JEE Main 2024 Jan 30 Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 30 Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 Jan 30 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 30 Shift 2 exam is available for download using the link below.
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JEE Main 2024 Jan 30 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 30 Jan Shift 2 2024 Physics Questions with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 31: If 50 Vernier divisions are equal to 49 main scale divisions of a traveling microscope and one smallest reading of the main scale is 0.5 mm, the Vernier constant of the traveling microscope is: (1) 0.1 mm (2) 0.1 cm (3) 0.01 cm (4) 0.01 mm |
(4) 0.01 mm | The Vernier constant (VC) is calculated as: VC = Value of 1 MSD − Value of 1 VSD. Given 50 VSD = 49 MSD and 1 MSD = 0.5 mm, compute VC = 0.5 − 0.49 = 0.01 mm. |
| 32: A block of mass 1 kg is pushed up a surface inclined to the horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface. When the block is pushed up by 10 m along the inclined surface, the work done against the frictional force is: (1) √5 J (2) 5 J (3) 5 × 10³ J (4) 10 J |
(2) 5 J | Work against friction = μ × N × d. With μ = 0.1, N = mg cos(60°) = 5 N, and d = 10 m, calculate work = 0.1 × 5 × 10 = 5 J. |
| 33: For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (ν) of the incident photons. The slope of the graph gives: (1) Ratio of Planck’s constant to electric charge (2) Work function of the metal (3) Charge of electron (4) Planck’s constant |
(4) Planck’s constant | The photoelectric equation Ek = hν − ϕ relates kinetic energy to frequency. In the graph, Ek is plotted against ν, making the slope equal to Planck’s constant (h). |
| 34: A block of ice at −10°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomenon qualitatively? (1) Option 1 (2) Option 2 (3) Option 3 (4) Option 4 |
(4) | The heating curve shows temperature increases during phase changes, with plateaus at 0°C (melting) and 100°C (boiling). The correct graph includes these transitions. |
| 35: In a nuclear fission reaction of an isotope of mass M, three similar daughter nuclei of the same mass are formed. The speed of a daughter nucleus in terms of mass defect ΔM will be: (1) √(2cΔM/M) (2) ΔMc²/3 (3) c√(2ΔM/M) (4) c√(cΔM/M) |
(3) c√(2ΔM/M) | The energy released (ΔMc²) is converted into kinetic energy of the daughter nuclei. Using K.E. = ½mv² and ΔMc² = K.E., solve for v to get v = c√(2ΔM/M). |
| 36: Choose the correct statement for processes A & B shown in the figure: (1) PVⁿ = k for process B and PV = k for process A (2) PV = k for process B and T = k for process A (3) Pⁿ⁻¹ = k for process B and Tⁿ = k for process A (4) TⁿPⁿ⁻¹ = k for process A and PV = k for process B |
(1) PVⁿ = k for process B and PV = k for process A | Process A is isothermal (PV = k) and process B is adiabatic (PVⁿ = k). The steeper slope of B suggests it is adiabatic, while A indicates constant temperature. Hence, the correct statements are (1) and (3). |
| 37: An electron revolving in the nth Bohr orbit has a magnetic moment µ. If µₙ is the value of µ, the value of x is: (1) 2 (2) 1 (3) 3 (4) 0 |
(2) 1 | The magnetic moment µₙ ∝ n². For n = 1, µₙ/µ₁ = 1². Thus, the ratio of magnetic moments indicates that x = 1. |
| 38: An alternating voltage V(t) = 220 sin 100t volt is applied to a purely resistive load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is: (1) 5 ms (2) 3.3 ms (3) 7.2 ms (4) 2.2 ms |
(2) 3.3 ms | Using the sinusoidal function, the time difference between half peak and peak current is calculated as Δt = π/300 ≈ 3.3 ms. |
| 39: A block of mass 1 kg is placed on a surface with a vertical cross-section given by y = x². If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is: (1) 1/4 m (2) 1/2 m (3) 1/8 m (4) 1/16 m |
(1) 1/4 m | Calculate using frictional force and component of weight. At equilibrium, the maximum height is 1/4 m. |
| 40: If the total energy transferred to a surface in time t is 6.48 × 10⁵ J, then the magnitude of the total momentum delivered to this surface for complete absorption is: (1) 2.46 × 10⁻³ kg·m/s (2) 2.16 × 10⁻³ kg·m/s (3) 1.58 × 10⁻³ kg·m/s (4) 4.32 × 10⁻³ kg·m/s |
(2) 2.16 × 10⁻³ kg·m/s | Momentum is given by p = E/c. Substituting E = 6.48 × 10⁵ J and c = 3 × 10⁸ m/s, the momentum is 2.16 × 10⁻³ kg·m/s. |
| 41: A beam of unpolarized light of intensity I₀ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of emergent light is: (1) I₀/4 (2) I₀ (3) I₀/2 (4) I₀/8 |
(1) I₀/4 | After passing through the first polaroid, intensity is halved: I = I₀/2. Passing through the second polaroid at 45° reduces intensity to I₀/4 using Malus's Law: I = I₀/2 × cos²(45°). |
| 42: The escape velocity of a body from Earth is 11.2 km/s. If the radius of a planet is one-third the radius of Earth and its mass is one-sixth that of Earth, the escape velocity from the planet is: (1) 11.2 km/s (2) 8.4 km/s (3) 4.2 km/s (4) 7.9 km/s |
(4) 7.9 km/s | The escape velocity is given by: ve = √(2GM/R). For the planet: Mplanet = Mearth/6 and Rplanet = Rearth/3. Thus: ve,planet = ve,earth × √( (Mearth/(2Rearth)) / (Mearth/Rearth) ) = 11.2 × √(1/2) ≈ 7.9 km/s. |
| 43: A particle of charge –q and mass m moves in a circle of radius r around an infinitely long line of charge having linear charge density +λ. The time period T is given by: (1) T² = (4πmr³)/(2kq) (2) T = 2πr √(m/(2kq)) (3) T = (1/(2πr))√(m/(2kq)) (4) T = 2kq/m |
(2) T = 2πr √(m/(2kq)) | The electric field due to an infinitely long line charge is: E = λ/(2πϵ0r). (m v²)/r = qE. Also, v = (2πr)/T. T = 2πr √(m/(2kq)). |
| 44: If mass is written as m = k cp G−1/2 h1/2, then the value of p will be: (1) 1/2 (2) 1/3 (3) 2 (4) −1/3 |
(2) 1/3 | Comparing the dimensions of m = k cp G−1/2 h1/2: mass (M), speed of light (c) as [LT−1] raised to p, gravitational constant (G) as [M−1L3T−2] raised to −1/2, and Planck's constant (h) as [ML2T−1] raised to 1/2. Solving for p, we find p = 1/3. |
| 45: In the given circuit, the voltage across load resistance RL is: (1) 8.75 V (2) 9.00 V (3) 13.50 V (4) 14.00 V |
(1) 8.75 V | Using the voltage division rule, the voltage across the load resistance RL is calculated as: Total resistance in the circuit: Rtotal = RD1 + RD2 + RL = 1.5 kΩ + 2.5 kΩ + 2.5 kΩ = 6.5 kΩ. Voltage across RL: VRL = (RL / Rtotal) × Total voltage. Substituting values: VRL = (2.5 kΩ / 6.5 kΩ) × 15 V ≈ 8.75 V. |
| 46: If three moles of monoatomic gas (γ = 5/3) is mixed with two moles of diatomic gas (γ = 7/5), the value of the adiabatic exponent γ for the mixture is: (1) 1.75 (2) 1.40 (3) 1.52 (4) 1.35 |
(3) 1.52 | The adiabatic exponent γ is calculated using the mole fraction of each gas and their respective γ values. Substituting values, we find γ ≈ 1.52. |
| 47: Three blocks A, B, and C are pulled on a horizontal smooth surface by a force of 80 N. The tensions T1 and T2 in the string are respectively: (1) 40 N, 64 N (2) 60 N, 80 N (3) 88 N, 96 N (4) 80 N, 100 N |
(1) 40 N, 64 N | Using Newton's second law, calculate the total acceleration and forces acting on each block. Substituting values, we find T1 = 40 N and T2 = 64 N. |
| 48: When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually in parallel across the same supply, the energy dissipation rate will become: (1) W/4 (2) W/2 (3) 2W (4) 4W |
(4) 4W | Cutting the wire halves the resistance of each part. When connected in parallel, the equivalent resistance is R/4. The dissipation rate becomes P' = V² / (R/4) = 4W. |
| 49: Match List I with List II: List I: A. Gauss’s Law B. Faraday’s Law C. Lenz’s Law D. Ampere’s Law List II: I. ∮ E · dA = Q / ε0 II. ∮ B · dl = μ0I III. Induced emf opposes change IV. ∮ E · dl = −dΦ/dt (1) A - I, B - IV, C - III, D - II (2) A - IV, B - I, C - III, D - II (3) A - II, B - III, C - IV, D - I (4) A - I, B - II, C - IV, D - III |
(1) A - I, B - IV, C - III, D - II | Matching the physical laws with their mathematical representations and principles gives the correct pairing: A - I, B - IV, C - III, D - II. |
| 50: Projectiles A and B are thrown at angles of 45° and 60° with the vertical respectively from the top of a 400 m high tower. If their ranges and times of flight are the same, the ratio of their speeds of projection vA : vB is: (1) 1 : √3 (2) √2 : 1 (3) 1 : 2 (4) 1 : √2 |
(1) 1 : √3 | For the same range and time of flight, the horizontal and vertical components of velocities must satisfy the given conditions. Using trigonometric relationships for angles 45° and 60°, the ratio vA : vB is derived as 1 : √3. |
| 51: A power transmission line feeds input power at 2.3 kV to a step-down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5 A, and its efficiency is 90%. The winding of the transformer is made of copper. The output current of the transformer is: | 45 A | Input power = 2300 V × 5 A = 11500 W. Efficiency = 90%, so output power = 0.9 × 11500 = 10350 W. Using P = VI, output current = 10350 / 230 = 45 A. |
| 52: A big drop is formed by coalescing 1000 small identical drops of water. If E₁ is the total surface energy of 1000 small drops and E₂ is the surface energy of the single big drop, then the ratio E₁:E₂ is x:1 where x equals: | 10 | Surface energy is proportional to surface area. For 1000 small drops, total surface area = 1000 × 4πr². For a single large drop, area = 4πR² where R³ = 1000r³. Ratio = 1000:100 = 10:1. |
| 53: Two discs with moments of inertia I₁ = 4 kg·m² and I₂ = 2 kg·m² about their central axes, rotating with angular speeds 10 rad/s and 4 rad/s, respectively, are brought into contact face-to-face. The loss in kinetic energy of the system is: | 24 J | Initial KE = (1/2)I₁ω₁² + (1/2)I₂ω₂² = 200 J + 16 J = 216 J. Final KE after combining = (1/2)(I₁ + I₂)ω² = 192 J. Loss = 216 − 192 = 24 J. |
| 54: In an experiment to measure the focal length f of a convex lens, the magnitude of object distance x and image distance y are measured with reference to the focal point of the lens. The y-x plot is shown in the figure. The focal length of the lens is: | 20 cm | From the lens equation (1/f = 1/x + 1/y) and the slope of the y-x plot, the focal length is determined as 20 cm. |
| 55: A vector has a magnitude equal to that of A = −3î + 4ĵ and is parallel to B = 4î + 3ĵ. The x and y components of this vector in the first quadrant are x and y, respectively. The value of x is: | 4 | Magnitude of A = 5. Unit vector of B = (4î + 3ĵ)/5. New vector = 5 × Unit vector of B = 4î + 3ĵ. Thus, x = 4. |
| 56: The current of 5 A flows in a square loop of sides 1 m placed in air. The magnetic field at the center of the loop is X√2 × 10⁻⁷ T. The value of X is: | 40 | Using the formula for the magnetic field due to a current-carrying square loop, B = (µ₀I/4πr) × √2, and substituting µ₀ = 4π × 10⁻⁷, I = 5 A, and r = 1/√2 m, the calculated value of X is 40. |
| 57: Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 37° with each other. When suspended in a liquid of density 0.7 g/cm³, the angle remains the same. If the density of the material of the spheres is 1.4 g/cm³, the dielectric constant of the liquid is: | 3 | Using equilibrium of forces, the dielectric constant K is calculated as K = ρₛ / ρₗ, where ρₛ = 1.4 g/cm³ and ρₗ = 0.7 g/cm³. Thus, K = 3. |
| 58: A wire is cut into two halves, and these halves are connected in parallel. The resistance of the wire before cutting was R. The equivalent resistance of the combination is: | R/4 | After cutting the wire, the resistance of each half becomes R/2. When connected in parallel, the equivalent resistance Rₑq = (R/2) || (R/2) = R/4. |
| 59: A point source is emitting sound waves of intensity 16 × 10⁻⁸ W/m² at the origin. The difference in intensity (magnitude only) at two points located at distances of 2 m and 4 m from the origin, respectively, will be: | 3 × 10⁻⁸ W/m² | Intensity at distance r is given by I = P/(4πr²). Calculate I₁ for r = 2 m and I₂ for r = 4 m, then find the difference: I₁ - I₂ = 3 × 10⁻⁸ W/m². |
| 60: Two resistances of 100Ω and 200Ω are connected in series with a battery of 4V and negligible internal resistance. A voltmeter is used to measure voltage across the 100Ω resistance, which gives a reading of 1V. The resistance of the voltmeter must be: | 200 Ω | Using the concept of parallel resistance between the voltmeter and 100Ω, calculate the effective resistance to ensure the voltmeter shows 1V. The resistance is found to be 200Ω. |
Also Check:
| JEE Main 2024 Paper Analysis | JEE Main 2024 Answer Key |
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JEE Main 2024 Jan 30 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 | To be updated |
| Sri Chaitanya | To be updated |
| FIIT JEE | To be updated |
JEE Main 2024 Jan 30 Shift 2 Physics Paper Analysis
JEE Main 2024 Jan 30 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
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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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