the right-hand rule. (Note: There's a slight discrepancy in the solution provided in the PDF. The final answer should be (μ₀I)/(4r) out of the plane, based on the current distribution and the diagram. )
Question 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. The direction of the magnetic field is 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	(2) 2 × 10⁻⁴ Nm	Torque (τ) = NIABsinθ, where N=1, I=10A, A=(0.01m)²=10⁻⁴m², B=0.2T, and θ=90° (since the field is parallel to the plane). τ = 1 * 10 * 10⁻⁴ * 0.2 * sin90° = 2 × 10⁻⁴ Nm.
Question 12: In an AC circuit, the current leads the voltage by π/2. The circuit is: (1) Purely resistive (2) Circuit elements with resistance equal to reactance (3) Purely inductive (4) Purely capacitive	(4) Purely capacitive	In a purely capacitive circuit, the current leads the voltage by 90° (π/2).
Question 13: In a pair of adjacent coils, for a change of current in one coil from 0 A to 10 A in 0.25 s, the magnetic flux in the adjacent coil changes by 15 Wb. The mutual inductance of the coils is: (1) 120 H (2) 12 H (3) 1.5 H (4) 0.75 H	(3) 1.5 H	Mutual inductance (M) = ΔΦ/ΔI = 15Wb / 10A = 1.5 H.
Question 14: A wire of irregular shape (figure a) and a circular loop of wire (figure b) are placed in different uniform magnetic fields. In figure (a), the magnetic field is perpendicular into the plane. In figure (b), the magnetic field is perpendicular out of the plane. The wire in figure (a) is turning into a circular loop, and that in figure (b) into a narrow straight wire. The direction of induced current will be: (See Diagrams in PDF) (1) Clockwise in both (a) and (b) (2) Anticlockwise in both (a) and (b) (3) Clockwise in (a) and anticlockwise in (b) (4) Anticlockwise in (a) and clockwise in (b)	(2) Anticlockwise in both (a) and (b)	Using Lenz's Law, the induced current flows anticlockwise in both (a) and (b) to oppose the respective changes in flux. In (a), the area increases, so the induced current creates a magnetic field opposing the increase (out of the plane). In (b), the area decreases, so the induced current creates a magnetic field opposing the decrease (out of the plane).
Question 15: Match List-I with List-II: (See Lists in PDF) (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)	(2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I)	Impedance depends on both reactance and resistance. Capacitive reactance decreases with frequency; inductive reactance increases with frequency; resistance is frequency independent.
Question 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	(1) 1:1	The energy density of an electromagnetic wave is equal for the electric and magnetic fields because E = cB.
Question 17: Of the following, 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	(3) Radio waves, Microwaves, Infrared waves, Visible waves, X-rays	The electromagnetic spectrum is ordered by wavelength from radio to gamma rays.
Question 18: Match the electromagnetic waves in Column I with their production methods in Column II: (See Columns in PDF) (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)	(2) (A) - (II), (B) - (III), (C) - (IV), (D) - (I)	Microwaves are produced by magnetrons. Infrared waves arise from vibrations of atoms/molecules. X-rays are generated by bombarding heavy nuclei with electrons. Radio waves are created by LC oscillators.
Question 19: In the figure given below, APB is a curved surface of radius of curvature 10 cm separating air and a transparent material (μ = 1.5). A point object O is placed in air on the principal axis of the surface 20 cm from P. The distance of the image of O from P will be: (See Diagram in PDF) (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	(1) 16 cm left of P in air	Using the refraction formula for curved surfaces: (μ₂/v) - (μ₁/u) = (μ₂ - μ₁)/R. Substituting the values, v = -16 cm, indicating the image is 16 cm to the left of P in air.
Question 20: For fixed values of radii of curvature of a lens, the power of the lens will be: (1) P ∝ (μ – 1) (2) P ∝ μ² (3) P ∝ μ/(μ -1) (4) P ∝ (μ - 2)	(1) P ∝ (μ – 1)	Lens power (P) = (μ - 1)[(1/R₁) + (1/R₂)]. For fixed radii, power is directly proportional to (μ - 1).
Question 21: The graph correctly representing the variation of image distance v for a convex lens of focal length f versus object distance u is: (See graphs in PDF)	Graph showing a hyperbola asymptotic to both axes.	The lens equation (1/v) - (1/u) = 1/f produces a hyperbolic relation between v and u, with asymptotes along the axes.
Question 22: Using light from a monochromatic source to study diffraction in a single slit of width 0.1 mm, the linear width of the central maximum is measured to be 5 mm on a screen held 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	(3) 5 × 10⁻⁷ m	Linear width of central maximum (Δx) = 2λL/a, where λ is wavelength, L is distance to screen, and a is slit width. Solving for λ: λ = (Δx * a)/(2L) = (5 × 10⁻³ m * 0.1 × 10⁻³ m) / (2 * 0.5 m) = 5 × 10⁻⁷ m.
Question 23: Radiation of frequency 2ν₀ is incident on a metal with threshold frequency ν₀. The correct statement is: (1) No photoelectrons will be emitted (2) All photoelectrons emitted will have kinetic energy equal to hν₀ (3) Maximum kinetic energy of photoelectrons emitted can be hν₀ (4) Maximum kinetic energy of photoelectrons emitted will be 2hν₀	(3) Maximum kinetic energy of photoelectrons emitted can be hν₀	Photoelectric equation: K_max = hf - hν₀. Since f = 2ν₀, K_max = 2hν₀ - hν₀ = hν₀.
Question 24: A point source causing photoelectric emission from a metallic plate is moved away from the plate. The variation of photoelectric current with distance from the source is correctly represented by the graph: (See graphs in PDF)	Horizontal line (constant current)	Photoelectric current depends on light intensity, which follows the inverse square law (I ∝ 1/d²). However, at saturation, all emitted electrons are collected, making the current independent of distance (as long as intensity is above the threshold).
Question 25: A proton accelerated through a potential difference V has a de Broglie wavelength λ. On doubling the accelerating potential, the de Broglie wavelength of the proton: (1) Remains unchanged (2) Becomes double (3) Becomes four times (4) Decreases	(4) Decreases	de Broglie wavelength (λ) = h/(√2m_pV), where m_p is proton mass and V is potential difference. Doubling V reduces λ by a factor of √2.
Question 26: The kinetic energy of an electron in the ground level of a hydrogen atom is K units. The values of its potential energy and total energy respectively are: (1) -2K; -K (2) +2K; -K (3) -K; +2K (4) +K; +2K	(1) -2K; -K	In the Bohr model, total energy (E) = -K (negative because it's a bound state). Potential energy (U) = 2E = -2K. Kinetic energy (K) = -E = K.
Question 27: Two nuclei have mass numbers A and B respectively. The density ratio of the nuclei is: (1) A:B (2) √A:√B (3) A²:B² (4) 1:1	(4) 1:1	Nuclear density is constant and independent of mass number because volume is proportional to A^1/3 and mass is proportional to A.
Question 28: The shortest wavelengths emitted in the hydrogen spectrum corresponding to different spectral series are as under: (See list in PDF) (1) (A), (B), (C), (D) (2) (A), (C), (B), (D) (3) (B), (A), (D), (C) (4) (A), (C), (D), (B)	(2) (A), (C), (B), (D)	Shortest wavelength corresponds to the transition to the lowest energy level. The order of series in decreasing wavelength is Pfund (n=5), Brackett (n=4), Balmer (n=2), Lyman (n=1).
Question 29: Silicon can be doped using one of the following elements as dopant: (See list in PDF) (1) (A) and (C) only (2) (B) and (C) only (3) (A), (B), (C), and (D) (4) (C) and (D) only	(1) (A) and (C) only	For n-type semiconductors, dopants from group 15 (Arsenic and Phosphorus) are used as they have one more valence electron than silicon.
Question 30: Given below are V versus I graphs for different types of p-n junction diodes marked (A), (B), (C), and (D): (See graphs in PDF) (1) (D), (C), (A), (B) (2) (A), (C), (B), (D) (3) (B), (A), (D), (C) (4) (C), (B), (D), (A)	(1) (D), (C), (A), (B)	Match the V-I characteristics with the diode type based on their typical operation in forward or reverse bias. Solar cell (D), Zener diode (C), forward-biased p-n junction (A), photodiode (B).
Question 31: A wire carrying current I, bent as shown in the figure, is placed in a uniform field B that emerges normally out of the plane of the figure. The force on this wire is: (See diagram in PDF) (1) 4BIR, directed vertically downward (2) 3BIR, directed vertically upward (3) BI(2R + πR), vertically downward (4) 2πBIR, from P to Q	(1) 4BIR, directed vertically downward	Force (F) = IL x B. For straight segments, F = ILB sinθ = ILB (θ = 90°). For the semicircular segment, the force is equivalent to that on a straight wire of length 2R. Total force = 2(ILB) + 2(ILB) = 4BIR, vertically downward.
Question 32: The refractive index of the material of an equilateral prism is √2. The angle of minimum deviation of that prism is: (1) 60° (2) 75° (3) 30° (4) 90°	(3) 30°	For an equilateral prism (A = 60°), μ = sin[(A + δ_m)/2] / sin(A/2). Substituting μ = √2 and A = 60°, δ_m = 30°.
Question 33: The transfer of an integral number of ______ is one of the evidence of quantization of electric charge. (1) Photons (2) Nuclei (3) Electrons (4) Neutrons	(3) Electrons	Electric charge is quantized in integral multiples of the elementary charge (e), carried by electrons.
Question 34: When a slab of insulating material 4 mm thick is introduced between the plates of a parallel plate capacitor of separation 4 mm, it is found that the distance between the plates has to be increased by 3.2 mm to restore the capacitance to its original value. The dielectric constant of the material is: (1) 2 (2) 5 (3) 3 (4) 7	(2) 5	With dielectric, effective separation (d_eff) = d - t + t/K, where d is initial separation, t is dielectric thickness, and K is dielectric constant. d_eff = 7.2 mm, d = 4 mm, t = 4 mm. Solving for K gives K = 5.
Question 35: A copper ball of density 8.0 g/cc and 1 cm in diameter is immersed in oil of density 0.8 g/cc. The charge on the ball, if it remains just suspended in oil in an electric field of intensity 600π V/m acting in the upward direction, is: (1) 2 × 10⁻⁶ C (2) 2 × 10⁻⁵ C (3) 1 × 10⁻⁵ C (4) 1 × 10⁻⁶ C	(2) 2 × 10⁻⁵ C	Upward electric force (qE) balances downward gravitational force (Vρg), where V is volume, ρ is density difference, and g is acceleration due to gravity. Solving for q gives q ≈ 2 × 10⁻⁵ C.
Question 36: A metal wire is subjected to a constant potential difference. When the temperature of the metal wire increases, the drift velocity of the electron in it: (1) Increases; thermal velocity of the electrons decreases (2) Decreases; thermal velocity of the electrons decreases (3) Increases; thermal velocity of the electrons increases (4) Decreases; thermal velocity of the electrons increases	(4) Decreases; thermal velocity of the electrons increases	Drift velocity (v_d) = (eEτ)/m, where τ is relaxation time. τ decreases with increasing temperature due to increased scattering, thus decreasing v_d. Thermal velocity increases with temperature.
Question 37: For the given mixed combination of resistors, calculate the total resistance between points A and B: (See diagram in PDF)	18Ω	Simplify the circuit step-by-step, combining parallel resistors first, then adding series resistances. The equivalent resistance between A and B is 18Ω.
Question 38: A cell of emf 1.1 V and internal resistance 0.5Ω is connected to a wire of resistance 0.5Ω. Another cell of the same emf is now connected in series with the intention of increasing the current, but the current in the wire remains the same. The internal resistance of the second cell is: (1) 1Ω (2) 2.5Ω (3) 1.5Ω (4) 2Ω	(1) 1Ω	Initially, I = 1.1V/(0.5Ω + 0.5Ω) = 1.1A. After adding the second cell, total emf = 2.2V. For the current to remain 1.1A, total resistance must be 2Ω. Therefore, 0.5Ω + 0.5Ω + r = 2Ω, where r is the internal resistance of the second cell; r = 1Ω.
Question 39: P, Q, R, and S are four wires of resistances 3Ω, 3Ω, 3Ω, and 4Ω respectively. They are connected to form the four arms of a Wheatstone bridge circuit. The resistance with which S must be shunted in order that the bridge may be balanced is: (1) 14Ω (2) 12Ω (3) 15Ω (4) 7Ω	(2) 12Ω	For a balanced Wheatstone bridge, P/Q = R/S_shunted. Solving for S_shunted with P=Q=3Ω and R=3Ω, and S=4Ω, we need to find the parallel resistance that satisfies the balance equation. The parallel combination of 4Ω and 12Ω resistors results in an equivalent resistance of 3Ω satisfying the balance condition.
Question 40: Magnetic moment of a thin bar magnet is M. If it is bent into a semicircular form, its new magnetic moment will be: (1) M/π (2) M/2 (3) M (4) 2M/π	(4) 2M/π	Magnetic moment (M) = m * l, where m is pole strength and l is length. When bent into a semicircle, effective length becomes 2r/π (where r is the radius). Therefore M' = m * (2r/π) = (2M)/π.
Question 41: Ferromagnetic material used in transformers must have: (1) Low permeability and high hysteresis loss (2) High permeability and low hysteresis loss (3) High permeability and high hysteresis loss (4) Low permeability and low hysteresis loss	(2) High permeability and low hysteresis loss	High permeability ensures efficient flux conduction, and low hysteresis loss minimizes energy waste during magnetization/demagnetization cycles.
Question 42: A conducting ring of radius r is placed in a varying magnetic field perpendicular to the plane of the ring. If the rate at which the magnetic field varies is x, the electric field intensity at any point of the ring is: (1) rx (2) rx/2 (3) 2rx (4) 4rx	(2) rx/2	Induced emf (ε) = -dΦ/dt = -πr²x. Electric field (E) = ε/(2πr) = (rx)/2.
Question 43: A 50 Hz AC current of crest value 1 A flows through the primary of a transformer. If the mutual inductance between the primary and secondary is 0.5 H, the crest voltage induced in the secondary is: (1) 75 V (2) 150 V (3) 100 V (4) 200 V	(3) 100 V	Induced emf (ε) = M(dI/dt). For a sinusoidal current, dI/dt = 2πfI₀. ε = 0.5 H * 2π * 50 Hz * 1 A ≈ 100 V.
Question 44: A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per meter. At the center of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π² Ω, then the total charge flowing through the coil during this time is: (1) 16µC (2) 32µC (3) 16πµC (4) 32πµC	(3) 32µC	Induced emf (ε) = -N(dΦ/dt) = -N * A * μ₀n(dI/dt). Total charge (Q) = εt/R. Substituting the values, Q ≈ 32µC.
Question 45: Lower half of a convex lens is made opaque. Which of the following statement describes the image of the object placed in front of the lens? (A) No change in image (B) Image will show only half of the object (C) Intensity of image gets reduced (1) (A) only (2) (B) only (3) (C) only (4) (B) and (C) only	(3) (C) only	Blocking part of the lens reduces the intensity (brightness) but doesn't affect the image formation itself.
Question 46: Two slits are made 0.1 mm apart, and the screen is placed 2 m away. The fringe separation when a light of wavelength 500 nm is used is: (1) 1 cm (2) 0.15 cm (3) 1.5 cm (4) 0.1 cm	(1) 1 cm	Fringe separation (β) = λD/d, where λ is wavelength, D is distance to screen, and d is slit separation. β = (500 × 10⁻⁹ m * 2 m) / (0.1 × 10⁻³ m) = 1 cm.
Question 47: For an astronomical telescope having an objective lens of focal length 10 m and an eyepiece lens of focal length 10 cm, the tube length and magnification respectively are: (1) 20 cm, 1 (2) 1000 cm, 1 (3) 1010 cm, 1 (4) 1010 cm, 100	(4) 1010 cm, 100	Tube length = f_o + f_e = 10 m + 0.1 m = 10.1 m = 1010 cm. Magnification = -f_o/f_e = -10 m / 0.1 m = -100. (The negative sign indicates an inverted image).
Question 48: According to Bohr's Model: (See statements in PDF) (1) (A), (B), and (C) only (2) (A), (B), and (D) only (3) (A), (B), (C), and (D) (4) (B), (C), and (D) only	(4) (B), (C), and (D) only	According to Bohr's model: radius ∝ n², speed ∝ 1/n, total energy ∝ -1/n².
Question 49: For a full-wave rectifier, if the input frequency is 50 Hz, the output frequency will be: (1) 50 Hz (2) 100 Hz (3) 25 Hz (4) 0 Hz	(2) 100 Hz	A full-wave rectifier utilizes both halves of the AC cycle, doubling the output frequency.
Question 50: For an electric dipole in a non-uniform electric field with dipole moment parallel to the direction of the field, the force F and torque τ on the dipole respectively are: (1) F = 0, τ = 0 (2) F ≠ 0, τ = 0 (3) F = 0, τ ≠ 0 (4) F ≠ 0, τ ≠ 0	(2) F ≠ 0, τ = 0	In a non-uniform field, there's a net force due to varying field strength, but no torque if the dipole is aligned with the field (τ = pEsinθ, θ = 0).

CUET Physics Questions with Solution Set B

Question	Answer	Solution
Question 1: 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	(1) 1:1	The energy density of an electromagnetic wave is equal for the electric and magnetic fields because E = cB.
Question 2: Match List-I with List-II: List-I: Opposition to ac... (A) Opposition to ac (B) Opposition to ac (C) Opposition to ac (D) Opposition to ac List-II: (I) Impedance, (II) Capacitive reactance, (III) Inductive reactance, (IV) Resistance (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (2) (A) - (IV), (B) - (II), (C) - (III), (D) - (I) (3) (A) - (I), (B) - (II), (C) - (IV), (D) - (III) (4) (A) - (III), (B) - (IV), (C) - (I), (D) - (II)	(2) (A) - (IV), (B) - (II), (C) - (III), (D) - (I)	Resistance is frequency independent. Capacitive reactance decreases with frequency. Inductive reactance increases with frequency. Impedance depends on both reactance and resistance.
Question 3: Of the following, the correct arrangement of the 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	(3) Radio waves, Microwaves, Infrared waves, Visible waves, X-rays	This is the correct order of decreasing wavelength for the electromagnetic spectrum.
Question 4: Match the electromagnetic waves in Column I with their production methods in Column II: Column I: (A) Microwaves, (B) Infrared, (C) X-rays, (D) Radio waves Column II: (I) LC oscillator, (II) Magnetron, (III) Vibration of atoms/molecules, (IV) Bombarding large atomic number metal target with fast-moving electrons (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)	(2) (A) - (II), (B) - (III), (C) - (IV), (D) - (I)	Magnetrons produce microwaves. Infrared waves result from atomic/molecular vibrations. X-rays are produced by bombarding a metal target. Radio waves are generated by LC oscillators.
Question 5: In the figure given below, APB is a curved surface of radius of curvature 10 cm separating air and a transparent material (μ = 1). A point object O is placed in air on the principal axis of the surface 20 cm from P. The distance of the image of O from P will be: (Diagram shown in PDF) (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	(1) 16 cm left of P in air	Using the refraction formula for curved surfaces and substituting values gives v = -16cm, indicating the image is 16cm to the left of P in air.
Question 6: For fixed values of radii of curvature of a lens, the power of the lens will be: (1) P ∝ (μ − 1) (2) P ∝ μ² (3) P ∝ 1/μ (4) P ∝ (μ − 2)	(1) P ∝ (μ − 1)	Lens power (P) is directly proportional to (μ -1) when radii of curvature are fixed.
Question 7: The graph correctly representing the variation of image distance v for a convex lens of focal length f versus object distance u is: (Graphs shown in PDF)	(Graph B) - Hyperbolic curve asymptotic to both axes.	The lens formula (1/f = 1/u + 1/v) represents a hyperbola.
Question 8: Using light from a monochromatic source to study diffraction in a single slit of width 0.1 mm, the linear width of the central maximum is measured to be 5 mm on a screen held 50 cm away. The wavelength of light used is: (1) 2.5 x 10^-7 m (2) 4 x 10^-7 m (3) 5 x 10^-7 m (4) 7.5 x 10^-7 m	(3) 5 x 10^-7 m	Using the formula for the width of the central maximum in single-slit diffraction and substituting values, λ = 5 x 10^-7 m.
Question 9: Radiation of frequency 2v_o is incident on a metal with threshold frequency v_o. The correct statement is: (1) No photoelectrons will be emitted (2) All photoelectrons emitted will have kinetic energy equal to hv_o (3) Maximum kinetic energy of photoelectrons emitted can be hv_o (4) Maximum kinetic energy of photoelectrons emitted will be 2hv_o	(3) Maximum kinetic energy of photoelectrons emitted can be hv_o	Using Einstein's photoelectric equation, K_max = h(2v_o - v_o) = hv_o
Question 10: A point source causing photoelectric emission from a metallic plate is moved away from the plate. The variation of photoelectric current with distance from the source is correctly represented by the graph: (Graphs shown in PDF)	A horizontal line (constant current)	Photoelectric current saturates when light intensity is above the threshold, remaining constant with distance.
Question 11: A proton accelerated through a potential difference V has a de Broglie wavelength λ. On doubling the accelerating potential, the de Broglie wavelength of the proton: (1) Remains unchanged (2) Becomes double (3) Becomes four times (4) Decreases	(4) Decreases	The de Broglie wavelength is inversely proportional to the square root of the accelerating potential. Doubling V reduces λ by a factor of √2.
Question 12: The kinetic energy of an electron in the ground level of a hydrogen atom is K units. The values of its potential energy and total energy respectively are: (1) -2K; -K (2) +2K; -K (3) -K; +2K (4) +K; +2K	(1) -2K; -K	In the Bohr model, total energy E = -K, and potential energy U = 2E = -2K. Kinetic energy K = -E.
Question 13: Two nuclei have mass numbers A and B respectively. The density ratio of the nuclei is: (1) A:B (2) √A:√B (3) A²:B² (4) 1:1	(4) 1:1	Nuclear density is approximately constant for all nuclei.
Question 14: The shortest wavelengths emitted in the hydrogen spectrum corresponding to different spectral series are as under: (A) Pfund series, (B) Balmer series, (C) Brackett series, (D) Lyman series. The wavelengths arranged correctly in decreasing order are: (1) (A), (B), (C), (D) (2) (A), (C), (B), (D) (3) (B), (A), (D), (C) (4) (A), (C), (D), (B)	(2) (A), (C), (B), (D)	Shortest wavelength corresponds to the transition to the lowest energy level. The order is Pfund (n=5), Brackett (n=4), Balmer (n=2), Lyman (n=1).
Question 15: Silicon can be doped using one of the following elements as dopant: (A) Arsenic, (B) Indium, (C) Phosphorus, (D) Boron. To get an n-type semiconductor, the dopants that can be used are: (1) (A) and (C) only (2) (B) and (C) only (3) (A), (B), (C), and (D) (4) (C) and (D) only	(1) (A) and (C) only	Arsenic and Phosphorus are group 15 elements, providing extra electrons for n-type doping.
Question 16: Given below are V versus I graphs for different types of p-n junction diodes marked (A), (B), (C), and (D): (Graphs shown in PDF) The correct sequence of graphs corresponding to forward biased p-n junction, Zener diode, Photodiode, and Solar cell in order is: (1) (D), (C), (A), (B) (2) (A), (C), (B), (D) (3) (B), (A), (D), (C) (4) (C), (B), (D), (A)	(1) (D), (C), (A), (B)	Match the characteristic I-V curves with the corresponding diode types based on their operational modes.
Question 17: A wire carrying current I, bent as shown in the figure, is placed in a uniform field B that emerges normally out of the plane of the figure. The force on this wire is: (Diagram shown in PDF) (1) 4BIR, directed vertically downward (2) 3BIR, directed vertically upward (3) BI(2R + πR), vertically downward (4) 2πBIR, from P to Q	(1) 4BIR, directed vertically downward	Calculate the force on each segment of the wire and sum the vectors. The straight segments contribute 2BIR each, and the semicircular segment contributes 2BIR.
Question 18: The refractive index of the material of an equilateral prism is √2. The angle of minimum deviation of that prism is: (1) 60° (2) 75° (3) 30° (4) 90°	(3) 30°	Use the formula relating refractive index, prism angle, and minimum deviation angle for an equilateral prism (A=60°).
Question 19: The transfer of an integral number of ______ is one of the evidence of quantization of electric charge. (1) Photons (2) Nuclei (3) Electrons (4) Neutrons	(3) Electrons	Electric charge is quantized in multiples of the elementary charge (e), carried by electrons.
Question 20: When a slab of insulating material 4mm thick is introduced between the plates of a parallel plate capacitor of separation 4mm, it is found that the distance between the plates has to be increased by 3.2mm to restore the capacitance to its original value. The dielectric constant of the material is: (1) 2 (2) 5 (3) 3 (4) 7	(2) 5	The effective separation with the dielectric is d_eff = d - t/k + t, where d is the initial separation, and t is the thickness of the dielectric. Solve for k.
Question 21: A copper ball of density 8.0 g/cc and 1 cm in diameter is immersed in oil of density 0.8 g/cc. The charge on the ball, if it remains just suspended in oil in an electric field of intensity 600π V/m acting in the upward direction, is: (1) 2 x 10^-6 C (2) 2 x 10^-5 C (3) 1 x 10^-5 C (4) 1 x 10^-6 C	(2) 2 x 10^-5 C	The upward electric force (qE) balances the downward gravitational force (V_sphere(ρ_copper - ρ_oil)g). Solve for q.
Question 22: A metal wire is subjected to a constant potential difference. When the temperature of the metal wire increases, the drift velocity of the electron in it: (1) Increases; thermal velocity of the electrons decreases (2) Decreases; thermal velocity of the electrons decreases (3) Increases; thermal velocity of the electrons increases (4) Decreases; thermal velocity of the electrons increases	(4) Decreases; thermal velocity of the electrons increases	Drift velocity is inversely proportional to relaxation time (τ), which decreases with temperature increase due to increased scattering. Thermal velocity increases with temperature.
Question 23: For the given mixed combination of resistors, calculate the total resistance between points A and B: (Diagram shown in PDF) (1) 9 Ω (2) 18 Ω (3) 4 Ω (4) 12 Ω	(2) 18 Ω	Simplify the circuit step-by-step using parallel and series resistor rules.
Question 24: A cell of emf 1.1 V and internal resistance 0.5 Ω is connected to a wire of resistance 0.5 Ω. Another cell of the same emf is now connected in series with the intention of increasing the current, but the current in the wire remains the same. The internal resistance of the second cell is: (1) 1 Ω (2) 2.5 Ω (3) 1.5 Ω (4) 2 Ω	(1) 1 Ω	Use Ohm's law to calculate the current initially and with the second cell in series. Set the currents equal and solve for the internal resistance of the second cell.
Question 25: P, Q, R, and S are four wires of resistances 3Ω, 3Ω, 3Ω, and 4Ω respectively. They are connected to form the four arms of a Wheatstone bridge circuit. The resistance with which S must be shunted in order that the bridge may be balanced is: (1) 14 Ω (2) 12 Ω (3) 15 Ω (4) 7 Ω	(2) 12 Ω	For a balanced Wheatstone bridge, the ratio of resistances in opposite arms must be equal: P/Q = R/S. Solve for the shunted resistance.
Question 26: Magnetic moment of a thin bar magnet is M. If it is bent into a semicircular form, its new magnetic moment will be: (1) M/π (2) 2M/π (3) M (4) 2M	(2) 2M/π	Magnetic moment (M = ml) changes because bending the magnet changes its effective length (l).
Question 27: Ferromagnetic material used in transformers must have: (1) Low permeability and high hysteresis loss (2) High permeability and low hysteresis loss (3) High permeability and high hysteresis loss (4) Low permeability and low hysteresis loss	(2) High permeability and low hysteresis loss	High permeability minimizes energy loss, and low hysteresis loss reduces energy waste during magnetization/demagnetization cycles.
Question 28: A conducting ring of radius r is placed in a varying magnetic field perpendicular to the plane of the ring. If the rate at which the magnetic field varies is x, the electric field intensity at any point of the ring is: (1) rx (2) rx/2 (3) 2rx (4) rx/4	(2) rx/2	Using Faraday's law and relating induced emf to electric field, E = rx/2.
Question 29: A 50 Hz AC current of crest value 1 A flows through the primary of a transformer. If the mutual inductance between the primary and secondary is 0.5 H, the crest voltage induced in the secondary is: (1) 75 V (2) 150 V (3) 100 V (4) 200 V	(3) 100 V	Induced emf in the secondary is given by e = M(dI/dt). For a sinusoidal current, dI/dt = I_o(2πf), where I_o is crest current and f is frequency.
Question 30: A long solenoid of diameter 0.1 m has 2 x 10⁴ turns per meter. At the center of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π² Ω, then the total charge flowing through the coil during this time is: (1) 16 μC (2) 32 μC (3) 16π μC (4) 32π μC	(2) 32 μC	Use Faraday's law to find the induced emf in the coil, then use Q = (emf)(t)/R to find the total charge.
Question 31: Lower half of a convex lens is made opaque. Which of the following statement describes the image of the object placed in front of the lens? (A) No change in image (B) Image will show only half of the object (C) Intensity of image gets reduced (1) (A) only (2) (B) only (3) (C) only (4) (B) and (C) only	(3) (C) only	Blocking half the lens reduces the amount of light forming the image, decreasing intensity. The image remains complete.
Question 32: Two slits are made 0.1 mm apart, and the screen is placed 2 m away. The fringe separation when a light of wavelength 500 nm is used is: (1) 1 cm (2) 0.15 cm (3) 1.5 cm (4) 0.1 cm	(1) 1 cm	Use the formula for fringe separation in Young's double-slit experiment: Δx = λL/d.
Question 33: For an astronomical telescope having an objective lens of focal length 10 m and an eyepiece lens of focal length 10 cm, the tube length and magnification respectively are: (1) 20 cm, 1 (2) 1000 cm, 1 (3) 1010 cm, 1 (4) 1010 cm, 100	(4) 1010 cm, 100	Tube length L = f_o + f_e. Magnification M = f_o/f_e.
Question 34: According to Bohr's Model: (A) The radius of the orbiting electron is directly proportional to n. (B) The speed of the orbiting electron is directly proportional to 1/n. (C) The magnitude of the total energy of the orbiting electron is directly proportional to 1/n². (D) The radius of the orbiting electron is directly proportional to n². Choose the correct answer: (1) (A), (B), and (C) only (2) (A), (B), and (D) only (3) (A), (B), (C), and (D) (4) (B), (C), and (D) only	(4) (B), (C), and (D) only	According to Bohr's model: r ∝ n², v ∝ 1/n, E ∝ 1/n²
Question 35: For a full-wave rectifier, if the input frequency is 50 Hz, the output frequency will be: (1) 50 Hz (2) 100 Hz (3) 25 Hz (4) 0 Hz	(2) 100 Hz	A full-wave rectifier utilizes both halves of the AC cycle, doubling the output frequency.
Question 36: For an electric dipole in a non-uniform electric field with dipole moment parallel to the direction of the field, the force F and torque τ on the dipole respectively are: (1) F = 0, τ = 0 (2) F ≠ 0, τ = 0 (3) F = 0, τ ≠ 0 (4) F ≠ 0, τ ≠ 0	(2) F ≠ 0, τ = 0	In a non-uniform field, there's a net force due to varying field strength, but no torque if the dipole is aligned with the field (τ = pEsinθ, θ = 0).
37. Two large plane parallel sheets with equal but opposite surface charge densities +σ and −σ have a point charge q placed at points P₁, P₂, and P₃. The forces F₁, F₂, and F₃ experienced by q are: (Diagram provided in PDF) (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	(2) F₁ = 0, F₂ ≠ 0, F₃ = 0	The electric field outside the parallel plates is zero due to cancellation of fields from the two plates. Inside the plates, the field is uniform, and the charge experiences a force.
38. 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) Q₁/Q₂ = R₂/R₁ (2) Q₁/Q₂ = R₁/R₂ (3) Q₁/R₁ > Q₂/R₂ (4) Q₁/R₁ = Q₂/R₂	(4) Q₁/R₁ = Q₂/R₂	When the spheres are in contact, charge distributes proportionally to their capacitances, which are proportional to their radii. Therefore, the final charge ratio is equal to the ratio of their radii.
39. 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	(2) 2d	Coulomb's law states F ∝ q₁q₂/r². Doubling both charges quadruples the force, so the distance must be doubled to maintain the original force.
40. Two parallel plate capacitors of capacitances 2µF and 3µF 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 U₁ and U₂ respectively are related as: (1) V₁/V₂ = 3/2; U₁/U₂ = 3/2 (2) V₁/V₂ = 2/3; U₁/U₂ = 2/3 (3) V₁/V₂ = 3/2; U₁/U₂ = 2/3 (4) V₁/V₂ = 2/3; U₁/U₂ = 3/2	(1) V₁/V₂ = 3/2; U₁/U₂ = 3/2	In a series combination, the charge is the same on both capacitors. The voltage divides inversely proportional to capacitance (V ∝ 1/C). The energy stored (U = 1/2CV²) also follows the same ratio.
41. Two resistances of 100Ω and 200Ω are connected in series across a 20 V battery. The reading in a 200Ω voltmeter connected across the 200Ω resistance is: (1) 4 V (2) 2/3 V (3) 10 V (4) 16 V	(3) 10 V	The voltage across a resistor in a series circuit is proportional to its resistance. The voltage across the 200Ω resistor is (200/(100+200)) * 20 V = 10 V.
42. The current through a 4/3 Ω external resistance connected to a parallel combination of two cells of 2 V and 1 V emf and internal resistances of 1Ω and 2Ω respectively is: (1) 1 A (2) 1/4 A (3) 3/4 A (4) 5/6 A	(4) 5/6 A	First, find the equivalent emf and internal resistance for the parallel combination of cells. Then, use Ohm's law to find the current in the circuit.
43. A metallic wire of uniform cross-sectional area has a resistance R, resistivity ρ, 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 are now denoted as R', ρ', and P'. The corresponding relations are: (1) ρ' = 2ρ, R' = 2R, P' = 2P (2) ρ' = ρ/2, R' = R/2, P' = P/2 (3) ρ' = ρ, R' = 16R, P' = P/16 (4) ρ' = ρ, R' = 16R, P' = 16P	(3) ρ' = ρ, R' = 16R, P' = P/16	Stretching the wire changes its length and area, affecting resistance (R = ρL/A). Resistivity (ρ) remains constant. Power (P = V²/R) changes accordingly.
44. Three magnetic materials are listed: (A) Paramagnetics, (B) Diamagnetics, and (C) Ferromagnetics. Choose the correct order of the materials in increasing magnetic susceptibility: (1) (A), (B), (C) (2) (C), (A), (B) (3) (B), (A), (C) (4) (B), (C), (A)	(3) (B), (A), (C)	Diamagnetic materials have the lowest susceptibility, followed by paramagnetic and then ferromagnetic materials.
45. Two infinitely long straight parallel conductors carrying currents I₁ and I₂ 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: (1) Proportional to L but independent of I₁ × I₂ (2) Proportional to I₁ × I₂ but independent of length L (3) Proportional to I₁ × I₂ × L (4) Proportional to L/(I₁ × I₂)	(3) Proportional to I₁ × I₂ × L	The force between two parallel current-carrying conductors is given by F = (μ₀I₁I₂L)/(2πd). The force is directly proportional to the product of currents (I₁I₂) and the length (L).
46. In the circuit shown below, a current 3I enters at A. The semicircular parts ABC and ADC have equal radii r but resistances 2R and R, respectively. The magnetic field at the center of the circular loop ABCD is: (Diagram provided in PDF) (1) (μ₀I)/(4r) out of the plane (2) (μ₀I)/(4r) into the plane (3) (3μ₀I)/(4r) out of the plane (4) (3μ₀I)/(4r) into the plane	(1) (μ₀I)/(4r) out of the plane	The current splits inversely proportional to the resistances. Calculate the magnetic field due to each semicircular part and add them vectorially using the right-hand rule.
47. 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: (1) Zero (2) 2 × 10^-4 Nm (3) 2 × 10^-2 Nm (4) 2 Nm	(2) 2 × 10^-4 Nm	The torque on a current loop in a magnetic field is given by τ = NIABsinθ. Since the magnetic field is parallel to the plane of the loop, θ = 90°, and the torque is maximum.
48. In an AC circuit, the current leads the voltage by π/2. The circuit is: (1) Purely resistive (2) Circuit elements with resistance equal to reactance (3) Purely inductive (4) Purely capacitive	(4) Purely capacitive	In a purely capacitive circuit, the current leads the voltage by 90° (π/2).
49. In a pair of adjacent coils, for a change of current in one coil from 0 A to 10 A in 0.25 s, the magnetic flux in the adjacent coil changes by 15 Wb. The mutual inductance of the coils is: (1) 120 H (2) 12 H (3) 1.5 H (4) 0.75 H	(3) 1.5 H	Mutual inductance (M) is defined as M = ΔΦ/ΔI. Substitute the given values to find M.
50. A wire of irregular shape (figure a) and a circular loop of wire (figure b) are placed in different uniform magnetic fields. In figure (a), the magnetic field is perpendicular into the plane. In figure (b), the magnetic field is perpendicular out of the plane. The wire in figure (a) is turning into a circular loop, and that in figure (b) into a narrow straight wire. The direction of induced current will be: (Diagrams provided in PDF) (1) Clockwise in both (a) and (b) (2) Anticlockwise in both (a) and (b) (3) Clockwise in (a) and anticlockwise in (b) (4) Anticlockwise in (a) and clockwise in (b)	(2) Anticlockwise in both (a) and (b)	Apply Lenz's law to determine the direction of induced current in each case. The induced current will oppose the change in magnetic flux.

CUET 2024 Physics Question Paper (SET C) with Solutions

Question	Answer	Solution
Q1. The transfer of an integral number of ______ is one of the evidence of quantization of electric charge. 1. Photons 2. Nuclei 3. Electrons 4. Neutrons	(3) Electrons	Electric charge is quantized and occurs in integral multiples of the elementary charge e. This is attributed to the transfer of electrons, which carry charge e.
Q2. When a slab of insulating material 4mm thick is introduced between the plates of a parallel plate capacitor of separation 4mm, it is found that the distance between the plates has to be increased by 3.2mm to restore the capacitance to its original value. The dielectric constant of the material is: 1. 2 2. 5 3. 3 4. 7	(2) 5	When the dielectric is introduced, the effective separation becomes: deff = d + t/k. Solving for k with d = 4mm and deff = 7.2mm (4 + 3.2), k = 5.
Q3. A copper ball of density 8.0g/cc and 1cm in diameter is immersed in oil of density 0.8g/cc. The charge on the ball, if it remains just suspended in oil in an electric field of intensity 600πV/m acting in the upward direction, is: 1. 2 × 10^-6 C 2. 2 × 10^-5 C 3. 1 × 10^-5 C 4. 1 × 10^-6 C	(2) 2 × 10^-5 C	The electric force (qE) must balance the gravitational force minus the buoyant force. Solving for q yields approximately 2 × 10^-5 C.
Q4. A metal wire is subjected to a constant potential difference. When the temperature of the metal wire increases, the drift velocity of the electron in it: 1. Increases; thermal velocity of the electrons decreases 2. Decreases; thermal velocity of the electrons decreases 3. Increases; thermal velocity of the electrons increases 4. Decreases; thermal velocity of the electrons increases	(4) Decreases; thermal velocity of the electrons increases	Drift velocity (v_d) is inversely proportional to relaxation time (τ), which decreases with temperature. Thermal velocity increases with temperature.
Q5. For the given mixed combination of resistors, calculate the total resistance between points A and B (Diagram provided in PDF). 1. 9Ω 2. 18Ω 3. 4Ω 4. 14Ω	(2) 18Ω	Solve using series and parallel resistor combination rules. The parallel combinations simplify to 3Ω and 3Ω which are then in series with 12Ω, for a total of 18Ω
Q6. A cell of emf 1.1 V and internal resistance 0.5Ω is connected to a wire of resistance 0.5Ω. Another cell of the same emf is now connected in series with the intention of increasing the current, but the current in the wire remains the same. The internal resistance of the second cell is: 1. 1Ω 2. 2.5Ω 3. 1.5Ω 4. 2Ω	(1) 1Ω	Use Ohm's law to calculate the initial current. With the second cell added in series, the total EMF is doubled, but to keep the current the same, the total resistance must also be doubled. This allows for calculation of the second cell's internal resistance.
Q7. P, Q, R, and S are four wires of resistances 3Ω, 3Ω, 3Ω, and 4Ω respectively. They are connected to form the four arms of a Wheatstone bridge circuit. The resistance with which S must be shunted in order that the bridge may be balanced is: 1. 14Ω 2. 12Ω 3. 15Ω 4. 7Ω	(2) 12Ω	For a balanced Wheatstone bridge, the ratio of resistances in opposite arms must be equal (P/Q = R/S). Solving for the shunt resistance required to satisfy this condition yields 12Ω.
Q8. Magnetic moment of a thin bar magnet is M. If it is bent into a semicircular form, its new magnetic moment will be: 1. M/π 2. M/2 3. M 4. 2M/π	(4) 2M/π	The magnetic moment (M) is the product of pole strength (m) and effective length (l). Bending changes the effective length to 2r/π (where r is the radius of the semicircle).
Q9. Ferromagnetic material used in transformers must have: 1. Low permeability and high hysteresis loss 2. High permeability and low hysteresis loss 3. High permeability and high hysteresis loss 4. Low permeability and low hysteresis loss	(2) High permeability and low hysteresis loss	High permeability minimizes energy loss during magnetization and demagnetization. Low hysteresis loss minimizes energy wastage during each cycle.
Q10. A conducting ring of radius r is placed in a varying magnetic field perpendicular to the plane of the ring. If the rate at which the magnetic field varies is x, the electric field intensity at any point of the ring is: 1. rx 2. rx/2 3. 2rx 4. 4rx	(2) rx/2	Use Faraday's law to find the induced EMF and then relate it to the electric field intensity around the ring.
Q11. A 50 Hz AC current of crest value 1 A flows through the primary of a transformer. If the mutual inductance between the primary and secondary is 0.5 H, the crest voltage induced in the secondary is: 1. 75 V 2. 150 V 3. 100 V 4. 200 V	(3) 100 V	Using e = M(dI/dt) and a sinusoidal current, the induced EMF is calculated as 100V.
Q12. A long solenoid of diameter 0.1m has 2 × 10⁴ turns per meter. At the center of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05s. If the resistance of the coil is 10π² Ω, then the total charge flowing through the coil during this time is: 1. 16 μC 2. 32 μC 3. 16π μC 4. 32π μC	(2) 32 μC	Calculate the induced EMF using Faraday's Law, then use Q = EMF*t/R to find the charge.
Q13. Lower half of a convex lens is made opaque. Which of the following statement describes the image of the object placed in front of the lens? (A) No change in image (B) Image will show only half of the object (C) Intensity of image gets reduced 1. (A) only 2. (B) only 3. (C) only 4. (B) and (C) only	(3) (C) only	Blocking half the lens reduces the intensity but doesn't affect the image formation itself.
Q14. Two slits are made 0.1mm apart, and the screen is placed 2m away. The fringe separation when a light of wavelength 500 nm is used is: 1. 1 cm 2. 0.15 cm 3. 1.5 cm 4. 0.1 cm	(1) 1 cm	Use the fringe separation formula: Δx = λL/d
Q15. For an astronomical telescope having an objective lens of focal length 10m and an eyepiece lens of focal length 10cm, the tube length and magnification respectively are: 1. 20 cm, 1 2. 1000 cm, 1 3. 1010 cm, 1 4. 1010 cm, 100	(4) 1010 cm, 100	Tube length = f_o + f_e; Magnification = f_o/f_e
Q16. According to Bohr's Model: (statements about radius, speed, and energy of orbiting electron given in PDF) Choose the correct answer. 1. (A), (B), and (C) only 2. (A), (B), and (D) only 3. (A), (B), (C), and (D) 4. (B), (C), and (D) only	(4) (B), (C), and (D) only	Apply Bohr's model relationships for radius, speed, and energy levels.
Q17. For a full-wave rectifier, if the input frequency is 50 Hz, the output frequency will be: 1. 50 Hz 2. 100 Hz 3. 25 Hz 4. 0 Hz	(2) 100 Hz	Full-wave rectifiers double the input frequency.
Q18. For an electric dipole in a non-uniform electric field with dipole moment parallel to the direction of the field, the force F and torque τ on the dipole respectively are: 1. F = 0, τ = 0 2. F ≠ 0, τ = 0 3. F = 0, τ ≠ 0 4. F ≠ 0, τ ≠ 0	(2) F ≠ 0, τ = 0	In a non-uniform field, there's a net force but no torque if the dipole is aligned with the field.
Q19. Two large plane parallel sheets with equal but opposite surface charge densities +σ and -σ have a point charge q placed at points P1, P2, and P3. The forces F1, F2, and F3 experienced by q are: (Diagram in PDF) 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	(2) F₁ = 0, F₂ ≠ 0, F₃ = 0	The field is zero outside the plates and uniform inside.
Q20. 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. Q₁/Q₂ = R₂/R₁ 2. Q₁/Q₂ < R₂/R₁ 3. Q₁/Q₂ > R₂/R₁ 4. Q₁/Q₂ = R₁/R₂	(4) Q₁/Q₂ = R₁/R₂	Charge distribution is proportional to the sphere's radius.
Q21. 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	(2) 2d	Using Coulomb's Law (F = kq₁q₂/r²), doubling both charges quadruples the force. To maintain the original force, the distance must be doubled (√4d = 2d).
Q22. Two parallel plate capacitors of capacitances 2µF and 3µF 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 U₁ and U₂ respectively are related as: 1. V₁/V₂ = 2/3; U₁/U₂ = 3/2 2. V₁/V₂ = 3/2; U₁/U₂ = 2/3 3. V₁/V₂ = 2/3; U₁/U₂ = 2/3 4. V₁/V₂ = 3/2; U₁/U₂ = 3/2	(1) V₁/V₂ = 2/3; U₁/U₂ = 3/2	In a series combination, charge is the same, and voltage is inversely proportional to capacitance (V ∝ 1/C). Energy is proportional to CV².
Q23. Two resistances of 100Ω and 200Ω are connected in series across a 20 V battery. The reading in a 200Ω voltmeter connected across the 200Ω resistance is: 1. 4 V 2. 20/3 V 3. 10 V 4. 16 V	(3) 10 V	Calculate the total resistance (300Ω), then the current (I = V/R = 20V/300Ω = 1/15 A). The voltage across the 200Ω resistor is V = IR = (1/15 A)(200Ω) = 10V.
Q24. The current through a 4/3Ω external resistance connected to a parallel combination of two cells of 2 V and 1 V emf and internal resistances of 1Ω and 2Ω respectively is: 1. 1 A 2. 1/4 A 3. 5/4 A 4. 5/3 A	(4) 5/3 A	First, find the equivalent EMF and internal resistance of the parallel combination of cells. Then, calculate the total resistance (external + internal) and use Ohm's law (I = V/R).
Q25. A metallic wire of uniform cross-sectional area has a resistance R, resistivity ρ, 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 are now denoted as R', ρ', and P'. The corresponding relations are: 1. ρ' = 2ρ, R' = 2R, P' = 2P 2. ρ' = ρ/2, R' = R/2, P' = P/2 3. ρ' = ρ, R' = 16R, P' = P/16 4. ρ' = ρ, R' = R/16, P' = 16P	(3) ρ' = ρ, R' = 16R, P' = P/16	Stretching changes length and area; resistivity remains constant. Resistance is proportional to length/area. Power (P = V²/R) changes accordingly.
Q26. Three magnetic materials are listed: (A) Paramagnetics, (B) Diamagnetics, and (C) Ferromagnetics. Choose the correct order of the materials in increasing magnetic susceptibility: 1. (A), (B), (C) 2. (C), (A), (B) 3. (B), (A), (C) 4. (B), (C), (A)	(3) (B), (A), (C)	Diamagnetic materials have the lowest susceptibility, followed by paramagnetic, and then ferromagnetic materials.
Q27. Two infinitely long straight parallel conductors carrying currents I₁ and I₂ 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: 1. Proportional to L but independent of I₁ × I₂ 2. Proportional to I₁ × I₂ but independent of length L 3. Proportional to I₁ × I₂ × L 4. Proportional to L/(I₁ × I₂)	(3) Proportional to I₁ × I₂ × L	The force between parallel conductors is given by F/L = (μ₀I₁I₂)/(2πd). Therefore, F is proportional to I₁, I₂, and L.
Q28. In the circuit shown below, a current 3I enters at A. The semicircular parts ABC and ADC have equal radii r but resistances 2R and R, respectively. The magnetic field at the center of the circular loop ABCD is: (Diagram in PDF) 1. (μ₀I) / (4r), out of the plane 2. (μ₀I) / (4r), into the plane 3. (3μ₀I) / (4r), out of the plane 4. (3μ₀I) / (4r), into the plane	(1) (μ₀I) / (4r), out of the plane	The current splits inversely proportional to the resistances. Calculate the magnetic field contribution from each semicircular segment and add them vectorially.
Q29. 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: 1. Zero 2. 2 × 10^-4 Nm 3. 2 × 10^-2 Nm 4. 2 Nm	(2) 2 × 10^-4 Nm	Use the torque formula (τ = NIABsinθ). Since the field is parallel to the plane, θ = 90°.
Q30. In an AC circuit, the current leads the voltage by π/2. The circuit is: 1. Purely resistive 2. Circuit elements with resistance equal to reactance 3. Purely inductive 4. Purely capacitive	(4) Purely capacitive	In a purely capacitive circuit, the current leads the voltage by π/2 (90°).
Q31. In a pair of adjacent coils, for a change of current in one coil from 0 A to 10 A in 0.25s, the magnetic flux in the adjacent coil changes by 15 Wb. The mutual inductance of the coils is: 1. 120 H 2. 12 H 3. 1.5 H 4. 0.75 H	(3) 1.5 H	Use the formula for mutual inductance: M = ΔΦ/ΔI
Q32. A wire of irregular shape (figure a) and a circular loop of wire (figure b) are placed in different uniform magnetic fields. In figure (a), the magnetic field is perpendicular into the plane. In figure (b), the magnetic field is perpendicular out of the plane. The wire in figure (a) is turning into a circular loop, and that in figure (b) into a narrow straight wire. The direction of induced current will be: (Diagrams in PDF) 1. Clockwise in both (a) and (b) 2. Anticlockwise in both (a) and (b) 3. Clockwise in (a) and anticlockwise in (b) 4. Anticlockwise in (a) and clockwise in (b)	(2) Anticlockwise in both (a) and (b)	Apply Lenz's Law. In (a), the increasing flux requires an induced current to oppose it (anticlockwise). In (b), the decreasing flux requires an induced current to maintain it (anticlockwise).
Q33. Match List-I with List-II: (Lists and diagrams provided in PDF) 1. (A)-(I), (B)-(II), (C)-(III), (D)-(IV) 2. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) 3. (A)-(I), (B)-(II), (C)-(IV), (D)-(III) 4. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)	(2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)	Match the type of opposition to AC (List I) with its corresponding frequency dependence (List II): Resistance is frequency independent, capacitive reactance decreases with frequency, inductive reactance increases with frequency, and impedance is a combination of resistance and reactance.
Q34. 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	(1) 1:1	The energy density is equal for both electric and magnetic fields in an electromagnetic wave.
Q35. Of the following, 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	(3) Radio waves, Microwaves, Infrared waves, Visible waves, X-rays	Order the electromagnetic spectrum by decreasing wavelength: Radio, Microwave, Infrared, Visible, X-ray.
Q36. Match the electromagnetic waves in Column I with their production methods in Column II: (Columns provided in PDF) 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)	(2) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)	Match the electromagnetic waves (microwaves, infrared, x-rays, radio waves) with their corresponding production methods (magnetron, vibration of atoms/molecules, bombarding metal target, LC oscillator).
Q37. In the figure given below, APB is a curved surface of radius of curvature 10 cm separating air and a transparent material (μ = 1.5). A point object O is placed in air on the principal axis of the surface 20 cm from P. The distance of the image of O from P will be: (Diagram in PDF) 1. 10 cm left of P in air 2. 10 cm right of P in water 3. 20 cm right of P in water 4. 20 cm left of P in air	(1) 10 cm left of P in air	Use the refraction formula for curved surfaces: (μ₂/v) - (μ₁/u) = (μ₂-μ₁)/R. Solve for v.
Q38. For fixed values of radii of curvature of a lens, the power of the lens will be: 1. P ∝ (μ – 1) 2. P ∝ μ² 3. P ∝ μ 4. P ∝ (μ – 2)	(1) P ∝ (μ – 1)	Lens power (P) is directly proportional to (μ-1) for fixed radii of curvature.
*Q39. The graph correctly representing the variation of image distance v* for a convex lens of focal length f versus object distance u is: (Graphs in PDF)**	(2) Graph B (hyperbolic curve asymptotic to both axes)	The lens equation (1/v - 1/u = 1/f) represents a hyperbola.
Q40. Using light from a monochromatic source to study diffraction in a single slit of width 0.1 mm, the linear width of the central maximum is measured to be 5mm on a screen held 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	(3) 5 × 10^-7 m	Use the formula for the width of the central maximum in single-slit diffraction: Δx = 2λL/a
Q41. Radiation of frequency 2ν₀ is incident on a metal with threshold frequency ν₀. The correct statement is: 1. No photoelectrons will be emitted 2. All photoelectrons emitted will have kinetic energy equal to hν₀ 3. Maximum kinetic energy of photoelectrons emitted can be hν₀ 4. Maximum kinetic energy of photoelectrons emitted will be 2hν₀	(3) Maximum kinetic energy of photoelectrons emitted can be hν₀	Use the photoelectric effect equation: KE_max = h(2ν₀ - ν₀) = hν₀
Q42. A point source causing photoelectric emission from a metallic plate is moved away from the plate. The variation of photoelectric current with distance from the source is correctly represented by the graph: (Graphs in PDF) 1. A straight line decreasing with distance 2. A curve decreasing non-linearly with distance 3. A horizontal line (constant current) 4. A curve increasing with distance	(3) A horizontal line (constant current)	Photoelectric current saturates at a certain intensity. Increasing distance reduces intensity but, as long as the intensity is above the threshold, the current remains constant at its saturation value.
*Q43. A proton accelerated through a potential difference V* has a de Broglie wavelength λ. On doubling the accelerating potential, the de Broglie wavelength of the proton:** 1. Remains unchanged 2. Becomes double 3. Becomes four times 4. Decreases	(4) Decreases	The de Broglie wavelength (λ = h/√(2mK)) is inversely proportional to the square root of the kinetic energy (K), which is directly proportional to the accelerating potential (V). Doubling V reduces λ by a factor of √2.
*Q44. The kinetic energy of an electron in the ground level of a hydrogen atom is K* units. The values of its potential energy and total energy respectively are:** 1. -2K; -K 2. +2K; -K 3. -K; +2K 4. +K; +2K	(1) -2K; -K	In the Bohr model, the total energy (E) is negative, the potential energy (U) is twice the total energy (-2K), and the kinetic energy (K) is half the magnitude of the total energy.
*Q45. Two nuclei have mass numbers A* and B respectively. The density ratio of the nuclei is:** 1. A: B 2. √A: √B 3. A²:B² 4. 1:1	(4) 1:1	Nuclear density is approximately constant and independent of the mass number.
Q46. The shortest wavelengths emitted in the hydrogen spectrum corresponding to different spectral series are as under: (Series listed in PDF) The wavelengths arranged correctly in decreasing order are: 1. (A), (B), (C), (D) 2. (A), (C), (B), (D) 3. (B), (A), (D), (C) 4. (A), (C), (D), (B)	(2) (A), (C), (B), (D)	The shortest wavelength in each series corresponds to the transition to the lowest energy level (n=1 for Lyman, n=2 for Balmer, etc.). Therefore, the order of decreasing wavelength is Lyman, Balmer, Brackett, Pfund.
Q47. Silicon can be doped using one of the following elements as dopant: (A) Arsenic (B) Indium (C) Phosphorus (D) Boron To get an n-type semiconductor, the dopants that can be used are: 1. (A) and (C) only 2. (B) and (C) only 3. (A), (B), (C), and (D) 4. (C) and (D) only	(1) (A) and (C) only	n-type doping requires elements with 5 valence electrons (group 15). Arsenic and Phosphorus fit this criteria.
Q48. Given below are V versus I graphs for different types of p-n junction diodes marked (A), (B), (C), and (D): (Graphs in PDF) The correct sequence of graphs corresponding to forward biased p-n junction, Zener diode, Photodiode, and Solar cell in order is: 1. (D), (C), (A), (B) 2. (A), (C), (B), (D) 3. (B), (A), (D), (C) 4. (C), (B), (D), (A)	(1) (D), (C), (A), (B)	Match the V-I characteristics to the operational behavior of each diode type: Solar cell (D), Zener diode (C), forward-biased p-n junction (A), and photodiode (B).
*Q49. A wire carrying current I, bent as shown in the figure, is placed in a uniform field B* that emerges normally out of the plane of the figure. The force on this wire is: (Diagram in PDF)** 1. 4BIR, directed vertically downward 2. 3BIR, directed vertically upward 3. BI(2R + πR), vertically downward 4. 2πBIR, from P to Q	(1) 4BIR, directed vertically downward	Calculate the force on each segment of the wire (straight and semicircular) using F = IL x B. The forces on the straight segments add, and the force on the semicircular segment is equivalent to the force on its chord (2R).
Q50. The refractive index of the material of an equilateral prism is √2. The angle of minimum deviation of that prism is: 1. 60° 2. 75° 3. 30° 4. 90°	(3) 30°	Use the formula relating refractive index (μ), prism angle (A), and minimum deviation angle (δ_m): μ = sin[(A+δ_m)/2] / sin(A/2)

CUET 2024 Physics Question Paper (SET D) with Solutions

Question	Answer	Solution
Q1: The kinetic energy of an electron in the ground level of a hydrogen atom is K units. The values of its potential energy and total energy respectively are: (1) -2K; -K (2) +2K; -K (3) -K; +2K (4) +K; +2K	(1) -2K; -K	Bohr model: Total energy E=-K; Potential energy U=2E=-2K; Kinetic energy K=-E.
Q2: Two nuclei have mass numbers A and B respectively. The density ratio of the nuclei is: (1) A:B (2) √A:√B (3) A²:B² (4) 1:1	(4) 1:1	Nuclear density is constant.
Q3: A point source causing photoelectric emission is moved away from a metallic plate. The variation of photoelectric current with distance is best represented by: (1) Straight line decreasing with distance (2) Curve decreasing non-linearly (3) Horizontal line (constant current) (4) Curve increasing with distance	(3) Horizontal line (constant current)	Saturation current is independent of distance above threshold intensity.
Q4: A proton's de Broglie wavelength is λ after acceleration through potential V. Doubling V changes the wavelength to: (1) Remains unchanged (2) Doubles (3) Quadruples (4) Decreases	(4) Decreases	λ is inversely proportional to √V.
Q5: The shortest wavelengths in the hydrogen spectrum for different series (Pfund, Balmer, Brackett, Lyman) are in decreasing order: (1) A, B, C, D (2) A, C, B, D (3) B, A, D, C (4) A, C, D, B	(2) A, C, B, D	Shortest wavelength corresponds to lowest energy level (Lyman, then Balmer, etc.).
Q6: To get an n-type semiconductor using Silicon, which dopants can be used? (1) Arsenic and Indium only (2) Indium and Phosphorus only (3) Arsenic and Phosphorus only (4) Phosphorus and Boron only	(3) Arsenic and Phosphorus only	n-type requires group 15 elements (one extra electron).
Q7: The correct sequence of graphs (forward biased p-n junction, Zener diode, Photodiode, Solar cell) is: (1) D, C, A, B (2) A, C, B, D (3) B, A, D, C (4) C, B, D, A	(1) D, C, A, B	Match V-I characteristics with diode operation modes.
Q8: The force on a bent wire carrying current I in a uniform magnetic field B is: (1) 4BIR, vertically downward (2) 3BIR, vertically upward (3) BI(2R + πR), vertically downward (4) 2πBIR, from P to Q	(1) 4BIR, vertically downward	Sum forces on straight and semicircular segments.
Q9: The angle of minimum deviation for an equilateral prism (refractive index √2) is: (1) 60° (2) 75° (3) 30° (4) 90°	(3) 30°	Use the prism formula relating refractive index and minimum deviation.
Q10: The transfer of an integral number of is evidence of: (1) Photons (2) Nuclei (3) Electrons (4) Neutrons	(3) Electrons	Charge quantization is due to electrons carrying charge 'e'.
Q11: A 4 mm thick slab increases the distance between capacitor plates by 3.2mm to restore original capacitance. Dielectric constant is: (1) 2 (2) 5 (3) 3 (4) 7	(2) 5	Use the formula for effective separation with a dielectric.
Q12: Charge on a copper ball (density 8 g/cc, diameter 1 cm) suspended in oil (density 0.8 g/cc) in an electric field of 600π V/m is: (1) 2 × 10^-6 C (2) 2 × 10^-5 C (3) 1 × 10^-5 C (4) 1 × 10^-6 C	(2) 2 × 10^-5 C	Balance electric force with gravitational and buoyant forces.
Q13: Increasing the temperature of a metal wire under constant potential difference affects drift velocity how? (1) Drift velocity increases, thermal velocity decreases (2) Drift velocity decreases, thermal velocity decreases (3) Drift velocity increases, thermal velocity increases (4) Drift velocity decreases, thermal velocity increases	(4) Drift velocity decreases, thermal velocity increases	Relaxation time decreases, reducing drift velocity; thermal velocity increases with temperature.
Q14: Total resistance between A and B in the given mixed resistor combination (diagram shown in PDF) is: (1) 9 Ω (2) 18 Ω (3) 4 Ω (4) 14 Ω	(2) 18 Ω	Combine parallel resistors, then series resistors.
Q15: Two identical cells (emf 1.1V, internal resistance 0.5Ω) are connected in series with a 0.5Ω wire, maintaining the same current. The second cell's internal resistance is: (1) 1 Ω (2) 2.5 Ω (3) 1.5 Ω (4) 2 Ω	(1) 1 Ω	Use Ohm's law with total emf and resistance.
Q16: In a Wheatstone bridge (P=3Ω, Q=3Ω, R=3Ω, S=4Ω), what resistance shunted with S balances the bridge? (1) 14 Ω (2) 12 Ω (3) 15 Ω (4) 7 Ω	(2) 12 Ω	Use the Wheatstone bridge balance condition.
Q17: A bar magnet with magnetic moment M is bent into a semicircle. The new magnetic moment is: (1) M/π (2) 2M/π (3) M (4) 2M	(2) 2M/π	Magnetic moment depends on the effective length.
Q18: Ferromagnetic materials used in transformers must 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	(2) High permeability, low hysteresis loss	High permeability for efficient flux; low hysteresis for minimal energy loss.
Q19: Electric field intensity at any point on a ring (radius r) in a varying magnetic field (rate x) perpendicular to the ring's plane is: (1) rx (2) rx/2 (3) 2rx (4) 4rx	(2) rx/2	Use Faraday's law to relate induced emf to electric field.
Q20: Crest voltage induced in a transformer secondary (mutual inductance 0.5H, primary current 1A crest, 50 Hz): (1) 75V (2) 150V (3) 100V (4) 200V	(3) 100V	Use the formula for induced emf in a transformer.
Q21: Total charge through a coil (100 turns, radius 0.01m) at the center of a solenoid (2x10⁴ turns/m, diameter 0.1m) when current drops from 4A to 0A in 0.05s (coil resistance 10π²Ω): (1) 16 µC (2) 32 µC (3) 16π µC (4) 32π µC	(2) 32 µC	Use Faraday's law and Q = emf*t/R.
Q22: Lower half of a convex lens is made opaque. The image shows: (1) No change in image (2) Only half the object (3) Reduced intensity (4) Only half the object and reduced intensity	(3) Reduced intensity	Blocking part of a lens reduces intensity but not image formation.
Q23: Two slits (0.1mm apart), screen (2m away), wavelength 500nm. Fringe separation is: (1) 1 cm (2) 0.15 cm (3) 1.5 cm (4) 0.1 cm	(1) 1 cm	Use the fringe separation formula.
Q24: Astronomical telescope (objective lens 10m, eyepiece 10cm). Tube length and magnification are: (1) 20cm, 1 (2) 100cm, 1 (3) 1010cm, 1 (4) 1010cm, 100	(4) 1010cm, 100	Tube length = f_o + f_e; Magnification = f_o/f_e
Q25: According to Bohr's model, which are true? (A) Radius α n, (B) Speed α 1/n, (C) Total energy α 1/n², (D) Radius α n² (1) A, B, C only (2) A, B, D only (3) A, B, C, D (4) B, C, D only	(4) B, C, D only	Correct relationships between radius, speed, energy, and n.
Q26: Output frequency of a full-wave rectifier with a 50 Hz input is: (1) 50 Hz (2) 100 Hz (3) 25 Hz (4) 0 Hz	(2) 100 Hz	Output frequency doubles the input frequency.
Q27: Force (F) and torque (τ) on a dipole parallel to a non-uniform field are: (1) F=0, τ=0 (2) F≠0, τ=0 (3) F=0, τ≠0 (4) F≠0, τ≠0	(2) F≠0, τ=0	Non-uniform field causes net force, but torque is zero if aligned.
Q28: Forces (F1, F2, F3) on a charge (q) at points P1, P2, P3 between two parallel oppositely charged sheets are: (1) F1=0, F2=0, F3=0 (2) F1=0, F2≠0, F3=0 (3) F1≠0, F2≠0, F3≠0 (4) F1=0, F2=0, F3≠0	(2) F1=0, F2≠0, F3=0	Field is zero outside, uniform inside the plates.
Q29: Two spheres (radii R1, R2) are charged and then separated. Ratio of final charges (Q1/Q2) is: (1) R2/R1 (2) R1/R2 (3) R1²/R2² (4) R1/R2	(1) R2/R1	Charge distributes proportionally to capacitance (radius).
Q30: Two charges exert force F at distance d. Doubling both charges, what distance maintains F? (1) 4d (2) 2d (3) d (4) d/2	(2) 2d	Force is inversely proportional to distance squared.
Q31: Two series capacitors (2µF, 3µF) connected to V volts. The relation between V1/V2 and U1/U2 is: (1) V1/V2 = U1/U2 = 3/2 (2) V1/V2 = U1/U2 = 2/3 (3) V1/V2 = 2/3, U1/U2 = 3/2 (4) V1/V2 = 3/2, U1/U2 = 2/3	(1) V1/V2 = U1/U2 = 3/2	Potential and energy are inversely proportional to capacitance in series.
Q32: Voltmeter (200Ω) reading across a 200Ω resistor in series with a 100Ω resistor (20V battery): (1) 4V (2) 7V (3) 10V (4) 16V	(3) 10V	Voltage is proportional to resistance in series.
Q33: Current through a 4/3Ω external resistance connected to parallel 2V and 1V cells (internal resistances 1Ω and 2Ω respectively) is: (1) 1A (2) 2/3A (3) 1/2A (4) 5/6A	(4) 5/6A	Find equivalent emf and resistance of the parallel cells.
Q34: A wire (resistance R, resistivity ρ, power rating P) is stretched, halving its radius. R', ρ', P' are: (1) ρ'=2ρ, R'=2R, P'=2P (2) ρ'=ρ/2, R'=R/2, P'=P/2 (3) ρ'=ρ, R'=16R, P'=P/16 (4) ρ'=ρ, R'=R/16, P'=16P	(3) ρ'=ρ, R'=16R, P'=P/16	Resistivity is unchanged; resistance changes with area and length; power α 1/R.
Q35: Increasing order of magnetic susceptibility for Paramagnetic (A), Diamagnetic (B), Ferromagnetic (C) materials is: (1) A, B, C (2) C, A, B (3) B, A, C (4) B, C, A	(3) B, A, C	Diamagnetic < Paramagnetic < Ferromagnetic susceptibility.
Q36: Force (F) on length L of a conductor due to another parallel conductor (currents I1, I2, distance d): (1) F α L, independent of I1I2 (2) F α I1I2, independent of L (3) F α I1I2L (4) F α L/(I1I2)	(3) F α I1I2L	Use the formula for force between parallel conductors.
Q37: Magnetic field at the center of a loop (current 3I, semicircular parts with resistances 2R and R) is: (1) (μ₀I)/(4r), out of plane (2) (μ₀I)/(4r), into plane (3) (3μ₀I)/(4r), out of plane (4) (3μ₀I)/(4r), into plane	(1) (μ₀I)/(4r), out of plane	Current splits inversely proportional to resistance; sum magnetic field contributions.
Q38: Torque on a square loop (side 1cm, current 10A) in a magnetic field (0.2T, parallel to the plane): (1) Zero (2) 2 × 10^-4 Nm (3) 2 × 10^-2 Nm (4) 2 Nm	(2) 2 × 10^-4 Nm	Use the torque formula for a current loop.
Q39: In an AC circuit, current leads voltage by π/2. The circuit is: (1) Purely resistive (2) Resistance = Reactance (3) Purely inductive (4) Purely capacitive	(4) Purely capacitive	Phase difference of π/2 indicates a purely capacitive circuit.
Q40: Mutual inductance of adjacent coils (current change 0A to 10A in 0.25s, flux change 15Wb): (1) 120 H (2) 12 H (3) 1.5 H (4) 0.75 H	(3) 1.5 H	Use the definition of mutual inductance.
Q41: Direction of induced current (figures a and b in PDF): (1) Clockwise in both (2) Anticlockwise in both (3) Clockwise in (a), Anticlockwise in (b) (4) Anticlockwise in (a), Clockwise in (b)	(2) Anticlockwise in both	Apply Lenz's law to determine the direction of induced current.
Q43: Ratio of energy densities of electric and magnetic fields in an electromagnetic wave is: (1) 1:1 (2) 1:c (3) c:1 (4) 1:c²	(1) 1:1	Energy densities of electric and magnetic fields are equal.
Q44: The correct arrangement of the electromagnetic spectrum in decreasing order of wavelength is: (1) Radio, X-rays, Infrared, Microwaves, Visible (2) Infrared, Microwaves, Radio, X-rays, Visible (3) Radio, Microwaves, Infrared, Visible, X-rays (4) X-rays, Visible, Infrared, Microwaves, Radio	(3) Radio, Microwaves, Infrared, Visible, X-rays	Wavelength decreases from radio to gamma rays.
Q45: Match the electromagnetic waves (Column I) with their production methods (Column II): (A) Microwaves, (B) Infrared, (C) X-rays, (D) Radio waves; (I) LC oscillator, (II) Magnetron, (III) Atomic/molecular vibrations, (IV) Bombarding a heavy metal target with electrons. (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	(2) A-II, B-III, C-IV, D-I	Match each wave with its typical generation method.
Q46: A curved surface (radius 10cm, refractive index 4/3) separates air and a transparent material. An object (20cm from P) forms an image at: (1) 16cm left of P in air (2) 16cm right of P in the material (3) 20cm right of P in the material (4) 20cm left of P in air	(1) 16cm left of P in air	Use the refraction formula for curved surfaces.
Q47: For fixed radii of curvature of a lens, its power (P) is proportional to: (1) (μ - 1) (2) μ² (3) μ (4) (μ - 2)	(1) (μ - 1)	Lens power formula shows direct proportionality to (μ - 1).
Q48: The correct graph (v vs u) for a convex lens with focal length f is: (Graphs shown in PDF)	(2) Hyperbolic curve	The lens equation gives a hyperbolic relationship.
Q49: Linear width of central maximum (5mm) for a single slit (0.1mm) diffraction on a screen (50cm away). Wavelength is: (1) 2.5 × 10^-7 m (2) 4 × 10^-7 m (3) 5 × 10^-7 m (4) 7.5 × 10^-7 m	(3) 5 × 10^-7 m	Use the single-slit diffraction formula.
Q50: Frequency 2ν₀ is incident on a metal (threshold frequency ν₀). Which statement is correct? (1) No photoelectrons emitted (2) All have kinetic energy hν₀ (3) Maximum kinetic energy can be hν₀ (4) Maximum kinetic energy is 2hν₀	(3) Maximum kinetic energy can be hν₀	Use the photoelectric equation (K_max = h(f - f₀)).