
JEE Main 2024 Jan 27 Shift 2 Physics Question Paper with Solution pdf is available for download here. Students found Physics easy and Mathematics hard. Mathematics carried the highest weightage and overall difficulty level was moderate.
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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:
Using dimensional analysis on a/b², we find it has the same dimensions as the pressure P.
The total kinetic energy of 1 mole of oxygen at 27°C is:
For a diatomic gas at 300 K, E = (5/2) n R T. Substituting n = 1 and T = 300 K gives 6232.5 J.
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:
The secondary voltage V₂ = (230/10) = 23 V. Power P = V₂² / R = 23² / 46 = 529/46 = 11.5 W.
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:
From the adiabatic relation P ∝ T^(γ/(γ−1)) and given P ∝ T³, we get γ/(γ−1) = 3, leading to γ = 3/2.
The threshold frequency of a metal with work function 6.63 eV is:
Using E = h ν₀ and converting 6.63 eV to joules gives ν₀ ≈ 1.6 × 10¹⁵ Hz.
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:
Current is proportional to the deflection angle. 60° is π/3 radians. For deflection π/10, current is (200 × (π/10)) / (π/3) = 60 µA.
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:
Mass defect = 13.003354 − (12.000000 + 1.008665) = −(some small value). Converting mass defect to energy via 931.5 MeV/u yields 4.95 MeV.
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:
Equating accelerations (centripetal and tangential) using energy conservation, the angle θ is found to be 2tan⁻¹(1/2).
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:
Applying Kirchhoff’s Voltage Law around the loop, we see that V₁ + V₂ = V₃.
The total kinetic energy of 1 mole of oxygen at 27°C is:
For a diatomic gas at 300 K, using E = (5/2) n R T, we get 6232.5 J.
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.
Angular speed ω is inversely proportional to the orbital period. The moon’s orbital period is shorter, so ω(moon) > ω(earth).
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:
Secondary voltage V₂ = 230/10 = 23 V, so power P = 23² / 46 = 11.5 W.
During an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The ratio of Cp/Cv is:
From P ∝ T^(γ/(γ−1)) = T³, we get γ/(γ−1) = 3, leading to γ = 3/2.
The threshold frequency of a metal with work function 6.63 eV is:
E = 6.63 eV → convert to joules. Then E = h ν₀. Solving gives ν₀ ≈ 1.6 × 10¹⁵ Hz.
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:
Current is proportional to deflection angle. 60° is π/3, so for π/10, current = 200 × (π/10) / (π/3) = 60 µA.
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:
Using the mass defect and the formula E = Δm × 931.5 MeV/u, the energy required is 4.95 MeV.
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:
By applying energy conservation and equating accelerations at the extreme and lowest positions, the angle θ is found to be 2tan⁻¹(1/2).
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.
Applying Kirchhoff’s Voltage Law around the loop shows that V₁ + V₂ = V₃.
The total kinetic energy of 1 mole of oxygen at 27°C is:
For a diatomic gas at 300 K, using E = (5/2) × n × R × T gives 6232.5 J for 1 mole of O₂.
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:
Secondary voltage V₂ = 230 / 10 = 23 V. Then power P = (23²) / 46 = 529 / 46 = 11.5 W.
The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1 = 2 m and R2 = 4 m carrying current I = 4 A as per figure given below is α×10⁻⁷ T. The value of α is:
The magnetic field at the center of a semicircular wire of radius R carrying current I is B = μ₀ I / (4 R). Adding fields from both semicircles of radii 2 m and 4 m gives 3π × 10⁻⁷ T total, so α = 4.
Two charges of −4µC and +4µC are placed at the points A(1, 0, 4)m and B(2, −1, 5)m in an electric field E = 0.20 i V/cm. The magnitude of the torque acting on the dipole is 8√α×10⁻⁵ Nm, where α = :
The electric dipole moment p = q × d, and torque τ = p × E. Calculation yields 8√2×10⁻⁵ Nm, hence α = 2.
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:
The beat frequency of 7 Hz is the difference of their fundamental frequencies. Using v = 4L × frequency for the closed pipe and v = 2L × frequency for the open pipe, solving yields 294 m/s.
A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is:
From the equations of motion under free fall, the body travels 80 m between A and B in 2 s. The distance from the start to A is found to be 45 m.
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:
Using Bernoulli’s principle, (P₁ − P₂) = (1/2)ρv². Substituting ΔP = 2.5×10⁴ N/m² and ρ for water, we get v = √50 m/s.
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:
For rolling without slipping, total kinetic energy = translational + rotational. The solid sphere has a different moment of inertia from the ring, giving a ratio of 7.
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:
For the first minima in single-slit diffraction, sinθ = λ/a. Here, λ = 5000×10⁻¹⁰ m, a = 1×10⁻⁶ m. Thus sinθ = 5×10⁻⁴, θ ≈ 30°.
The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9×10⁻¹³ cm is ×10⁶ V:
Potential V = kZe/R. With Z = 50, R = 9×10⁻¹³ cm, and k = 9×10⁹, we get 8×10⁶ V.
If Rydberg’s constant is R, the longest wavelength of radiation in Paschen series will be α×7R, where α = :
The Paschen series starts at n=3. The longest wavelength transition is n=4 to n=3. Using the Rydberg formula, α is 144.
A series LCR circuit with L = 100π mH, C = 10⁻³ F, and R = 10 Ω is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be:
At resonance in a series LCR circuit, inductive and capacitive reactances cancel out, leaving only R. Therefore, the power factor is 1.
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