JEE Main 2024 Jan 31 Shift 1 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 31 Shift 1 exam from 9 AM to 12 PM. The Physics question paper for JEE Main 2024 Jan 31 Shift 1 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 1 is available for download using the link below.
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JEE Main 2024 Jan 31 Shift 1 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 Jan 31 Shift 1 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
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Question 31: The parameter that remains the same for molecules of all gases at a given temperature is:
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(1) Kinetic energy | The kinetic energy of gas molecules at a given temperature is the same for all gases according to the equipartition theorem. This is because the average kinetic energy is solely dependent on the temperature. |
| Question 32: Identify the logic operation performed by the given circuit. (1) NAND (2) NOR (3) OR (4) AND |
(3) OR | The circuit simplifies to Y = A + B (OR operation). By applying De Morgan's law, we identify the operation as OR. |
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Question 33: The relation between time t and distance x is t = αx² + βx, where α and β are constants. The relation between acceleration (a) and velocity (v) is:
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(1) a = −2αv³ | By differentiating t = αx² + βx with respect to x and then applying the chain rule, we obtain the relationship between acceleration and velocity as a = −2αv³. |
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Question 34: The refractive index of a prism with apex angle A is cot(A/2). The angle of minimum deviation is:
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(4) δm = 180° − 2A | Using the relation for the refractive index and applying Snell's law, we find that the angle of minimum deviation is δm = 180° − 2A. |
| Question 35: A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B⃗ = B₀ ĵ. The magnetic force on the wire if it has a current i is: (1) −iBR ĵ (2) 2iBR ĵ (3) iBR ĵ (4) −2iBR ĵ |
(4) −2iBR ĵ | The magnetic force on the wire is given by F = i l × B. After calculating the forces from the semicircular arc and the straight sections, we find that the total magnetic force is −2iBR ĵ. |
| Question 36: If the wavelength of the first member of the Lyman series of hydrogen is λ, the wavelength of the second member will be: (1) 27/32λ (2) 32/27λ (3) 27/5λ (4) 5/27λ |
(1) 27/32λ | Using the Rydberg formula for spectral lines, we find that the wavelength of the second member of the Lyman series is 27/32λ. |
| Question 37: Four identical particles of mass m are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is (2√2 + 1) Gm² / L², the length of the sides of the square is: (1) L/2 (2) 4L (3) 3L (4) 2L |
(2) 4L | By calculating the gravitational forces and applying vector addition, we find that the length of the sides of the square is 4L. |
| Question 38: The given figure represents two isobaric processes for the same mass of an ideal gas. Then: (1) P₂ ≥ P₁ (2) P₂ > P₁ (3) P₁ = P₂ (4) P₁ > P₂ |
(4) P₁ > P₂ | From the ideal gas law and the slope of the V-T graph, we determine that P₁ > P₂ due to the relationship between the slope and pressure. |
| Question 39: If the percentage errors in measuring the length and the diameter of a wire are 0.1% each, the percentage error in measuring its resistance will be: (1) 0.2% (2) 0.3% (3) 0.1% (4) 0.144% |
(2) 0.3% | Using the formula for resistance R = ρL / A, the total percentage error in resistance is 0.3%, considering the errors in both length and diameter. |
| Question 40: In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ Hz and an amplitude of 50 V/m. The total average energy density of the electromagnetic field of the wave is: (1) 1.106 × 10⁻⁸ J/m³ (2) 4.425 × 10⁻⁸ J/m³ (3) 2.212 × 10⁻⁸ J/m³ (4) 2.212 × 10⁻¹⁰ J/m³ |
(1) 1.106 × 10⁻⁸ J/m³ | The average energy density of the electric field in an EM wave is given by: U = (1/2) ε₀ E² Substituting the given values of the electric field amplitude (E = 50 V/m) and the permittivity of free space (ε₀ = 8.85 × 10⁻¹² C²/N·m²), we get the total average energy density as 1.106 × 10⁻⁸ J/m³. |
| Question 41: A force is represented by F = ax² + bt¹/² where x = distance and t = time. The dimensions of b² / a are: (1) [ML³T⁻³] (2) [MLT⁻²] (3) [ML²T⁻¹] (4) [MLT²] |
(1) [ML³T⁻³] | By applying dimensional analysis, we can derive the dimensions of a and b from the equation F = ax² + bt¹/². The dimensions of b² / a result in [ML³T⁻³]. |
| Question 42: Two charges q and 3q are separated by a distance r in air. At a distance x from charge q, the resultant electric field is zero. The value of x is: (1) (1 + √3) r / 3 (2) r / 3 (1 + √3) (3) r (1 + √3) (4) r (1 + √3) / 3 |
(3) r (1 + √3) | The electric field at a point x from charge q where the net electric field is zero is given by equating the fields due to both charges and solving for x. We find that the value of x is r(1 + √3). |
| Question 43: In the given arrangement of a doubly inclined plane, two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M = 10 kg will move down with an acceleration of 2 m/s², is: (1) 9 kg (2) 4.5 kg (3) 6.5 kg (4) 2.25 kg |
(2) 4.5 kg | By applying Newton's second law of motion to the blocks and considering the forces acting along the inclined planes, we determine that the value of m is 4.5 kg. |
| Question 44: A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is: (1) 7 (2) 70 (3) 35 (4) 140 |
(2) 70 | The induced emf in the coil is related to the change in magnetic flux and the number of turns by the formula: ε = N(Δφ / Δt). Using the given values and solving, we find that the number of turns in the coil is 70. |
| Question 45: The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is 60 cm, the length of the closed pipe will be: (1) 60 cm (2) 45 cm (3) 30 cm (4) 15 cm |
(4) 15 cm | The fundamental frequency of a closed pipe corresponds to a quarter wavelength, while the first overtone of an open pipe corresponds to a half wavelength. Equating these, we find the length of the closed pipe to be 15 cm. |
| Question 46: A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity-time graph for the transit of the ball? (1) Linear (2) Exponential (3) Parabolic (4) Constant |
(2) Exponential | The motion of the ball in glycerine is subject to a viscous drag force, leading to an exponential increase in velocity until terminal velocity is reached. |
| Question 47: A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ. If the distance of the coin from the center of the disc is r, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is: (1) μg / r (2) √r μg (3) π μg / r (4) μ √rg |
(3) π μg / r | The maximum angular velocity is derived from the frictional force preventing the coin from slipping. This results in ω = μg / r, where r is the distance from the center of the disc. |
| Question 48: Two conductors have the same resistances at 0°C but their temperature coefficients of resistance are α₁ and α₂. The respective temperature coefficients for their series and parallel combinations are: (1) α₁ + α₂, α₁ + α₂ / 2 (2) α₁ + α₂ / 2, α₁ + α₂ / 2 (3) α₁ + α₂, α₁ α₂ / α₁ + α₂ (4) α₁ + α₂ / 2, α₁ + α₂ |
(2) α₁ + α₂ / 2, α₁ + α₂ / 2 | The equivalent temperature coefficients for series and parallel combinations are both the average of the individual temperature coefficients. |
| Question 49: An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is: (1) M₁ / (M₁ + M₂) (2) M₂ / M₁ (3) M₁ / (M₁ + M₂) (4) M₁ / M₂ |
(2) M₂ / M₁ | Using the conservation of momentum and kinetic energy formulas, the ratio of the kinetic energies of the artillery and the shell is found to be M₂ / M₁. |
| Question 50: When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8V. When the same surface is illuminated by light of wavelength 3λ, the stopping potential is 2V. The threshold wavelength for this surface is: (1) 5λ (2) 3λ (3) 9λ (4) 4.5λ |
(3) 9λ | Using the photoelectric equation and solving for the threshold wavelength, we find that λ₀ = 9λ. |
| Question 51: An electron moves through a uniform magnetic field B⃗ = B₀ ĩ + 2B₀ ĵ T. At a particular instant of time, the velocity of the electron is ⃗u = 3ĩ + 5ĵ m/s. If the magnetic force acting on the electron is F⃗ = 5e ˆk N, where e is the charge of the electron, then the value of B₀ is T: (1) 1 T (2) 3 T (3) 5 T (4) 10 T |
(3) 5 T | Using the magnetic force equation F = q(⃗v × B⃗), we compute the value of B₀ and find that B₀ = 5 T. |
| Question 52: A parallel plate capacitor with plate separation 5 mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mm, while keeping the battery connections intact, the capacitor draws 25% more charge from the battery than before. The dielectric constant of the sheet is: (1) 1 (2) 2 (3) 3 (4) 4 |
(2) 2 | By using the equation for the capacitance with and without the dielectric, we find the dielectric constant K = 2. |
| Question 53: The equivalent resistance of the following network is Ω: (1) 1Ω (2) 2Ω (3) 3Ω (4) 4Ω |
(1) 1Ω | After simplifying the given circuit, we find that the equivalent resistance is 1Ω. |
| Question 54: A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is J: (1) 6 J (2) 12 J (3) 18 J (4) 24 J |
(6) 6 J | By using the work-energy theorem, we account for both translational and rotational kinetic energy. The work done to stop the disc is 6 J. |
| Question 55: A body starts falling freely from height H and hits an inclined plane at height h. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of H / h for which the body will take the maximum time to reach the ground is: (1) 1 (2) 2 (3) 3 (4) 4 |
(2) 2 | By differentiating the time of flight with respect to h and setting the derivative equal to zero, we find that H/h = 2 maximizes the time to reach the ground. |
| Question 56: Two waves of intensity ratio 1:9 cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is I₁ and (b) Waves are coherent is I₂ and differ in phase by 60°. If I₁ / I₂ = 10 / x, then x = . (1) 13 (2) 12 (3) 14 (4) 15 |
(13) 13 | By applying the formulas for coherent and incoherent wave intensities and using phase difference, we calculate that x = 13. |
| Question 57: A small square loop of wire of side l is placed inside a large square loop of wire of side L (L = l²). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is √x × 10⁻⁷ H, where x = . (1) 128 (2) 64 (3) 32 (4) 16 |
(128) 128 | Using symmetry and proportionality principles, the mutual inductance between the loops is calculated to be √128 × 10⁻⁷ H. |
| Question 58: The depth below the surface of the sea to which a rubber ball can be taken so as to decrease its volume by 0.02% is: (1) 18 m (2) 20 m (3) 25 m (4) 30 m |
(18) 18 m | Using the bulk modulus formula, we calculate that the depth required to decrease the volume by 0.02% is 18 m. |
| Question 59: A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A / 3. The new amplitude of motion is nA / 3. The value of n is . (1) 7 (2) 6 (3) 5 (4) 4 |
(7) 7 | Using the relationship between velocity, amplitude, and displacement for SHM, we find that the new amplitude is 7A/3. |
| Question 60: The mass defect in a particular reaction is 0.4 g. The amount of energy liberated is n × 10⁷ kWh, where n = . (1) 1 (2) 2 (3) 3 (4) 4 |
(1) 1 | Using the mass-energy equivalence formula E = ∆mc², we calculate the energy released as 1 × 10⁷ kWh. |
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 Jan 31 Shift 1 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 1 Physics Paper Analysis
JEE Main 2024 Jan 31 Shift 1 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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