JEE Main 2024 question paper pdf with solutions- Download Jan 29 Shift 2 Physics Question Paper pdf

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JEE Main 2024 Jan 29 Shift 2 Physics question paper with solutions and answers pdf is available here. NTA conducted JEE Main 2024 Jan 29 Shift 2 exam from 3 PM to 6 PM. The Physics question paper for JEE Main 2024 Jan 29 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 29 Shift 2 exam is available for download using the link below.

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JEE Main 29 Jan Shift 2 2024 Physics Questions with Solutions

Question Answer Detailed Solution

31: Two sources of light emit with a power of 200 W. The ratio of the number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be:

(1) 1:5
(2) 1:3
(3) 5:3
(4) 3:5

(4) 3:5 Using the formula E = hc/λ, the energy of photons at 300 nm and 500 nm is calculated. The ratio of the number of photons emitted is inversely proportional to the wavelength. Hence, the ratio is 3:5.
32: The truth table for the given circuit is:

(1) 1
(2) 2
(3) 3
(4) 4
(2) 2 The output Y = (A·B + A·B). Simplifying the circuit and constructing the truth table, the output matches option 2.
33: A physical quantity Q is found to depend on quantities a, b, c by the relation Q = a⁴b³/c². The percentage error in a, b, c are 3%, 4%, and 5% respectively. Then, the percentage error in Q is:

(1) 66%
(2) 43%
(3) 34%
(4) 14%
(3) 34% Using the formula for percentage error: ΔQ/Q = 4(Δa/a) + 3(Δb/b) + 2(Δc/c), the total error is calculated as 34%.
34: In an a.c. circuit, voltage and current are given by V = 100 sin(100t) V and I = 100 sin(100t + π/3) mA, respectively. The average power dissipated in one cycle is:

(1) 5 W
(2) 10 W
(3) 2.5 W
(4) 25 W
(3) 2.5 W The formula for average power is Pavg = Vrms·Irms·cos(ϕ). Substituting values: Vrms = 100/√2, Irms = 0.1/√2, and cos(ϕ) = 1/2. The calculated power is 2.5 W.
35: The temperature of a gas having 2.0×10²⁵ molecules per cubic meter at 1.38 atm (Given, k = 1.38×10⁻²³ JK⁻¹) is:

(1) 500 K
(2) 200 K
(3) 100 K
(4) 300 K
(1) 500 K Using the ideal gas law PV = NkT, rearranged to find T, and substituting the given values, the temperature is calculated as 500 K.

36: A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point, is:

(1) 97 N
(2) 9.8 N
(3) 8.82 N
(4) 17.8 N

(2) 9.8 N Using the formula for tension at the lowest point: T = mg + mrω², where ω = 2πn/60, and substituting the given values:
T = 0.9 × 9.8 + 0.9 × 1 × (π/3)², the calculated tension is 9.8 N.

37: The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

(1) 6√5 m/s
(2) 5√6 m/s
(3) 5√5 m/s
(4) 2√5 m/s

(1) 6√5 m/s Using energy conservation, with 10% energy dissipated: v² = 2 × 0.9 × g × ℓ, where ℓ = 10 m. Substituting values, v = √180 = 6√5 m/s.

38: If the distance between an object and its two-times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be:

(1) 15 cm
(2) −12 cm
(3) −10 cm
(4) 10/3 cm

(3) −10 cm Using magnification m = −v/u and mirror equation: 1/f = 1/v − 1/u, with v = −2u and |u| + |v| = 15 cm. Solving gives f = −10 cm.

39: Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter, they enter normally in a region of uniform magnetic field and describe circular paths of radii R₁ and R₂, respectively. The mass ratio of X and Y is:

(1) (R₂/R₁)²
(2) (R₁/R₂)²
(3) R₁/R₂
(4) R₂/R₁

(2) (R₁/R₂)² Using the relation R ∝ √m, the mass ratio is proportional to the square of the ratio of the radii: m₁/m₂ = (R₁/R₂)².

40: In Young’s double slit experiment, light from two identical sources is superimposing on a screen. The path difference between the two lights reaching a point on the screen is 7λ/4. The ratio of intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

(1) 1/2
(2) 3/4
(3) 1/3
(4) 1/4

(1) 1/2 The phase difference corresponding to the path difference is ϕ = 2π × (7λ/4λ) = 7π/2. Using I/Imax = cos²(ϕ/2), the calculated intensity ratio is 1/2.

41: A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be:

(1) 8πR²T
(2) 3πR²T
(3) 1/8πR²T
(4) 4πR²T

(1) 8πR²T The initial surface area is A₁ = 4πR². After division, the radius of each small drop is R/3, and the total surface area becomes A₂ = 27 × 4π(R/3)² = 12πR². The work done is TΔA = T(A₂ − A₁) = T(12πR² − 4πR²) = 8πR²T.

42: A bob of mass ‘m’ is suspended by a light string of length ‘L’. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes a half-circle reaching the topmost position B. The ratio of kinetic energies (K.E.)A : (K.E.)B is:

(1) 3:2
(2) 5:1
(3) 2:5
(4) 1:5

(2) 5:1 Using energy conservation: K.E.B = K.E.A − 2mgL, and K.E.A = 1/2m(5gL), K.E.B = 1/2m(gL). The ratio is K.E.A : K.E.B = 5:1.

43: A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force are both reduced to half of their original values, keeping the original length constant, the increase in length will become:

(1) 3 times
(2) 3/2 times
(3) 4 times
(4) 2 times

(4) 2 times Using the formula for elongation: Δl = Fℓ / AY, where A = πr². Halving both F and r increases the elongation by a factor of 2.

44: A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from the Sun is reduced to one-fourth of the original distance, how many days will it take to complete one revolution?

(1) 25
(2) 50
(3) 100
(4) 20

(1) 25 Using Kepler's third law: T² ∝ r³, reducing the distance to r/4 gives T₂ = T₁/8. Substituting T₁ = 200 days, the new period is 25 days.

45: A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time), E = 9.6 j V/m. The value of the magnetic field at this point is:

(1) 3.2×10⁻⁸ kT
(2) 3.2×10⁻⁸ iT
(3) 9.6 jT
(4) 9.6×10⁻⁸ kT

(1) 3.2×10⁻⁸ kT Using the relationship E/B = c, where c = 3×10⁸ m/s, the magnetic field is calculated as B = E/c = 9.6/(3×10⁸) = 3.2×10⁻⁸ T. The direction of B is along the k-axis.

46: In the given circuit, the current in resistance R₃ is:

(1) 1 A
(2) 1.5 A
(3) 2 A
(4) 2.5 A

(1) 1 A The equivalent resistance of R₂ and R₃ in parallel is Req = 2 Ω. The total resistance of the circuit is Rtotal = 5 Ω. The total current is 2 A. Current through R₃ is 1 A.

47: A particle is moving in a straight line. The variation of position x as a function of time t is given as x = (t³ − 6t² + 20t + 15) m. The velocity of the body when its acceleration becomes zero is:

(1) 4 m/s
(2) 8 m/s
(3) 10 m/s
(4) 6 m/s

(2) 8 m/s The velocity is v = dx/dt = 3t² − 12t + 20, and the acceleration is a = dv/dt = 6t − 12. Setting a = 0, we get t = 2 s. Substituting t = 2 in the velocity equation gives v = 8 m/s.

48: N moles of a polyatomic gas (f = 6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is:

(1) 6
(2) 3
(3) 4
(4) 2

(3) 4 Using the formula for effective degrees of freedom: feq = (n₁f₁ + n₂f₂) / (n₁ + n₂), with feq = 5 for a diatomic gas, substituting f₁ = 6, f₂ = 3, and n₂ = 2, we solve for N = 4.

49: Given below are two statements:
Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford’s model.
Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford’s model.


(1) Both Statement I and Statement II are false
(2) Statement I is false but Statement II is true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true

(3) Statement I is true but Statement II is false Statement I is correct as it describes Rutherford’s atomic model. Statement II is incorrect as it describes Thomson’s model, not Rutherford’s. Hence, the correct option is (3).

50: An electric field is given by E = (6i + 5j + 3k) N/C. The electric flux through a surface area 30i m² lying in the YZ-plane (in SI units) is:

(1) 90
(2) 150
(3) 180
(4) 60

(3) 180 Using the formula for electric flux: Φ = E · A, where A = 30i. Substituting E = 6i + 5j + 3k, the flux is Φ = 6 × 30 = 180 Nm²/C.

51: Two metallic wires P and Q have the same volume and are made up of the same material. If their areas of cross-section are in the ratio 4:1 and force F₁ is applied to P, an extension of Δℓ is produced. The force required to produce the same extension in Q is F₂. The value of F₁/F₂ is:

16 Using the formula for Young's modulus: Δℓ ∝ F / A, the force ratio is F₁/F₂ = (A₂/A₁). With the cross-sectional areas in a 4:1 ratio, F₁/F₂ = 16.

52: A horizontal straight wire 5 m long extending from east to west falls freely at a right angle to the horizontal component of Earth’s magnetic field 0.60 × 10⁻⁴ Wb/m². The instantaneous value of emf induced in the wire when its velocity is 10 m/s is:

3 × 10⁻³ V The induced emf is calculated using E = Bℓv. Substituting the values: E = 0.60 × 10⁻⁴ × 5 × 10, the emf is 3 × 10⁻³ V.

53: Hydrogen atom is bombarded with electrons accelerated through a potential difference V, which causes excitation of hydrogen atoms. If the experiment is performed at T = 0 K, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be α/10 V, where α is:

121 To observe Balmer series lines, electrons must be excited from n = 1 to n = 3. The required energy is ΔE = 13.6 × (1 − 1/9) = 12.1 eV. Thus, α = 121.

54: A charge of 4.0 µC is moving with a velocity of 4.0 × 10⁶ m/s along the positive y-axis under a magnetic field B of strength 2k̂ T. The force acting on the charge is xî N. The value of x is:

32 Using the formula F = q(v × B), where v × B = î, the force is F = 4 × 10⁻⁶ × 4 × 10⁶ × 2 = 32 N.

55: A simple harmonic oscillator has an amplitude A and a time period 6π seconds. Assuming the oscillation starts from its mean position, the time required by it to travel from x = A to x = √3/2 A will be π/x seconds, where x is:

2 Using the SHM equation: cosϕ = √3/2, ϕ = π/6. The time taken is t = Tϕ/2π, substituting values: t = π/2. Hence, x = 2.

56: In the given figure, the charge stored in a 6µF capacitor, when points A and B are joined by a connecting wire, is µC:

36 µC When A and B are connected, the 3µF capacitors are in parallel, forming a 6µF equivalent capacitor. The voltage across this capacitor is 6V. Using Q = CV, the charge stored is 36 µC.

57: In a single slit diffraction pattern, a light of wavelength 6000 Å is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from the slits. The width of the slit is ×10⁻⁴ m:

2 Using the diffraction formula Δy = 2λD/b, substituting the values Δy = 3 mm, λ = 6000 × 10⁻¹⁰ m, D = 0.5 m, we solve for b = 2 × 10⁻⁴ m.

58: In the given circuit, the current flowing through the resistance 20Ω is 0.3A, while the ammeter reads 0.9A. The value of R₁ is Ω:

30 Using Ohm's law, the voltage across R₁ is V = IR = 0.3 × 20 = 6V. From the circuit, the current through R₁ is 0.6A. Substituting into R = V/I, we get R₁ = 30 Ω.

59: A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution will be 1/α [1 − e⁻²π] seconds, where α is:

8 Equating the normal and tangential accelerations: v²/r = dv/dt. Solving the differential equation and using the given conditions, the value of α is found to be 8.

60: A body of mass 5 kg moving with a uniform speed 3√2 m/s in the X-Y plane along the line y = x + 4. The angular momentum of the particle about the origin will be kg·m²/s:

60 The perpendicular distance from the origin to the line is d = 2√2 m. The angular momentum is L = mvd. Substituting values: L = 5 × 3√2 × 2√2 = 60 kg·m²/s.


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JEE Main 2024 Jan 29 Shift 2 Physics Paper Analysis

JEE Main 2024 Jan 29 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

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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

JEE Main Previous Year Question Paper

JEE Main Questions

  • 1.

    A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is: 
     


      • 2.

        The internal energy of air in $ 4 \, \text{m} \times 4 \, \text{m} \times 3 \, \text{m} $ sized room at 1 atmospheric pressure will be $ \times 10^6 \, \text{J} $. (Consider air as a diatomic molecule)


          • 3.
            An AC current is represented as: $ i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right) \text{ Amp} $ The RMS value of the current is:

              • 50 Amp
              • 100 Amp
              • 10 Amp
              • \( 5\sqrt{2} \) Amp

            • 4.
              A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be \( \frac{x}{100} \), where \( x \) is:


                • 5.

                  Match the LIST-I with LIST-II

                  LIST-ILIST-II
                  A.Boltzmann constantI.\( \text{ML}^2\text{T}^{-1} \)
                  B.Coefficient of viscosityII.\( \text{MLT}^{-3}\text{K}^{-1} \)
                  C.Planck's constantIII.\( \text{ML}^2\text{T}^{-2}\text{K}^{-1} \)
                  D.Thermal conductivityIV.\( \text{ML}^{-1}\text{T}^{-1} \)

                  Choose the correct answer from the options given below :

                    • A-III, B-IV, C-I, D-II
                    • A-II, B-III, C-IV, D-I
                    • A-III, B-II, C-I, D-IV
                    • A-III, B-IV, C-II, D-I

                  • 6.

                    In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:

                      • \( \frac{Kq_0^2}{a} (4\sqrt{2} - 2) \)
                      • \( \frac{Kq_0^2}{a} (4 - \sqrt{2}) \)
                      • \( \frac{Kq_0^2}{a} (3\sqrt{2} - 2) \)
                      • \( \frac{Kq_0^2}{a} (3 - \sqrt{2}) \)

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