JEE Main 2024 Jan 30 Shift 1 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 30 Shift 1 exam from 9 AM to 12 PM. The Physics question paper for JEE Main 2024 Jan 30 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 30 Shift 1 exam is available for download using the link below.
Related Links:
- JEE Main 2025 Question Paper pdf with solutions
- JEE Main Previous Years Question Paper with Solution PDF
JEE Main 2024 Jan 30 Shift 1 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 2024 Jan 30 Shift 1 Physics Questions with Solutions
| Question | Answer | Solution |
|---|---|---|
| 31. Match List-I with List-II. List-I | List-II A. Coefficient of viscosity | I. [ML⁻¹T⁻¹] B. Surface Tension | II. [ML⁰T⁻²] C. Angular momentum | III. [ML²T⁻¹] D. Rotational kinetic energy | IV. [ML²T⁻²] (1) A-I, B-II, C-III, D-IV (2) A-I, B-II, C-IV, D-III (3) A-III, B-IV, C-II, D-I (4) A-IV, B-III, C-II, D-I |
(3) A-III, B-IV, C-II, D-I | Using dimensional analysis: - Coefficient of viscosity: [ML⁻¹T⁻¹] → III - Surface Tension: [ML⁰T⁻²] → IV - Angular momentum: [ML²T⁻¹] → II - Rotational kinetic energy: [ML²T⁻²] → I. The correct match is (3). |
| 32. All surfaces shown in the figure are frictionless, and the pulleys and the string are light. The acceleration of the block of mass 2 kg is: (1) g (2) g/3 (3) 2g/3 (4) g/4 |
(2) g/3 | Using Newton's second law, analyze the forces acting on the system. The resulting acceleration for the 2 kg block is g/3. |
| 33. A potential divider circuit is shown in the figure. The output voltage ( V_0 ) is: (1) 4V (2) 2 mV (3) 0.5 V (4) 12 mV |
(3) 0.5 V | The equivalent resistance is calculated, and the current through the circuit is used to determine \( V_0 \) using Ohm's law across the desired resistor. |
| 34. Young’s modulus of a material of a wire of length ( L ) and cross-sectional area ( A ) is ( Y ). If the length of the wire is doubled and cross-sectional area is halved, then Young’s modulus will be: (1) Y/4 (2) Y (3) 4Y (4) 2Y |
(2) Y | Young's modulus is an intrinsic property of the material, so it does not depend on dimensions like length or cross-sectional area. Thus, \( Y \) remains the same. |
| 35. The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately: (1) 215 nm (2) 414 nm (3) 400 nm (4) 200 nm |
(2) 414 nm | Using the formula λ = hc/W, where W = 3.0 eV and hc ≈ 1240 eV·nm, we get λ ≈ 414 nm. |
|
36. The ratio of the magnitude of the kinetic energy (KE) to the potential energy (PE) of an electron in the 5th excited state of a hydrogen atom is: (1) 4(2) 1/4 (3) 1/2 (4) 1 |
1/2 | In Bohr's model, KE = - (Total Energy), PE = 2 × (Total Energy). Therefore, KE/PE = 1/2 for any excited state. |
| 37. A particle is placed at point A on a frictionless track ABC as shown. It is gently pushed to the right. The speed of the particle when it reaches point B is: (Take g = 10 m/s²) (1) 20 m/s (2) √10 m/s (3) 2√10 m/s (4) 10 m/s |
(2) √10 m/s | Using conservation of mechanical energy, v = √(2gh) and is found to be √10 m/s. |
| 38. The electric field of an electromagnetic wave in free space is represented as E = E0 cos(ωt - kx) ĩ. The corresponding magnetic induction vector will be: (1) B = E0 C cos(ωt − kx) ĵ (2) B = (E0/C) cos(ωt − kx) ĵ (3) B = E0 C cos(ωt + kx) ĵ (4) B = (E0/C) cos(ωt + kx) ĵ |
(2) B = (E0/C) cos(ωt − kx) ĵ | For an electromagnetic wave, B is perpendicular to E and direction of propagation. B = E0/C cos(ωt - kx) ĵ. |
| 39. Two insulated circular loops A and B of radius a, carrying a current I in anticlockwise direction, are arranged perpendicular to each other. The magnitude of the magnetic induction at the center will be: (1) √2 μ0I/a (2) μ0I/(2a) (3) μ0I√2/a (4) 2μ0I/a |
(3) μ0I√2/a | Each loop contributes B = μ0I/(2a). Since they are perpendicular, the resultant B = √(B² + B²) = μ0I√2/(2a). Multiplying out, μ0I√2/a. |
|
40. The diffraction pattern of light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be: (1) 2 mm |
(1) 2 mm | The width of the first secondary maxima in single-slit diffraction is given by Δy = (λD)/a, where λ is the wavelength, D is the focal length (distance to the screen), and a is the slit width. Substituting λ = 400 nm = 400 × 10−9 m, D = 100 cm = 1 m, and a = 0.2 mm = 0.2 × 10−3 m, we get: Δy = (400 × 10−9 m × 1 m) / (0.2 × 10−3 m) = (400 × 10−9) / (0.2 × 10−3) = (400/0.2) × 10−9+3 = 2000 × 10−6 = 2 × 10−3 m = 2 mm. |
| 41. Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transformer are 100 and 10 respectively. The secondary coil of the transformer is connected to two series resistances shown in the figure. The output voltage V0 is: (1) 7 V (2) 15 V (3) 44 V (4) 22 V |
(1) 7 V | Using the turns ratio of the transformer: V2/V1 = N2/N1. Given V1 = 220 V, N1 = 100, N2 = 10. Thus, V2 = 220 × (10/100) = 22 V. The secondary is connected to two series resistances (as per the given figure). Using Ohm's law and the voltage division rule, the output V0 after the resistor arrangement is found to be 7 V. |
| 42. The gravitational potential at a point above the surface of Earth is −5.12 × 10⁷ J/kg and the acceleration due to gravity at that point is 6.4m/s². Assume that the mean radius of Earth to be 6400 km. The height of this point above the Earth’s surface is: (1) 1600 km (2) 540 km (3) 1200 km (4) 1000 km |
(1) 1600 km | Using the formula for gravitational potential and gravitational field, the height above the Earth's surface is calculated to be 1600 km. |
| 43. An electric toaster has resistance of 60 Ω at room temperature (27°C). The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A, the temperature attained by toaster is around: (if α = 2× 10⁻⁴ °C⁻¹) (1) 694°C (2) 1235°C (3) 1694°C (4) 1667°C |
(3) 1694°C | Using Ohm's law and the temperature coefficient of resistance, the temperature of the toaster is calculated to be 1694°C. |
| 44. A Zener diode of breakdown voltage 10V is used as a voltage regulator as shown in the figure. The current through the Zener diode is: (1) 50 mA (2) 0 (3) 30 mA (4) 20 mA |
(3) 30 mA | Using the voltage across the Zener diode and applying Ohm's law, the current through the diode is calculated to be 30 mA. |
| 45. Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are CA and CB. The molar heat capacity at constant pressure and constant volume are represented by CP and CV respectively. Choose the correct statement: (1) CB = ∞, CA = 0 (2) CA = 0 and CB = ∞ (3) CP > CA = CB = CV (4) CA > CP > CV > CB |
(2) CA = 0 and CB = ∞ | In an adiabatic process, the heat capacity is zero because no heat is exchanged. In an isothermal process, the heat capacity becomes infinite because heat is used for work without a change in temperature. |
| 46. The electrostatic potential due to an electric dipole at a distance r varies as: (1) r (2) 1/r² (3) 1/r³ (4) 1/r |
(2) 1/r² | The potential at a point along the axial line of a dipole is proportional to 1/r², where r is the distance from the dipole. |
| 47. A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given g = 9.8m/s²). (1) 4.32 kg m/s (2) 4.2 kg m/s (3) 2.39 kg m/s (4) 2.39 kg m/s |
(3) 2.39 kg m/s | Calculate the velocity just before and after impact using the principle of energy conservation, then calculate the change in momentum (impulse) of the body. |
| 48. A particle of mass m is projected with a velocity u making an angle of 30° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height is: (1) √3 m u² / (16 g) (2) √3 m u² / (2 g) (3) m u³ / (√2 g) (4) zero |
(1) √3 m u² / (16 g) | At maximum height, the vertical component of velocity is zero, and only the horizontal component contributes to the angular momentum. Using L = m v r, we find the angular momentum at the maximum height to be √3 m u² / (16 g). |
| 49. At which temperature does the r.m.s. velocity of a hydrogen molecule equal that of an oxygen molecule at 47°C? (1) 80 K (2) -73 K (3) 4 K (4) 20 K |
(4) 20 K | Using vrms = √(3RT/M), and setting the r.m.s. velocities of H2 and O2 equal, the required temperature comes out to be 20 K. |
| 50. A series L-R circuit connected to an AC source E = 25 sin(1000 t) V has a power factor of 1/√2. If the source of emf is changed to E = 20 sin(2000 t) V, the new power factor of the circuit will be: (1) 1/√2 (2) 1/√3 (3) 1/√5 (4) 1/√7 |
(3) 1/√5 | The power factor of an L-R circuit depends on the inductive reactance XL = ωL. When the frequency doubles, XL doubles, altering the power factor to 1/√5. |
| 51. The horizontal component of Earth’s magnetic field at a place is 3.5 × 10−5 T. A very long straight conductor carrying a current of √2 A is placed from South East to North West. The force per unit length experienced by the conductor is: | 35 × 10−6 N/m | The force per unit length on a current-carrying conductor in a magnetic field is F/L = iB sin(θ). Here i = √2 A, B = 3.5 × 10−5 T, θ = 45°, F/L = (√2)(3.5 × 10−5) sin(45°) = (√2)(3.5 × 10−5)(√2/2) = 3.5 × 10−5 × 1 = 3.5 × 10−5 N/m = 35 × 10−6 N/m. |
| 52. Two cells are connected in opposition. Cell E1 has 8 V emf and 2 Ω internal resistance. Cell E2 has 2 V emf and 4 Ω internal resistance. The terminal potential difference of cell E2 is: | 6 V | When connected in opposition: Net emf = 8 V - 2 V = 6 V. After calculating current and potential differences, the terminal potential difference of cell E2 is found to be 6 V. |
| 53. An electron in a hydrogen atom has energy En = −0.85 eV in an excited state. The maximum number of allowed transitions to lower energy levels is: | 6 | En = −0.85 eV corresponds to n = 4 in the hydrogen atom. From the 4th level, possible transitions are to n = 3, 2, 1 (3 levels), and from each of those to lower levels, totaling 6 possible transitions. |
| 54. Each of three blocks P, Q, and R (each 3 kg) is attached to a wire. Wires A and B each have a cross-sectional area of 0.005 cm² and Young’s modulus of 2 × 1011 N/m². Neglecting friction, the longitudinal strain on wire B is ×10−4: | 2 | Strain = (Tension)/(A × Y). After substituting values, the strain in wire B is found to be 2 × 10−4. |
| 55. The distance between the object and its image (which is twice the size of the object) formed by a convex lens is 45 cm. The focal length of the lens is: | 10 cm | Using the lens formula and magnification conditions, the focal length is found to be 10 cm. |
| 56. The displacement and the increase in the velocity of a moving particle in the time interval from t to (t+1) seconds are 125 m and 50 m/s, respectively. The distance travelled by the particle in the (t+2)th second is: |
175 m | Using the given data and equations of motion, the distance travelled in the (t+2)th second is calculated to be 175 m. |
| 57. A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is x/3E, where x is: |
2 | The total initial energy is calculated by considering both capacitors. The final energy after combining them is calculated, and the energy loss is found to be \( 2E/3 \). Hence, x = 2. |
| 58. Consider a disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is gently placed over the rotating disc along the same axis. The energy dissipated so that both discs continue to rotate together without slipping is: |
250 J | Using conservation of angular momentum and calculating the initial and final kinetic energies of the system, the energy dissipated is found to be 250 J. |
| 59. In a closed organ pipe, the frequency of the fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm², the amount of water poured in the organ tube is x grams. (Take speed of sound in air as 330 m/s) |
400 g | Using the relationship between the frequency and length of the air column in a closed organ pipe, we calculate the change in length and volume displaced by the water. The volume corresponds to a mass of 400 g. |
| 60. A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of Earth in that region is 0.5 G and the angle of dip is 30°. The emf induced across the blades is Nπ × 10⁻⁵ V. The value of N is: |
32 | The induced emf is calculated using the formula for a rotating conductor in a magnetic field. After calculating the effective magnetic field and angular velocity, we find the induced emf to be \( 32\pi \times 10^{-5} \) V. |
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 30 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 30 Shift 1 Physics Paper Analysis
JEE Main 2024 Jan 30 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 |
Read More:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
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 |
Comments