JEE Main 2024 4 April Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 4 April Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 4 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 4 April Shift 2 exam is available for download using the link below.
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JEE Main 2024 4 April Shift 2 Physics Question Paper PDF Download
| JEE Main 2024 Physics Question Paper | JEE Main 2024 Physics Answer Key | JEE Main 2024 Physics Solution |
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JEE Main 2024 4 April Shift 2 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 31: The translational degrees of freedom (ft) and rotational degrees of freedom (fr) of CH4 molecule are: (1) ft = 2 and fr = 2 (2) ft = 3 and fr = 3 (3) ft = 3 and fr = 2 (4) ft = 2 and fr = 3 |
(2) ft = 3 and fr = 3 | CH4 is a polyatomic molecule with a non-linear structure. For such molecules: - The translational degrees of freedom (ft) are 3, corresponding to motion along x, y, and z axes. - The rotational degrees of freedom (fr) are also 3, as the molecule can rotate about three mutually perpendicular axes. Thus, ft = 3 and fr = 3. |
| 32: A cyclist starts from the point P of a circular ground of radius 2 km and travels along its circumference to the point S. The displacement of the cyclist is: (1) 6 km (2) √8 km (3) 4 km (4) 8 km |
(2) √8 km | The displacement of the cyclist is the straight-line distance between P and S. Since P and S are at opposite ends of the diameter of the circle: Displacement = √(Diameter²) = √(2×2)² = √8 km. |
| 33: The magnetic moment of a bar magnet is 0.5 Am2. It is suspended in a uniform magnetic field of 8 × 10-2 T. The work done in rotating it from its most stable to most unstable position is: (1) 16 × 10-2 J (2) 8 × 10-2 J (3) 4 × 10-2 J (4) Zero |
(2) 8 × 10-2 J | The work done in rotating the magnet is the change in potential energy: W = 2 × m × B = 2 × 0.5 × 8 × 10-2 = 8 × 10-2 J. |
| 34: Which of the diode circuits shows correct biasing used for the measurement of dynamic resistance of a p-n junction diode: (1) Option 1 (2) Option 2 (3) Option 3 (4) Option 4 |
(2) Option 2 | The dynamic resistance of a p-n junction diode is measured under forward bias. Option 2 correctly shows the diode in forward bias for dynamic resistance measurement. |
| 35: Arrange the following in ascending order of wavelength: (A) Gamma rays (λ1) (B) X-rays (λ2) (C) Infrared waves (λ3) (D) Microwaves (λ4) (1) λ4 < λ3 < λ1 < λ2 (2) λ4 < λ3 < λ2 < λ1 (3) λ1 < λ2 < λ3 < λ4 (4) λ2 < λ1 < λ4 < λ3 |
(3) λ1 < λ2 < λ3 < λ4 | Gamma rays (λ1) have the shortest wavelength, followed by X-rays (λ2), then infrared waves (λ3), and microwaves (λ4) with the longest wavelength. The correct order is λ1 < λ2 < λ3 < λ4. |
| 36: Identify the logic gate given in the circuit: (1) NAND gate (2) OR gate (3) AND gate (4) NOR gate |
(2) OR gate | The circuit contains NOT gates followed by a NAND gate. Using De Morgan's law, the output behaves as an OR gate. Therefore, the correct answer is OR gate. |
| 37: The width of one of the two slits in a Young’s double-slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is: (1) 9:1 (2) 16:1 (3) 1:1 (4) 4:1 |
(1) 9:1 | The intensity ratio is determined by the amplitude ratio, which is proportional to the square root of the slit width. For widths w1 and w2 (with w2 = 4w1), the ratio of maximum to minimum intensity is 9:1. |
| 38: Correct formula for height of a satellite from Earth's surface is: (1) (T²R²g/4π²)1/2 - R (2) (T²R²g/4π²)1/3 - R (3) (T²R²/4π²g)1/3 - R (4) (T²R²/4π²)1/3 + R |
(2) (T²R²g/4π²)1/3 - R | Using the formula for orbital mechanics, the height is calculated as h = [(T²R²g/4π²)1/3] - R, where T is the orbital period, R is the Earth's radius, and g is the gravitational acceleration. |
| 39: Match List I with List II: List I: A. Purely capacitive circuit B. Purely inductive circuit C. LCR series circuit at resonance D. General LCR series circuit List II: I. Voltage lags current by 90° II. Voltage leads current by 90° III. Voltage and current are in phase IV. Voltage and current are out of phase Choose the correct answer: (1) A-I, B-IV, C-III, D-II (2) A-IV, B-I, C-III, D-II (3) A-IV, B-I, C-II, D-III (4) A-I, B-II, C-III, D-IV |
(4) A-I, B-II, C-III, D-IV | A purely capacitive circuit causes voltage to lag current by 90° (A-I). In a purely inductive circuit, voltage leads current by 90° (B-II). At resonance in an LCR circuit, voltage and current are in phase (C-III). In a general LCR series circuit, voltage and current are out of phase (D-IV). |
| 40: Given below are two statements: Statement I: The contact angle between a solid and a liquid depends on the material properties of the solid and liquid. Statement II: The rise of a liquid in a capillary tube does not depend on the radius of the tube. Choose the correct answer: (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true |
(3) Statement I is true but Statement II is false | Statement I is true as the contact angle depends on cohesive and adhesive forces of the materials. Statement II is false because the capillary rise is inversely proportional to the radius of the tube, as given by h = 2Tcosθ/ρgr. |
| 41: A body of mass m kg slides from rest along the curve of a vertical circle from point A to B in a frictionless path. The velocity of the body at B is: Given: R = 14 m, g = 10 m/s2, √2 = 1.4 (1) 19.8 m/s (2) 21.9 m/s (3) 16.7 m/s (4) 10.6 m/s |
(2) 21.9 m/s | Using conservation of energy: PE at A = KE at B. Initial potential energy is m × g × (R + R/√2), and kinetic energy at B is (1/2)mv². Solving for v, we get v = √(2gR(1 + 1/√2)) = 21.9 m/s. |
| 42: An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is: (1) 12.5 W (2) 25 W (3) 50 W (4) 100 W |
(1) 12.5 W | The resistance of the bulb is R = V²/P = 200²/50 = 800 Ω. Power at 100 V is P = V²/R = 100²/800 = 12.5 W. |
| 43: A 2 kg brick begins to slide over a surface inclined at an angle of 45° with respect to the horizontal. The coefficient of static friction between their surfaces is: (1) 1 (2) 1/√3 (3) 0.5 (4) 1.7 |
(1) 1 | At the angle of repose, tanθ = µs. Since θ = 45°, µs = tan45° = 1. |
| 44: In simple harmonic motion, the total mechanical energy of a system is E. If the mass of the oscillating particle is doubled, the new energy of the system for the same amplitude is: (1) E/√2 (2) E (3) E√2 (4) 2E |
(2) E | Total mechanical energy in SHM is given by E = (1/2)kA², which is independent of the mass of the oscillating particle. Hence, the energy remains E. |
| 45: Assertion (A): Number of photons increases with increase in frequency of light. Reason (R): Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation. Choose the correct answer: (1) Both A and R are correct, and R is NOT the correct explanation of A. (2) A is correct but R is not correct. (3) Both A and R are correct, and R is the correct explanation of A. (4) A is not correct but R is correct. |
(4) A is not correct but R is correct | A is incorrect because the number of photons decreases with an increase in frequency for a constant intensity. R is correct as per the photoelectric effect: Kmax = hν - ϕ. |
| 46: According to Bohr’s theory, the moment of momentum of an electron revolving in the 4th orbit of a hydrogen atom is: (1) 8h/π (2) h/π (3) 2h/π (4) h/2π |
(3) 2h/π | According to Bohr’s theory, angular momentum L = n(h/2π). For n = 4, L = 4(h/2π) = 2h/π. |
| 47: A sample of gas at temperature T is adiabatically expanded to double its volume. The adiabatic constant for the gas is γ = 3/2. The work done by the gas in the process is: (1) RT[√2 - 2] (2) RT[1 - 2√2] (3) RT[2√2 - 1] (4) RT[2 - √2] |
(4) RT[2 - √2] | Using the adiabatic relation TVγ-1 = constant and the formula W = (R∆T)/(1-γ), the work done is calculated to be RT[2 - √2]. |
| 48: A charge q is placed at the center of one surface of a cube. The flux linked with the cube is: (1) q/4ε0 (2) q/2ε0 (3) q/8ε0 (4) Zero |
(2) q/2ε0 | The charge q contributes equally to two adjacent cubes. By Gauss’s law, total flux is q/ε0, so flux through one cube is q/2ε0. |
| 49: Applying the principle of homogeneity of dimensions, determine which one is correct: (1) T² = 4π²r/GM² (2) T² = 4π²r²r³ (3) T² = 4π²r³/GM (4) T² = 4π²r²/GM |
(3) T² = 4π²r³/GM | Dimensional analysis shows that T² = 4π²r³/GM is dimensionally consistent. The left-hand side and right-hand side both have the dimension of time squared. |
| 50: A 90 kg body placed at 2R distance from the surface of the Earth experiences a gravitational pull of: Given: R = Radius of Earth, g = 10 m/s² (1) 300 N (2) 225 N (3) 120 N (4) 100 N |
(4) 100 N | At 2R from the Earth's surface, gravitational acceleration is g/9. The force F = mg/9 = (90 × 10)/9 = 100 N. |
| 51: The displacement of a particle executing SHM is given by x = 10 sin(ωt + π/3) m. The time period of motion is 3.14 s. The velocity of the particle at t = 0 is: | 10 m/s | The angular frequency is ω = 2π/T = 2 rad/s. The velocity is v = dx/dt = 10ω cos(ωt + π/3). At t = 0, v = 10 × 2 × cos(π/3) = 10 m/s. |
| 52: A bus moving along a straight highway with a speed of 72 km/h is brought to a halt within 4 seconds after applying the brakes. The distance travelled by the bus during this time is: | 40 m | The initial velocity is u = 72 km/h = 20 m/s. Using the equation v = u + at and v = 0, we find a = -5 m/s². Distance is s = ut + (1/2)at² = 20 × 4 - (1/2) × 5 × 4² = 40 m. |
| 53: A parallel plate capacitor of capacitance 12.5 pF is charged by a 12.0 V battery. After disconnecting the battery, a dielectric slab (εr = 6) is inserted between the plates. The change in potential energy after inserting the dielectric slab is: | 750 × 10-12 J | Initial energy is Ui = (1/2)C0V². After inserting the dielectric, Uf = (1/2)CfV² = Ui/εr. The change in energy is ΔU = Ui(1 - 1/εr) = 750 × 10-12 J. |
| 54: In a system of two particles, m1 = 3 kg and m2 = 2 kg are placed at a certain distance. m1 is moved 2 cm towards the center of mass. To keep the center of mass at the original position, m2 should move by: | 3 cm | Using the center of mass condition, m1Δx1 = m2Δx2. Substituting m1 = 3, Δx1 = 2 cm, m2 = 2, we find Δx2 = 3 cm. |
| 55: The disintegration energy Q for the nuclear fission 235U → 140Ce + 94Zr + n is (in MeV): Given: Atomic masses - 235U = 235.0439 u, 140Ce = 139.9054 u, 94Zr = 93.9063 u, n = 1.0086 u, c² = 931 MeV/u |
208 MeV | The mass defect is Δm = mreactants - mproducts = 235.0439 - (139.9054 + 93.9063 + 1.0086) = 0.2236 u. Disintegration energy Q = Δm × c² = 0.2236 × 931 = 208 MeV. |
| 56: A light ray is incident on a glass slab of thickness 4√3 cm and refractive index √2. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of the ray after passing through the glass slab is: | 2√3 cm | The critical angle is given by sin C = 1/√2. The angle of refraction is 45°. Lateral displacement is calculated as d = t tan r = 4√3 × tan 45° = 2√3 cm. |
| 57: A rod of length 60 cm rotates with a uniform angular velocity 20 rad/s about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of the magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is: | 0.18 V | The potential difference is calculated using the formula V = (1/2)BωL². Substituting B = 0.5 T, ω = 20 rad/s, and L = 0.6 m: V = (1/2) × 0.5 × 20 × (0.6)² = 0.18 V. |
| 58: Two wires A and B are made up of the same material and have the same mass. Wire A has a radius of 2.0 mm and wire B has a radius of 4.0 mm. The resistance of wire B is 2 Ω. The resistance of wire A is: | 32 Ω | Resistance is inversely proportional to the square of the radius when the mass is constant (R ∝ 1/r²). Given RB = 2 Ω, radius rA/rB = 1/2. Thus, RA = RB × (rB/rA)² =RA /2 = 16, which gives: RA = 32 Ω.. |
| 59: Two parallel long current-carrying wires separated by a distance 2r are shown in the figure. The ratio of magnetic field at A to the magnetic field produced at C is x/7. The value of x is: | 5 | The magnetic field at a point due to a long straight wire is B = μ₀I/(2πr). By calculating the fields at points A and C and considering their directions, the ratio is found to be x/7, where x = 5. |
| 60: Mercury is filled in a tube of radius 2 cm up to a height of 30 cm. The force exerted by mercury on the bottom of the tube is: Given: Atmospheric pressure = 10⁵ N/m², density of mercury = 1.36 × 10⁴ kg/m³, g = 10 m/s², π = 22/7 |
1830 N | The pressure exerted by mercury is P = ρgh = (1.36 × 10⁴) × 10 × 0.3 = 40800 N/m². The total pressure is Ptotal = Atmospheric pressure + Pressure due to mercury = 10⁵ + 40800 = 140800 N/m². The force is F = P × Area = 140800 × πr² = 140800 × (22/7) × (0.02)² = 1830 N. |
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 4 April Shift 2 Physics Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Reliable Institute | To be updated |
| Resonance | To be updated |
| Vedantu | Download PDF |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 4 April Shift 2 Physics Paper Analysis
JEE Main 2024 4 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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