JEE Main 2024 5 April Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 5 April Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 5 April 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 5 April Shift 2 exam is available for download using the link below.
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JEE Main 2024 5 April Shift 2 Physics Question Paper PDF Download
| JEE Main 2024 Physics Question Paper | JEE Main 2024 Physics Answer Key | JEE Main 2024 Physics Solutions |
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JEE Main 2024 5 April Shift 2 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 31: Given below are two statements: Statement I: When white light passes through a prism, red light bends less than yellow and violet. Statement II: The refractive indices are different for different wavelengths in dispersive media. In the light of the above statements, choose the correct answer: (1) Both Statement I and Statement II are true. (2) Statement I is true but Statement II is false. (3) Both Statement I and Statement II are false. (4) Statement I is false but Statement II is true. |
(1) Both Statement I and Statement II are true | White light dispersion through a prism occurs because refractive indices vary with wavelength. Red light, having the longest wavelength, bends the least, while violet bends the most. Both statements are correct and consistent with this phenomenon. |
| 32: Which of the following statements is NOT true about the stopping potential (V0)? (1) It depends on the nature of the emitter material. (2) It depends upon the frequency of the incident light. (3) It increases with the increase in intensity of the incident light. (4) It is 1/e times the maximum kinetic energy of the emitted electrons. |
(3) It increases with the increase in intensity of the incident light | The stopping potential depends on the frequency of light and the material’s work function but not on light intensity. Light intensity only affects the number of photoelectrons emitted, not their energy or the stopping potential. |
| 33: The angular momentum of an electron in a hydrogen atom is proportional to (where r is the radius of the orbit): (1) √r (2) 1/r (3) r (4) 1/√r |
(1) √r | According to Bohr’s model, angular momentum (L) is proportional to the quantum number n, and the orbit radius (r) is proportional to n². Hence, L is proportional to √r. |
| 34: A galvanometer of resistance 100Ω is connected in series with a 400Ω resistor to measure up to 10V. The value of resistance required to convert the galvanometer into an ammeter to read up to 10A is x × 10-2Ω. The value of x is: (1) 2 (2) 800 (3) 20 (4) 200 |
(3) 20 | To convert the galvanometer into an ammeter, a shunt resistance is calculated as S = igRg / (I - ig). Substituting the given values, S is found to be 0.2Ω, or 20 × 10-2Ω, giving x = 20. |
| 35: The vehicles carrying inflammable fluids usually have metallic chains touching the ground: (1) To conduct excess charge due to air friction to the ground and prevent sparking. (2) To alert other vehicles. (3) To protect tires from catching dirt from the ground. (4) It is a custom. |
(1) To conduct excess charge due to air friction to the ground and prevent sparking | Metallic chains are used to discharge static electricity generated due to friction with air while moving, preventing sparking that could ignite inflammable fluids. |
| 36: If n is the number density and d is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e., mean free path) is represented by: (1) 1/√(2πd2) (2) √(2πd2) (3) 1/√(2πd2n) (4) 1/√(2nπ2d2) |
(3) 1/√(2πd2n) | The mean free path (λ) of a gas molecule is derived from kinetic theory and is given by λ = 1/√(2πd2n), where n is the number density and d is the molecular diameter. |
| 37: A particle moves in the x-y plane under the influence of a force F such that its linear momentum is P⃗(t) = î cos(kt) − ĵ sin(kt). If k is constant, the angle between F⃗ and P⃗ will be: (1) π/2 (2) π/6 (3) π/4 (4) π/3 |
(1) π/2 | Differentiating P⃗(t) to find the force F⃗ shows that F⃗ is perpendicular to P⃗ because their dot product is zero. Hence, the angle between F⃗ and P⃗ is π/2. |
| 38: The electrostatic force (F1) and magnetic force (F2) acting on a charge q moving with velocity v can be written as: (1) F1 = qv · E, F2 = q(B · v) (2) F1 = qB, F2 = q(B × v) (3) F1 = qE, F2 = q(v × B) (4) F1 = qE, F2 = q(B × v) |
(3) F1 = qE, F2 = q(v × B) | The electrostatic force is F1 = qE, and the magnetic force is F2 = q(v × B), as per the Lorentz force law. |
| 39: A man carrying a monkey on his shoulder cycles smoothly on a circular track of radius 9m and completes 120 revolutions in 3 minutes. The magnitude of the centripetal acceleration of the monkey is (in m/s2): (1) Zero (2) 16π2 m/s2 (3) 4π2 m/s2 (4) 57600π2 m/s2 |
(2) 16π2 m/s2 | The angular velocity is ω = 4π/3 rad/s. Using ac = ω2R, the centripetal acceleration is ac = 16π2 m/s2. |
| 40: A series LCR circuit is subjected to an AC signal of 200V, 50Hz. If the voltage across the inductor (L = 10mH) is 31.4V, then the current in this circuit is: (1) 68 A (2) 63 A (3) 10 A (4) 10 mA |
(3) 10 A | The inductive reactance is XL = 2πfL = 3.14Ω. Using V = IXL, the current is I = 31.4/3.14 = 10 A. |
| 41: What is the dimensional formula of ab-1 in the equation: (P + a/V2)(V − b) = RT, where letters have their usual meaning? (1) [M0L3T-2] (2) [ML2T-2] (3) [M1L5T-2] (4) [M4L7T4] |
(2) [ML2T-2] | Using the dimensional formula of pressure ([ML-1T-2]) and volume ([L3]), the formula for ab-1 simplifies to [ML2T-2]. |
| 42: The output (Y) of the logic circuit given below is 0 only when: (1) A = 1, B = 0 (2) A = 0, B = 0 (3) A = 1, B = 1 (4) A = 0, B = 1 |
(2) A = 0, B = 0 | Analyzing the logic gates, the output Y is 0 only when both inputs A and B are 0, as the OR gate and AND gate conditions are not satisfied simultaneously. |
| 43: A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to: (1) t2 (2) t2/3 (3) t3/2 (4) t |
(3) t3/2 | For a body moving under constant power, velocity is proportional to √t, and displacement (integral of velocity) is proportional to t3/2. |
| 44: Match List-I with List-II: List-I (EM-Wave) List-II (Wavelength Range) (A) Infra-red (I) < 10-3 nm (B) Ultraviolet (II) 400 nm to 1 nm (C) X-rays (III) 1 mm to 700 nm (D) Gamma rays (IV) 1 nm to 10-3 nm Choose the correct option: (1) (A)-(III), (B)-(I), (C)-(IV), (D)-(III) (2) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (3) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (4) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) |
(2) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) | Infra-red corresponds to 1 mm to 700 nm, ultraviolet corresponds to 400 nm to 1 nm, X-rays correspond to 1 nm to 10-3 nm, and gamma rays correspond to wavelengths less than 10-3 nm. |
| 45: During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of CP/CV for the gas is: (1) 5/3 (2) 9/7 (3) 7/5 (4) 3/2 |
(3) 7/5 | For an adiabatic process, P ∝ T3. Using the relation P ∝ V-γ, γ = 7/5 is obtained, where γ = CP/CV. |
| 46: Choose the answer from the options given below: (1) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) |
(3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) | The matching between List-I and List-II is as follows: (A) A force that restores an elastic body of unit area to its original state corresponds to Stress (III). (B) Two equal and opposite forces parallel to opposite faces correspond to Shear modulus (IV). (C) Forces perpendicular everywhere to the surface per unit area correspond to Bulk modulus (I). (D) Two equal and opposite forces perpendicular to opposite faces correspond to Young’s modulus (II). |
| 47: A vernier calipers has 20 divisions on the vernier scale, which coincides with the 19th division on the main scale. The least count of the instrument is 0.1mm. One main scale division is equal to: (1) 1 (2) 0.5 (3) 2 (4) 5 |
(3) 2 | From the given data, 20 vernier scale divisions coincide with 19 main scale divisions. The least count is given as 1 main scale division minus 1 vernier scale division. Substituting the values, one main scale division is calculated to be 2mm. |
| 48: A heavy box of mass 50kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the surface is 0.3, then the force of kinetic friction is: (1) 14.7N (2) 147N (3) 1.47N (4) 1470N |
(2) 147N | The normal force acting on the box is equal to its weight, calculated as 50 × 9.8 = 490N. The force of kinetic friction is given by µk × N = 0.3 × 490 = 147N. |
| 49: A satellite revolving around a planet in a stationary orbit has a time period of 6 hours. The mass of the planet is one-fourth the mass of Earth. The radius of the orbit of the planet is: (1) 1.4×10⁴ km (2) 8.4×10⁴ km (3) 1.68×10⁵ km (4) 1.05×10⁴ km |
(4) 1.05×10⁴ km | Using Kepler’s third law and the given ratio of the masses and time periods, the radius of the orbit is calculated as 1.05×10⁴ km, using the proportional relationship r³ ∝ T²/M. |
| 50: The ratio of heat dissipated per second through the resistances 5Ω and 10Ω in the circuit given below is: (1) 1 : 2 (2) 2 : 1 (3) 4 : 1 (4) 1 : 1 |
(2) 2 : 1 | The 5Ω and 10Ω resistors are connected in parallel. The current through each resistor is inversely proportional to its resistance. The power dissipated in a resistor is proportional to the square of the current and the resistance. This gives a ratio of heat dissipation of 2:1. |
| 51: A solenoid of length 0.5m has a radius of 1 cm and is made up of m number of turns. It carries a current of 5A. If the magnitude of the magnetic field inside the solenoid is 6.28×10⁻³T, then the value of m is: | 500 | The magnetic field inside a solenoid is given by B = µ₀ni, where n is the number of turns per unit length. Rearranging to find m, we use the given values of B, µ₀, and i to calculate m as 500. |
| 52: The shortest wavelength of the spectral lines in the Lyman series of the hydrogen spectrum is 915 Å. The longest wavelength of spectral lines in the Balmer series will be: | 6588 Å | The longest wavelength in the Balmer series corresponds to the transition from n=3 to n=2. Using the formula for energy levels and the given Lyman series data, the wavelength is calculated as 6588 Å. |
| 53: In a single-slit experiment, a parallel beam of green light of wavelength 550 nm passes through a slit of width 0.20 mm. The transmitted light is collected on a screen 100 cm away. The distance of the first-order minima from the central maximum will be x×10⁻⁵ m. The value of x is: | 275 | The distance of the first-order minima is given by y = λD/d. Substituting the given values, the distance is calculated as 275×10⁻⁵ m, where x=275. |
| 54: A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with a fundamental frequency of 600 Hz under the same tension is: | 60 cm | The length of the sonometer wire is inversely proportional to the frequency. Using the ratio L₁/L₂ = f₂/f₁, the resonating length for 600 Hz is calculated as 60 cm. |
| 55: A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is x/5. The value of x is: | 2 | The total kinetic energy is the sum of rotational and translational kinetic energy. For a hollow sphere, the ratio of rotational to total kinetic energy is 2/5, giving x=2. |
| 56: A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of 10 N is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is: | 1000 N | Using Pascal's law, the pressures in both arms are equal. The force is proportional to the cross-sectional area of each arm. Given the ratio of diameters, the force on the thicker arm is calculated to be 1000 N. |
| 57: The electric field at point P due to an electric dipole is E. The electric field at point R on the equatorial line will be E/x. The value of x is: | 16 | For an electric dipole, the field on the axial line is twice as strong as the field on the equatorial line at the same distance. The ratio E_axial/E_equatorial gives x = 16. |
| 58: The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is: | 16 m | The maximum height of a projectile is proportional to the square of the initial velocity. Halving the velocity reduces the height to one-fourth. Thus, the new height is 64/4 = 16 m. |
| 59: A wire of resistance 20 Ω is divided into 10 equal parts. A combination of two parts is connected in parallel, and so on. Now the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is: | 5 Ω | Each part of the wire has a resistance of 2 Ω. When two parts are connected in parallel, the equivalent resistance is 1 Ω. Five such pairs connected in series result in a total resistance of 5 Ω. |
| 60: The current in an inductor is given by I = (3t + 8), where t is in seconds. The magnitude of the induced emf produced in the inductor is 12 mV. The self-inductance of the inductor is: | 4 mH | The induced emf is given by |ε| = L (dI/dt). Differentiating I = 3t + 8 gives dI/dt = 3. Substituting |ε| = 12×10⁻³ V, L is calculated as 4 mH. |
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 | Physics Question Paper PDF |
|---|---|
| JEE Main 2024 Physics Question Paper Jan 27 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 27 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 29 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 29 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 30 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 30 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 31 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Jan 31 Shift 2 | Check Here |
| JEE Main 2024 Physics Question Paper Feb 1 Shift 1 | Check Here |
| JEE Main 2024 Physics Question Paper Feb 1 Shift 2 | Check Here |
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 5 April Shift 2 Physics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Vedantu | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 5 April Shift 2 Physics Paper Analysis
JEE Main 2024 5 April 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 |
Read More:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
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