JEE Main 2024 Jan 27 Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 27 Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 Jan 27 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 27 Shift 2 exam is available for download using the link below.

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JEE Main 27 Jan Shift 2 2024 Physics Questions with Solution

Question Answer Solution
31. The equation of state of a real gas is given by P + (a/V²)(V - b) = RT, where P, V, and T are pressure, volume, and temperature. The dimensions of a/b² are similar to:
(1) P
(2) PV
(3) RT
(4) R		 P Using dimensional analysis on a/b², we find it has the same dimensions as pressure P.
32. The total kinetic energy of 1 mole of oxygen at 27°C is:
(1) 6845.5 J
(2) 5942.0 J
(3) 6232.5 J
(4) 5670.5 J		 6232.5 J Using the formula E = (5/2)nRT for a diatomic gas at 300 K, we find the total kinetic energy as 6232.5 J.
33. The primary side of a transformer is connected to a 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10:1. Load resistance on the secondary side is 46 Ω. The power consumed in it is:
(1) 12.5 W
(2) 10.0 W
(3) 11.5 W
(4) 12.0 W		 11.5 W Using the turns ratio to find V₂ and calculating power with P = V₂²/R, we find it to be 11.5 W.
34. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of Cp/Cv for the gas is:
(1) 5/3
(2) 3/2
(3) 7/5
(4) 9/7		 3/2 Using the adiabatic relation and the given condition, we derive that the ratio of Cp/Cv (γ) is 3/2.
35. The threshold frequency of a metal with work function 6.63 eV is:
(1) 16 ×10¹⁵ Hz
(2) 16 ×10¹² Hz
(3) 1.6 × 10¹² Hz
(4) 1.6 × 10¹⁵ Hz		 1.6 × 10¹⁵ Hz Using E = hν₀ with work function converted to joules, we find ν₀ as 1.6 × 10¹⁵ Hz.
36. A current of 200 µA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π/10 radians is:
(1) 30 µA
(2) 120 µA
(3) 60 µA
(4) 180 µA		 60 µA Using proportionality of current and angle, we find the required current as 60 µA.
37. The atomic mass of 6C¹² is 12.000000 u and that of 6C¹³ is 13.003354 u. The required energy to remove a neutron from 6C¹³, if the mass of the neutron is 1.008665 u, will be:
(1) 62.5 MeV
(2) 6.25 MeV
(3) 4.95 MeV
(4) 49.5 MeV		 4.95 MeV Using mass defect and E = ∆m x 931.5 MeV/u, we find the energy required as 4.95 MeV.
38. A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme and lowest position are equal. The angle (θ) of deflection in the extreme position is:
(1) tan⁻¹(√2)
(2) 2tan⁻¹(1/2)
(3) tan⁻¹(1/2)
(4) 2tan⁻¹(1/√5)		 2tan⁻¹(1/2) Using energy conservation and solving for θ, we find it as 2tan⁻¹(1/2).
39. Three voltmeters are joined as shown. When a potential difference is applied across A and B, their readings are V₁, V₂, and V₃. Choose the correct option:
(1) V₁ = V₂
(2) V₁ = V₃ - V₂
(3) V₁ + V₂ > V₃
(4) V₁ + V₂ = V₃		 V₁ + V₂ = V₃ Applying Kirchhoff’s Voltage Law across the loop, we find that V₁ + V₂ = V₃.
40. The total kinetic energy of 1 mole of oxygen at 27°C is:
(1) 6845.5 J
(2) 5942.0 J
(3) 6232.5 J
(4) 5670.5 J		 6232.5 J Using the formula E = (5/2)nRT for a diatomic gas at 300 K, we find the total kinetic energy as 6232.5 J.
41. Given that the angular speed of the moon in its orbit about the earth is greater than that of the earth around the sun, identify the reason.
(1) Shorter period of orbit
(2) Larger radius
(3) Higher mass of the moon
(4) Higher gravitational pull		 Shorter period of orbit Angular speed ω is inversely proportional to time period. Since the moon's period is shorter, ωmoon > ωearth.
42. The primary side of a transformer is connected to 230 V, 50 Hz supply with a turns ratio of 10:1. Load resistance is 46 Ω. The power consumed is:
(1) 12.5 W
(2) 10 W
(3) 11.5 W
(4) 12 W		 11.5 W Using turns ratio and calculating V₂, then power with P = V₂²/R, we find it as 11.5 W.
43. During an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The ratio of Cp/Cv is:
(1) 5/3
(2) 3/2
(3) 7/5
(4) 9/7		 3/2 Using the adiabatic relation and condition given, we derive γ = Cp/Cv as 3/2.
44. The threshold frequency of a metal with work function 6.63 eV is:
(1) 16 ×10¹⁵ Hz
(2) 16 ×10¹² Hz
(3) 1.6 × 10¹² Hz
(4) 1.6 × 10¹⁵ Hz		 1.6 × 10¹⁵ Hz Using E = hν₀ with work function converted to joules, we find ν₀ as 1.6 × 10¹⁵ Hz.
45. A current of 200 µA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π/10 radians is:
(1) 30 µA
(2) 120 µA
(3) 60 µA
(4) 180 µA		 60 µA Using proportionality of current and angle, we find the required current as 60 µA.
46. The atomic mass of 6C¹² is 12.000000 u and that of 6C¹³ is 13.003354 u. The required energy to remove a neutron from 6C¹³, if the mass of the neutron is 1.008665 u, will be:
(1) 62.5 MeV
(2) 6.25 MeV
(3) 4.95 MeV
(4) 49.5 MeV		 4.95 MeV Using mass defect and E = ∆m x 931.5 MeV/u, we find the energy required as 4.95 MeV.
47. A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme and lowest position are equal. The angle (θ) of deflection in the extreme position is:
(1) tan⁻¹(√2)
(2) 2tan⁻¹(1/2)
(3) tan⁻¹(1/2)
(4) 2tan⁻¹(1/√5)		 2tan⁻¹(1/2) Using energy conservation and solving for θ, we find it as 2tan⁻¹(1/2).
48. Three voltmeters are joined as shown. When a potential difference is applied across A and B, their readings are V₁, V₂, and V₃. Choose the correct option.
(1) V₁ = V₂
(2) V₁ = V₃ - V₂
(3) V₁ + V₂ > V₃
(4) V₁ + V₂ = V₃		 V₁ + V₂ = V₃ Applying Kirchhoff’s Voltage Law across the loop, we find that V₁ + V₂ = V₃.
49. The total kinetic energy of 1 mole of oxygen at 27°C is:
(1) 6845.5 J
(2) 5942.0 J
(3) 6232.5 J
(4) 5670.5 J		 6232.5 J Using the formula E = (5/2)nRT for a diatomic gas at 300 K, we find the total kinetic energy as 6232.5 J.
50. The primary side of a transformer is connected to 230 V, 50 Hz supply with a turns ratio of 10:1. Load resistance is 46 Ω. The power consumed is:
(1) 12.5 W
(2) 10 W
(3) 11.5 W
(4) 12 W		 11.5 W Using turns ratio and calculating V₂, then power with P = V₂²/R, we find it as 11.5 W.
51: The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1 = 2m and R2 = 4m carrying current I = 4A as per figure given below is α× 10−7 T. The value of α is: 4 The magnetic field at the center of a semicircular wire carrying current I and radius R is given by B = µ₀I/4R. Adding the fields from both semicircular wires with radii R1 and R2 gives the total magnetic field at the center as 3π × 10−7 T, so α = 4.
52: Two charges of −4µC and +4µC are placed at the points A(1, 0, 4)m and B(2,−1, 5)m located in an electric field E⃗ = 0.20iV/cm. The magnitude of the torque acting on the dipole is 8√α× 10−5 Nm, where α = :
 		 2 The electric dipole moment is calculated by p⃗ = q × d⃗. The torque on the dipole is τ⃗ = p⃗ × E⃗, and the calculation gives a value of 8√2 × 10−5 Nm. Therefore, α = 2.
53: A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is:
 		 294 m/s The velocity of sound is determined using the fundamental frequency equations for both the closed and open organ pipes, and the difference in frequencies gives a velocity of sound equal to 294 m/s.
54: A body falling under gravity covers two points A and B separated by 80 m in 2s. The distance of upper point A from the starting point is:
 		 45 m Using the equations of motion for free fall, the distance covered by the body between points A and B is 80 m in 2 seconds. The distance from the starting point to point A is calculated to be 45 m.
55: The reading of a pressure meter attached with a closed pipe is 4.5×104 N/m2. On opening the valve, water starts flowing and the reading of pressure meter falls to 2.0×104 N/m2. The velocity of water is found to be √V m/s. The value of V is:
 		 50 Using Bernoulli’s equation for the flow of an ideal fluid, the difference in pressure readings is used to calculate the velocity of water, which is found to be √50 m/s.
56: A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is:
 		 7 In rolling motion, the total kinetic energy is a sum of translational and rotational components. The solid sphere has more rotational kinetic energy than the ring, which results in the ratio of their kinetic energies being 7.
57: A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of:
 		 30° For the first minima in a single-slit diffraction pattern, the angle is given by sin θ = λ/a, where a is the width of the slit and λ is the wavelength. Substituting the given values, θ = 30°.
58: The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9×10−13 cm is ×106 V:
 		 8 The electric potential at the surface of the nucleus is calculated using V = kZe/R, where k = 9×109 N·m2/C2, Z = 50, and R = 9×10−13 cm. The potential is found to be 8×106 V.
59: If Rydberg’s constant is R, the longest wavelength of radiation in Paschen series will be α×7R, where α = :
 		 144 The longest wavelength in the Paschen series corresponds to the transition from n = 4 to n = 3. Using the Rydberg formula, the value of α is 144.
60: A series LCR circuit with L = 100π mH, C = 10−3 F, and R = 10Ω, is connected across an ac source of 220V, 50Hz supply. The power factor of the circuit would be:
 		 1 At resonance in a series LCR circuit, the inductive and capacitive reactances cancel out, leaving only the resistance. Therefore, the impedance is equal to the resistance, and the power factor is 1.

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JEE Main 2024 Jan 27 Shift 2 Physics Paper Analysis

JEE Main 2024 Jan 27 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 Paper Analysis Jan 27 Shift 2 (Out)

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.

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