JEE Main 2024 Jan 29 Shift 1 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 29 Shift 1 exam from 9 AM to 12 PM. The Physics question paper for JEE Main 2024 Jan 29 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 29 Shift 1 exam is available for download using the link below.

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JEE Main 29 Jan Shift 1 2024 Physics Questions with Solution

Question Correct Answer Solution
31. In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of IZ?
(1) 3.3 mA
(2) 5.5 mA
(3) 10 mA
(4) 7 mA		 (2) 5.5 mA Given VZ = 3 V.
Let the potential at point B = 0 V. - 
I = (10 − 3) / 1000 = 7 / 1000 = 7 mA
I1 = 3 / 2000 = 1.5 mA
IZ = I − I1 = 7 mA − 1.5 mA = 5.5 mA
So, the correct option is : 5.5 mA
32. The electric current through a wire varies with time as I = I0 + βt, where I0 = 20 A and β = 3 A/s. The amount of electric charge that crosses through a section of the wire in 20 s is:
(1) 80 C
(2) 1000 C
(3) 800 C
(4) 1600 C		 (2) 1000 C I = I0 + βt = 20 + 3t
Using I = dq/dt, we write dq = (20 + 3t) dt.
(q) from t = 0 to t = 20:
q = ∫020 (20 + 3t) dt
q = ∫020 20 dt + ∫020 3t dt
q = [20t]020 + [(3t²)/2]020
q = (20 × 20) + (3 × 20²)/2 = 400 + 600 = 1000 C.
33. Given below are two statements:
Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water.
Statement II: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water.		 (3) Statement I is true but Statement II is false Statement I is correct because the height of capillary rise decreases in hot water due to reduced surface tension. Statement II is false because the height is larger in cold water due to higher surface tension.
34. A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is:
(1) −15 cm
(2) 45 cm
(3) −45 cm
(4) 15 cm		 (1) −15 cm Using the magnification formula for a convex mirror and the given magnification, the object distance is found to be −15 cm.
35. Two charges of 5Q and −2Q are situated at the points (3a, 0) and (−5a, 0) respectively. The electric flux through a sphere of radius 4a having its center at the origin is:
(1) 2Q/ε0
(2) 5Q/ε0
(3) 7Q/ε0
(4) 3Q/ε0		 (2) 5Q/ε0 The flux through the sphere is determined by Gauss's Law. Only the 5Q charge is inside the sphere, so the electric flux is 5Q/ε0.
36. A body starts moving from rest with constant acceleration and covers displacement S1 in the first (p−1) seconds and S2 in the first p seconds. The displacement S1 + S2 will be made in time:
(1) (2p + 1) s
(2) √(2p² − 2p + 1) s
(3) (2p − 1) s
(4) (2p² − 2p + 1) s		 (2) √(2p² − 2p + 1) s By using the equations for displacement in uniformly accelerated motion, the time taken to cover the total displacement is √(2p² − 2p + 1) s.
37. The potential energy function (in J) of a particle in a region of space is given as U = (2x² + 3y³ + 2z). Here x, y, and z are in meters. The magnitude of the x-component of force (in N) acting on the particle at point P (1, 2, 3) m is:
(1) 2
(2) 6
(3) 4
(4) 8		 (3) 4 The x-component of force is calculated as Fx = −∂U/∂x. Differentiating the potential function U with respect to x and evaluating at x = 1 gives Fx = −4 N.
38. The resistance R = V/I where V = (200± 5)V and I = (20± 0.2)A. The percentage error in the measurement of R is:
(1) 3.5%
(2) 7%
(3) 3%
(4) 5.5%		 (1) 3.5% The percentage error in resistance is calculated by summing the relative errors of voltage and current, resulting in a total of 3.5%.
39. A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the coefficient of friction between the surfaces is 0.4, then the work done against friction (in J) is:
(1) 4200 J
(2) 3900 J
(3) 4000 J
(4) 4500 J		 (3) 4000 J The work done against friction is calculated as the product of the frictional force (f = µmg) and the displacement (s = 10 m), yielding 4000 J.
40. Match List I with List II:
List I
A. ∮ B· dl = µ₀ic + µ₀ε₀ dΦE/dt
B. ∮ E· dl = −dΦB/dt
C. ∮ E· dA = Q/ε₀
D. ∮ B· dA = 0
List II
I. Gauss' law for electricity
II. Gauss' law for magnetism
III. Faraday's law
IV. Ampere–Maxwell law		 (3) A-IV, B-III, C-I, D-II A corresponds to Ampere–Maxwell law, B corresponds to Faraday's law, C corresponds to Gauss' law for electricity, and D corresponds to Gauss' law for magnetism.
41. If the radius of curvature of the path of two particles of the same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:
(1) √3 : 2
(2) 1 : √3
(3) √3 : 1
(4) 2 : √3		 (1) √3 : 2 Using the centripetal force formula (F = mv²/r) and keeping the force constant, the ratio of velocities is √3 : 2, derived from the ratio of their radii of curvature.
42. A galvanometer having coil resistance 10 Ω shows a full scale deflection for a current of 3 mA. For it to measure a current of 8 A, the value of the shunt should be:
(1) 3× 10⁻³ Ω
(2) 4.85× 10⁻³ Ω
(3) 3.75× 10⁻³ Ω
(4) 2.75× 10⁻³ Ω		 (3) 3.75× 10⁻³ Ω The shunt resistance is calculated using the formula S = (Ig × G) / (I − Ig), where Ig is the full scale deflection current and I is the desired current, giving a value of 3.75× 10⁻³ Ω.
43. The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
(1) 1/1
(2) 1/8
(3) 8 : 1
(4) 1/4		 (2) 1/8 By equating the de-Broglie wavelengths of the electron and the photon, and using the relationship between kinetic energy and velocity, the ratio of the electron's kinetic energy to the photon’s is 1/8.
44. The deflection in a moving coil galvanometer falls from 25 divisions to 5 divisions when a shunt of 24Ω is applied. The resistance of the galvanometer coil will be:
(1) 12 Ω
(2) 96 Ω
(3) 48 Ω
(4) 100 Ω		 (2) 96 Ω Using the condition of equal voltage across the galvanometer and the shunt, the resistance of the galvanometer is calculated as 96 Ω.
45. A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:
(1) −16 cm
(2) −160 cm
(3) +160 cm
(4) +16 cm		 (2) −160 cm Using the lens formula for different media, the focal length of the lens immersed in a liquid of higher refractive index is calculated as −160 cm.
46. A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. Its volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be:
(1) 33,800 J
(2) 2,200 J
(3) 600 J
(4) 1,200 J		 (BONUS) 0 J The work done is calculated using the area under the curve in the PV diagram. In this case, the total work done is zero because the area under the curve for the entire process cancels out.
47. At what distance above and below the surface of the earth a body will have the same weight (take radius of earth as R)?
(1) √5R−R
(2) √5R−R/2
(3) R/2
(4) R/2(√5−1)		 (4) R/2(√5−1) The distance is calculated by equating the gravitational accelerations above and below the Earth's surface and solving for h, resulting in h = R/2(√5−1).
48. A capacitor of capacitance 100 µF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be:
(1) 3.2 A
(2) 1.5 A
(3) 2.0 A
(4) 1.2 A		 (2) 1.5 A Using energy conservation in the LC circuit, the maximum current is calculated as Imax = V√(C/L) = 1.5 A.
49. The explosive in a Hydrogen bomb is a mixture of 1H2, 1H3, and 3Li6 in some condensed form. The chain reaction is given by:
3Li6 + 0 n1 → 2 He4 + 1 H3
1H2 + 1H3 → 2 He4 + 0 n1
During the explosion, the energy released is approximately:
(1) 28.12 MeV
(2) 12.64 MeV
(3) 16.48 MeV
(4) 22.22 MeV		 (4) 22.22 MeV The energy released during the explosion is calculated using mass-energy equivalence, and the energy is approximately 22.22 MeV.
50. Two vessels A and B are of the same size and are at the same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. PA and PB are the pressures of the gases in A and B respectively, then PA/PB is:
(1) 16
(2) 8
(3) 4
(4) 32		 (1) 16 Using the ideal gas equation and calculating the moles of hydrogen and oxygen, the ratio of their pressures is found to be 16.
51: When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is x/5 m/s. Where x = . (Use: mass of hydrogen atom = 1.6 × 10−27 kg) (17) The energy difference for the transition from n = 2 to n = 1 in a hydrogen atom is 10.2 eV. Using conservation of momentum and energy, the recoil speed of the atom is calculated as 17/5 m/s.
52: A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which the ball just hits the step 5 of the stairway will be √x m/s, where x = . (Use g = 10 m/s²) (2) The horizontal range to hit step 5 is 0.4 m. Using the equation for horizontal motion, the required velocity is calculated to be √2 m/s. Therefore, x = 2.
53: A square loop of side 10 cm and resistance 0.7 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in the northeast direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, the magnitude of induced emf is √x × 10−3 V. The value of x is . (2) Using the formula for induced emf, the emf induced in the loop is 2 × 10−3 V. Therefore, x = 2.
54: A cylinder is rolling down on an inclined plane of inclination 60°. Its acceleration during rolling down will be x√3 m/s², where x = . (Use g = 10 m/s²) (10) The acceleration of the rolling cylinder is calculated using the formula a = g sin θ / (1 + Icm/MR²). Substituting the values, we find x = 10.
55: The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5 × 10−5 Tm. The magnetic moment of the dipole is Am². Given: µ₀ / 4π = 10−7 Tm/A (6) Using the formula for magnetic potential on the axis of a dipole, the magnetic moment M is calculated as 6 Am².
56: In a double slit experiment shown in the figure, when light of wavelength 400 nm is used, a dark fringe is observed at P. If D = 0.2 m, the minimum distance between the slits S1 and S2 is mm. (0.20 mm) Using the path difference condition for minima, the distance between the slits is calculated to be 0.20 mm.
57: A 16 Ω wire is bent to form a square loop. A 9 V battery with internal resistance 1 Ω is connected across one of its sides. If a 4 µF capacitor is connected across one of its diagonals, the energy stored by the capacitor will be x² µJ, where x = . (81) The total energy in the circuit is used to calculate the current, and the voltage across the capacitor is used to find the energy stored. The final answer is 81 µJ.
58: When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x/8, where x = . (9) Using the energy relations for simple harmonic motion, the ratio of total energy to kinetic energy when displacement is one-third of the amplitude is 9/8, so x = 9.
59: An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density +σ. The electron at t = 0 is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of σ if the electron strikes S at t = 1 s is α[mϵ₀/e] C/m². The value of α is . (8) By using kinematic equations and the electric field due to the charged plane sheet, the value of α is found to be 8.
60: In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 m/s and 65 m/s respectively. If the wing area is 2 m², the lift of the wing is N. Given: density of air = 1.2 kg/m³ (810 N) Using Bernoulli's equation for the difference in pressures on the upper and lower surfaces of the wing, the lift force is calculated as 810 N.

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JEE Main 2024 Jan 29 Shift 1 Physics Paper Analysis

JEE Main 2024 Jan 29 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 Paper Analysis Jan 29 Shift 1 (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.

Exam Date and Shift Question Paper PDF
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