JEE Main 2024 Feb 1 Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Feb 1 Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 Feb 1 Shift 2 includes 30 questions divided into 2 sections, Section 1 with 20 MCQs and Section 2 with 10 numerical questions. Candidates must attempt any 5 numerical questions out of 10. The memory-based JEE Main 2024 question paper pdf for the Feb 1 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Feb 1 Shift 2 Physics Question Paper PDF Download
| JEE Main 2024 Physics Question Paper | JEE Main 2024 Physics Answer Key | JEE Main 2024 Physics Solution |
|---|---|---|
| Download PDF | Download PDF | Download PDF |
JEE Main 1 Feb Shift 2 2024 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 31: In an ammeter, 5% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be: (1) G/200 (2) G/199 (3) 199G (4) 200G |
BONUS | Use the relation for the effective resistance of the ammeter. With 5% of the current through the galvanometer, calculate the shunt resistance to find the total resistance as G/199. |
| 32: To measure the temperature coefficient of resistivity α of a semiconductor, an electrical arrangement is prepared. Arm BC is made of the semiconductor, with an initial resistance of 3 mΩ. If the galvanometer shows no deflection after 10 seconds as BC is cooled at 2°C/s, then α is: (1) −2 × 10⁻² °C⁻¹ (2) −1.5 × 10⁻² °C⁻¹ (3) −1 × 10⁻² °C⁻¹ (4) −2.5 × 10⁻² °C⁻¹ |
(2) −1.5 × 10⁻² °C⁻¹ | Using the Wheatstone bridge principle, calculate the temperature coefficient by equating the change in resistance due to cooling. |
| 33: From the statements given below: (A) The angular momentum of an electron in the nth orbit is an integral multiple of h. (B) Nuclear forces do not obey inverse square law. (C) Nuclear forces are spin-dependent. (D) Nuclear forces are central and charge independent. (E) Stability of nucleus is inversely proportional to the value of packing fraction. Choose the correct answer: (1) (A), (B), (C), (D) only (2) (A), (C), (D), (E) only (3) (A), (B), (C), (E) only (4) (B), (C), (D), (E) only |
(3) (A), (B), (C), (E) only | Analyze each statement: (A), (B), (C), and (E) are correct, while (D) is incorrect because nuclear forces are not completely central. |
| 34: A diatomic gas (γ = 1.4) does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is: (1) 850 J (2) 800 J (3) 600 J (4) 700 J |
(4) 700 J | Using Q = ΔU + W and γ = 1.4, calculate ΔU and substitute to find Q = 700 J. |
| 35: A disc of radius R and mass M is rolling horizontally without slipping with speed v. It then moves up an inclined smooth surface. The maximum height h the disc can go up the incline is: (1) v²/g (2) 3v²/4g (3) v²/2g (4) 2v²/3g |
(3) v²/2g | Apply conservation of energy: Total initial kinetic energy = mgh. Simplify to find h = v²/2g. |
| 36: Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of the photodiode is found to be X/8 eV. The value of X is: (1) 15 (2) 11 (3) 13 (4) 21 |
(1) 15 | Calculate the energy of the photon using E = hc/λ, where λ = 660 nm, h = 6.63×10⁻³⁴ Js, and c = 3×10⁸ m/s. The energy is 1.88 eV, and X/8 = 1.88. Solving gives X = 15. |
| 37: A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become: (1) 100 times (2) 10 times (3) 1/100 (4) 1/10 |
(4) 1/10 | Surface energy is proportional to surface area. Coalescing 1000 droplets reduces the surface area ratio to 1/10. |
| 38: If the frequency of an electromagnetic wave is 60 MHz and it travels in air along the z-direction, then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other, and the wavelength of the wave (in m) is: (1) 2.5 (2) 10 (3) 5 (4) 2 |
(3) 5 | The wavelength is calculated using the relation λ = c/f, where c = 3 × 10⁸ m/s and f = 60 × 10⁶ Hz. Substituting these values gives λ = 5 m. |
| 39: A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s, then the magnitude of force exerted by the ball on the hand of the player will be (in SI unit): (1) 24 (2) 12 (3) 25 (4) 30 |
(4) 30 | The force is calculated using the impulse-momentum theorem: F = Δp/Δt. Here, Δp = mv = 0.12 × 25 = 3 kg·m/s and Δt = 0.1 s. Substituting these values gives F = 30 N. |
| 40: Monochromatic light of frequency 6 × 10¹⁴ Hz is produced by a laser. The power emitted is 2 × 10⁻³ W. How many photons per second, on average, are emitted by the source? (1) 9 × 10¹⁸ (2) 6 × 10¹⁵ (3) 5 × 10¹⁵ (4) 7 × 10¹⁶ |
(3) 5 × 10¹⁵ | The energy of one photon is E = hf, where h = 6.63 × 10⁻³⁴ J·s and f = 6 × 10¹⁴ Hz. This gives E = 3.978 × 10⁻¹⁹ J. The number of photons per second is n = P/E = (2 × 10⁻³) / (3.978 × 10⁻¹⁹) ≈ 5 × 10¹⁵ photons per second. |
| 41: A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm. The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit, will be: (1) 30° (2) 15° (3) 60° (4) 45° |
(3) 60° | Using the formula θ = 2sin⁻¹(λ/a), where λ = 0.02 m and a = 0.04 m, the angular spread θ = 2 × sin⁻¹(0.5) = 2 × 30° = 60°. |
| 42: C1 and C2 are two hollow concentric cubes enclosing charges 2Q and 3Q, respectively. The ratio of electric flux passing through C1 and C2 is: (1) 2 : 5 (2) 5 : 2 (3) 2 : 3 (4) 3 : 2 |
(1) 2 : 5 | Using Gauss's Law, flux through C1 is proportional to the enclosed charge 2Q, and flux through C2 is proportional to the total enclosed charge 5Q. The ratio is 2:5. |
| 43: If the root mean square velocity of a hydrogen molecule at a given temperature and pressure is 2 km/s, the root mean square velocity of oxygen at the same condition in km/s is: (1) 2.0 (2) 0.5 (3) 1.5 (4) 1.0 |
(2) 0.5 | The ratio of velocities is inversely proportional to the square root of molar masses. Given MH₂ = 2 and MO₂ = 32, vrms,O₂ = vrms,H₂/√16 = 2/4 = 0.5 km/s. |
| 44: Train A is moving along two parallel rail tracks towards north with speed 72 km/h and train B is moving towards south with speed 108 km/h. The velocity of train B with respect to A and velocity of ground with respect to B are (in m/s): (1) -30 and 50 (2) -50 and -30 (3) -50 and 30 (4) 50 and -30 |
(3) -50 and 30 | Convert speeds to m/s: vA = 20 m/s, vB = -30 m/s. Velocity of B relative to A is vB - vA = -50 m/s, and ground relative to B is -vB = 30 m/s. |
| 45: A galvanometer G of 2 Ω resistance is connected in a circuit. The ratio of charge stored in C1 and C2 is: (1) 2/3 (2) 3/2 (3) 1 (4) 1/2 |
(4) 1:2 | Calculate the equivalent capacitance of C1 and G in series, and use the charge distribution formula. Voltage across C2 is higher, giving a charge ratio of 1:2. |
| 46: In a metre-bridge, when a resistance in the left gap is 2Ω and an unknown resistance in the right gap, the balance length is found to be 40 cm. On shunting the unknown resistance with 2Ω, the balance length changes by: (1) 22.5 cm (2) 20 cm (3) 62.5 cm (4) 65 cm |
(1) 22.5 cm | Using the metre-bridge formula and shunting resistance calculations, the change in balance length is determined as 22.5 cm. |
| 47: Match List-I with List-II. Choose the correct answer from the options given below: (1) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(III) (4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) |
(3) (A)-(III), (B)-(I), (C)-(IV), (D)-(III) | Use the properties of significant figures and scientific notations to determine the matches. The correct mapping aligns with option (3). |
| 48: A transformer has an efficiency of 80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is: (1) 1.59 A (2) 13.33 A (3) 1.33 A (4) 15.1 A |
(2) 13.33 A | The output power of the transformer is given by P = η × Pin, where η = 0.8 and Pin = 4 kW. Output power = 0.8 × 4000 = 3200 W. Using P = VI, current I = P/V = 3200/240 = 13.33 A. |
| 49: A light planet is revolving around a massive star in a circular orbit of radius R with a period T. If the force of attraction between the planet and the star is proportional to R⁻³/², then T² is proportional to: (1) R⁵/² (2) R⁷/² (3) R³/² (4) R³ |
(1) R⁵/² | Using the centripetal force relation and Kepler's law, F ∝ R⁻³/² implies T² ∝ R⁵/². This is derived by equating centripetal force with the gravitational force and solving for T. |
| 50: A body of mass 4 kg experiences two forces F₁ = 5î + 8ĵ + 7k̂ and F₂ = 3î − 4ĵ − 3k̂. The acceleration acting on the body is: (1) −2î − ĵ − k̂ (2) 4î + 2ĵ + 2k̂ (3) 2î + ĵ + k̂ (4) 4î + 3ĵ + 3k̂ |
(3) 2î + ĵ + k̂ | The net force is F = F₁ + F₂ = (5 + 3)î + (8 − 4)ĵ + (7 − 3)k̂ = 8î + 4ĵ + 4k̂. Using F = ma, acceleration a = F/m = (8/4)î + (4/4)ĵ + (4/4)k̂ = 2î + ĵ + k̂. |
| 51: A mass m is suspended from a spring of negligible mass, and the system oscillates with a frequency f₁. The frequency of oscillations if a mass 9m is suspended from the same spring is f₂. The value of f₁/f₂ is: | 3 | The frequency of oscillation is inversely proportional to the square root of the mass. Using the relation f₁/f₂ = √(m/9m), f₁/f₂ = 3. |
| 52: A particle initially at rest starts moving from the reference point x = 0 along the x-axis, with velocity v that varies as v = 4√x m/s. The acceleration of the particle is m/s²: | 8 | Acceleration is given by a = dv/dt. Using v = 4√x, differentiate with respect to x and multiply by dx/dt (v) to find a = 8 m/s². |
| 53: A moving coil galvanometer has 100 turns, and each turn has an area of 2.0 cm². The magnetic field produced by the magnet is 0.01T, and the deflection in the coil is 0.05 rad when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x × 10⁻⁵ N-m/rad. The value of x is: | 4 | Torque is given by τ = nBAI = kθ. Substituting n = 100, B = 0.01T, A = 2 × 10⁻⁴ m², I = 0.01A, and θ = 0.05 rad, solve for k = 4 × 10⁻⁵ N-m/rad. |
| 54: One end of a metal wire is fixed to a ceiling, and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load, and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of the upper wire to that of the lower wire will be: | 3 | The strain is proportional to the force. The force on the upper wire is due to both loads (3 kg), while the lower wire experiences a force from 1 kg. The strain ratio is 3:1. |
| 55: A particular hydrogen-like ion emits radiation of frequency 3 × 10¹⁵ Hz when it makes a transition from n = 2 to n = 1. The frequency of radiation emitted in the transition from n = 3 to n = 1 is x × 9 × 10¹⁵ Hz. The value of x is: | 32 | The frequency is proportional to the energy difference. Using the energy levels for n = 2 to 1 and n = 3 to 1 transitions, calculate x = 32. |
| 56: In the electrical circuit drawn below, the amount of charge stored in the capacitor is µC: | 60 | Using the formula Q = CV, calculate the charge stored in the capacitor as 60 µC based on the given circuit parameters. |
| 57: A coil of 200 turns and area 0.20 m² is rotated at half a revolution per second in a uniform magnetic field of 0.01T perpendicular to the axis of rotation of the coil. The maximum voltage generated in the coil is 2π/β volts. The value of β is: | 5 | The maximum induced EMF is given by E = NABω. Substituting the given values, solve for β to find β = 5. |
| 58: In Young’s double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are 1.0 mm apart, and the screen is placed at 1.0 m away from the slits. The distance from the center of the screen where intensity becomes half of the maximum intensity for the first time is ×10⁻⁶ m: | 125 | Using the condition for intensity I = I₀ cos²(πdx/λD), calculate the distance where intensity becomes half, resulting in 125 × 10⁻⁶ m. |
| 59: A uniform rod AB of mass 2 kg and length 30 cm is at rest on a smooth horizontal surface. An impulse of 0.2 Ns is applied to end B. The time taken by the rod to turn through a right angle will be π/x seconds, where x = : | 4 | Using the relation for angular acceleration and rotational dynamics, calculate the time taken for the rod to turn through a right angle, resulting in x = 4. |
| 60: Suppose a uniformly charged wall provides a uniform electric field of 2 × 10⁴ N/C normally. A charged particle of mass 2 g is suspended through a silk thread of length 20 cm and remains at a distance of 10 cm from the wall. The charge on the particle will be √1/x µC, where x = : | 3 | Equate the forces acting on the particle and solve for the charge. The charge is found to be √1/3 µC, giving x = 3. |
Also Check:
| JEE Main 2024 Paper Analysis | JEE Main 2024 Answer Key |
| JEE Main 2024 Cutoff | JEE Main 2024 Marks vs Rank |
JEE Main 2024 Feb 1 Shift 2 Physics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | To be updated |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Vedantu | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 Feb 1 Shift 2 Physics Paper Analysis
JEE Main 2024 Feb 1 Shift 2 Physics paper analysis is updated here with details on the difficulty level of the exam, topics with the highest weightage in the exam, section-wise difficulty level, etc.
JEE Main 2024 Physics Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Sectional Time Duration | None |
| Total Marks | 100 marks |
| Total Number of Questions Asked | 30 Questions |
| Total Number of Questions to be Answered | 25 questions |
| Type of Questions | MCQs and Numerical Answer Type Questions |
| Section-wise Number of Questions | 20 MCQs and 10 numerical type, |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
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JEE Main 2024 Question Paper Session 1 (January)
Those appearing for JEE Main 2024 can use the links below to practice and keep track of their exam preparation level by attempting the shift-wise JEE Main 2024 question paper provided below.
| Exam Date and Shift | Question Paper PDF |
|---|---|
| JEE Main 2024 Question Paper Jan 24 Shift 2 | Check Here |
| JEE Main 2024 Question Paper Jan 27 Shift 1 | Check Here |
| JEE Main 2024 Question Paper Jan 27 Shift 2 | Check Here |
| JEE Main 2024 Question Paper Jan 29 Shift 1 | Check Here |
| JEE Main 2024 Question Paper Jan 29 Shift 2 | Check Here |
| JEE Main 2024 Question Paper Jan 30 Shift 1 | Check Here |
| JEE Main 2024 Question Paper Jan 30 Shift 2 | Check Here |
| JEE Main 2024 Question Paper Jan 31 Shift 1 | Check Here |
| JEE Main 2024 Question Paper Jan 31 Shift 2 | Check Here |
| JEE Main 2024 Question Paper Feb 1 Shift 1 | Check Here |
| JEE Main 2024 Question Paper Feb 1 Shift 2 | Check Here |
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