JEE Main 2024 Jan 31 Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 31 Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 Jan 31 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 Jan 31 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 31 Shift 2 Physics Question Paper PDF Download
| JEE Main 2024 Physics Question Paper | JEE Main 2024 Physics Answer Key | JEE Main 2024 Physics Answer Key |
|---|---|---|
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JEE Main 31 Jan Shift 2 2024 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 31. A light string passing over a smooth light fixed pulley connects two blocks of masses m1 and m2. If the acceleration of the system is g/8, then the ratio of masses is: (1) 9/7 (2) 8/1 (3) 4/3 (4) 5/3 |
(1) 9/7 | Using the equation of motion for the pulley system and the given acceleration, the ratio of masses is calculated as 9/7. |
| 32. A uniform magnetic field of 2 × 10-3 T acts along the positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with a current of 5 A lies in the Y-Z plane. The current is in an anticlockwise sense with reference to the negative X-axis. The magnitude and direction of the torque are: (1) 2 × 10-4 N m along positive Z-direction (2) 2 × 10-4 N m along negative Z-direction (3) 2 × 10-4 N m along positive X-direction (4) 2 × 10-4 N m along positive Y-direction |
(2) 2 × 10-4 N m along negative Z-direction | Torque is calculated as τ = M × B, where M = IA. Substituting the values, the torque direction is along the negative Z-axis with magnitude 2 × 10-4 N m. |
| 33. The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is: (1) 4% (2) 8% (3) 6% (4) 5% |
(3) 6% | Percentage error in g is Δg/g = ΔL/L + 2 ΔT/T. Substituting the values, the total percentage error is 6%. |
| 34. Force between two point charges q1 and q2 placed in a vacuum at r cm apart is F. Force between them when placed in a medium having dielectric K = 5 at r/5 cm apart will be: (1) F/25 (2) 5F (3) F/5 (4) 25F |
(2) 5F | The force in the medium is given by Fmedium = F / K. Adjusting for the reduced distance, F' = (F / K) × 25 = 5F. |
| 35. An AC voltage V = 20 sin(200πt) is applied to a series LCR circuit which drives a current I = 10 sin(200πt + π/3). The average power dissipated is: (1) 21.6 W (2) 200 W (3) 173.2 W (4) 50 W |
(4) 50W | The average power is calculated using P = Vrms Irms cos(ϕ), where Vrms = V0/√2, Irms = I0/√2, and cos(ϕ) = 1/2. Substituting the values, P = 50 W. |
| 36. When unpolarized light is incident at an angle of 60° on a transparent medium from air, the reflected ray is completely polarized. The angle of refraction in the medium is: (1) 30° (2) 60° (3) 90° (4) 45° |
(1) 30° | According to Brewster's law, tan θp = n. Here, θp = 60°, so n = √3. Using Snell's law, sin θ2 = sin 60° / √3 = 1/2, which gives θ2 = 30°. |
| 37. The speed of sound in oxygen at S.T.P. will be approximately: (1) 310 m/s (2) 333 m/s (3) 341 m/s (4) 325 m/s |
(1) 310 m/s | The speed of sound is given by v = √(γRT/M). Substituting γ = 1.4, R = 8.3 J/(K·mol), T = 273 K, and M = 32 × 10-3 kg/mol, we calculate v ≈ 310 m/s. |
| 38. A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is: (1) 29RT (2) 20RT (3) 27RT (4) 21RT |
(3) 27RT | For argon (monatomic), CV = 3R/2; for oxygen (diatomic), CV = 5R/2. Using U = nCVT, the total internal energy is U = (8 × 3R/2 + 6 × 5R/2)T = 27RT. |
| 39. The resistance per centimeter of a meter bridge wire is r, with XΩ resistance in the left gap. The balancing length from the left end is at 40 cm with 25Ω resistance in the right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for the same settings will be at: (1) 20 cm (2) 10 cm (3) 80 cm (4) 40 cm |
(4) 40cm | The balancing length depends only on the ratio of resistances, which remains unchanged even if the resistance per unit length of the wire changes. Therefore, the new balancing length is still 40 cm. |
| 40. Given below are two statements: Statement I: Electromagnetic waves carry energy as they travel through space, and this energy is equally shared by the electric and magnetic fields. Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface. In the light of the above statements, choose the most appropriate answer from the options given below: (1) Statement I is incorrect but Statement II is correct (2) Both Statement I and Statement II are correct (3) Both Statement I and Statement II are incorrect (4) Statement I is correct but Statement II is incorrect |
(2) Both Statement I and Statement II are correct | Statement I is correct because electromagnetic waves carry energy shared equally between electric and magnetic fields. Statement II is correct because electromagnetic waves exert radiation pressure when they strike a surface. |
| 41. In a photoelectric effect experiment, a light of frequency 1.5 times the threshold frequency is made to fall on the surface of a photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photoelectrons emitted will be: (1) Doubled (2) Quadrupled (3) Zero (4) Halved |
(3) Zero | Photoelectron emission occurs only if the light's frequency is above the threshold frequency. Halving the frequency makes it less than the threshold, resulting in no emission regardless of intensity. Hence, the number of photoelectrons emitted is zero. |
| 42. A block of mass 5 kg is placed on a rough inclined surface. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent the block from sliding down, then the value of |F1 - F2| is: (1) 25√3 N (2) 5√3 N (3) 5√3/2 N (4) 10 N |
(2) 5√3 N | The forces are calculated using F1 = mgsinθ + friction and F2 = mgsinθ - friction. Substituting values and simplifying, |F1 - F2| = 5√3 N. |
| 43. By what percentage will the illumination of the lamp decrease if the current drops by 20%? (1) 46% (2) 26% (3) 36% (4) 56% |
(3) 36% | Illumination is proportional to power, and power depends on the square of the current (P ∝ I²). If current decreases by 20%, power decreases by (0.2)² = 0.04 or 4%. Therefore, illumination decreases by 36%. |
| 44. If two vectors A and B having equal magnitude R are inclined at an angle θ, then: (1) |A - B| = √2R sin(θ/2) (2) |A + B| = 2R sin(θ/2) (3) |A + B| = 2R cos(θ/2) (4) |A - B| = 2R cos(θ/2) |
(3) |A + B| = 2R cos(θ/2) | The resultant of two vectors is given by R' = √(A² + B² + 2AB cosθ). For equal magnitudes, R' = 2R cos(θ/2), which matches the given expression. |
| 45. The mass number of a nucleus having radius equal to half of the radius of a nucleus with mass number 192 is: (1) 24 (2) 32 (3) 40 (4) 20 |
(1) 24 | The radius of a nucleus is proportional to the cube root of its mass number (R ∝ A1/3). If the radius is halved, the mass number becomes (1/2)3 × 192 = 24. |
| 46. The mass of the moon is 1/144 times the mass of a planet and its diameter is 1/16 times the diameter of the planet. If the escape velocity on the planet is v, the escape velocity on the moon will be: (1) v/3 (2) v/4 (3) v/12 (4) v/6 |
(1) v/3 | Escape velocity is given by vescape = √(2GM/R). Substituting the moon's mass (M/144) and radius (R/16), we find vmoon = v/3. |
| 47. A small spherical ball of radius r, falling through a viscous medium of negligible density, has terminal velocity v. Another ball of the same mass but of radius 2r, falling through the same medium, will have terminal velocity: (1) v/2 (2) v/4 (3) 4v (4) 2v |
(1) v/2 | The terminal velocity for a sphere is inversely proportional to its radius (v ∝ 1/r). For a sphere of radius 2r, the terminal velocity becomes v/2. |
| 48. A body of mass 2 kg begins to move under the action of a time-dependent force given by F = (6t î + 6t2 ĵ) N. The power developed by the force at time t is: (1) (6t4 + 9t3) W (2) (3t3 + 6t5) W (3) (9t5 + 6t3) W (4) (9t3 + 6t5) W |
(4) (9t3 + 6t5) W | Power is given by P = F · v, where v is obtained by integrating acceleration. Substituting the force components, P = (9t3 + 6t5) W. |
| 49. The output of the given circuit diagram is: 1. A B Y 0 0 0 1 0 0 0 1 0 1 1 1 2. A B Y 0 0 0 1 0 1 0 1 1 1 1 0 3. A B Y 0 0 0 1 0 0 0 1 0 1 1 0 4. A B Y 0 0 0 1 0 0 0 1 1 1 1 0 |
(3) A B Y 0 0 0 1 0 0 0 1 0 1 1 0 |
Analyzing the circuit's logic gates, the truth table matches option 3, where the output Y = 0 for all inputs except when A = 1 and B = 1. |
| 50. Consider two physical quantities A and B related to each other as E = (B - x2) / (At), where E, x, and t have dimensions of energy, length, and time respectively. The dimension of AB is: (1) L-2M T0 (2) L2M-1T (3) L-3M T-1 (4) L0M-1T |
(2) L2M-1T | Rearranging the formula, AB has dimensions derived from E = ML2T-2, x = L, and t = T. Substituting, AB = L2M-1T. |
| 51. In the following circuit, the battery has an emf of 2 V and an internal resistance of 2/3 Ω. The power consumption in the entire circuit is: | 3 W | The equivalent resistance of the circuit is calculated by combining parallel and series resistances. Using the formula P = V² / Req, the total power consumption is 3 W. |
| 52. Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at a distance of: | 200 cm | Using the formula for refraction at a spherical surface: (μ2/v) - (μ1/u) = (μ2 - μ1)/R Here, μ1 = 1 (air), μ2 = 1.5, R = 20 cm, and u = -100 cm (object distance). Substituting these values and solving gives v = 200 cm. |
| 53. The magnetic flux Φ (in weber) linked with a closed circuit of resistance 8 Ω varies with time t (in seconds) as Φ = 5t² - 36t + 1. The induced current in the circuit at t = 2 s is: | 2 A | The induced emf is calculated as ε = -dΦ/dt. At t = 2 s, ε = 16 V. Using Ohm's law, I = ε/R, we find the current to be 2 A. |
| 54. Two blocks of mass 2 kg and 4 kg are connected by a metal wire going over a smooth pulley. The radius of the wire is 4.0 × 10-5 m and Young’s modulus of the metal is 2.0 × 1011 N/m². The longitudinal strain developed in the wire is 1/(απ). The value of α is: | 12 | The tension in the wire is T = (2 × 4 × g) / (2 + 4) = 80/3 N. Using the formula Strain = T / (A × Y), the longitudinal strain is 1/(12π), giving α = 12. |
| 55. A body of mass m is projected with a speed u making an angle of 45° with the ground. The angular momentum of the body about the point of projection, at the highest point, is expressed as √2mu³ / (Xg). The value of X is: | 8 | The angular momentum is given by L = m × ux × h, where ux = u/√2 and h = u² / (4g). Substituting, L = √2mu³ / (8g), hence X = 8. |
| 56. Two circular coils P and Q of 100 turns each have the same radius of π cm. The currents in P and Q are 1 A and 2 A, respectively. P and Q are placed with their planes mutually perpendicular with their centers coinciding. The resultant magnetic field induction at the center of the coils is √x mT. The value of x is: | 20 | The magnetic field for each coil is calculated using B = (μ0NI)/(2r). Adding the perpendicular fields vectorially, Bnet = √(BP² + BQ²) gives √20 mT. Thus, x = 20. |
| 57. The distance between charges +q and -q is 2l and between +2q and -2q is 4l. The electrostatic potential at point P at a distance r from center O is -α(q l / r²) × 10⁹ V. The value of α is: | 27 | Using the formula for dipole potential, V = K(P cosθ)/r² and summing contributions from both dipoles, we find α = 27. |
| 58. Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. The moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x/20 kg m². The value of x is: | 53 | Using the parallel axis theorem, Itotal = 2(Isphere + m d²). Substituting values, the total moment of inertia is 53/20 kg m². Hence, x = 53. |
| 59. The time period of simple harmonic motion of mass M in the given figure is π√(αM / 5k). The value of α is: | 12 | The equivalent spring constant is calculated as keq = 5k/3. Using the formula for time period T = 2π√(M/keq), we find α = 12. |
| 60. A nucleus has mass number A1 and volume V1. Another nucleus has mass number A2 and volume V2. If the relation between mass numbers is A2 = 4A1, then V2 / V1 is: | 4 | Volume is proportional to the mass number. Given A2 = 4A1, the ratio of volumes is V2 / V1 = 4. |
Also Check:
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JEE Main 2024 Jan 31 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 Jan 31 Shift 2 Physics Paper Analysis
JEE Main 2024 Jan 31 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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