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Simran Zutshi

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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.

JEE Main 2024 Physics Question Paper with Answer Key PDF

JEE Main 2024 Question Paper Download download icon Download Check Solution

JEE Main Physics 2024 Jan 27 Shift 2 Questions with Solution

Question 1:

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
Correct Answer: P
View Solution

Using dimensional analysis on a/b², we find it has the same dimensions as the pressure P.


Question 2:

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
Correct Answer: 6232.5 J
View Solution

For a diatomic gas at 300 K, E = (5/2) n R T. Substituting n = 1 and T = 300 K gives 6232.5 J.


Question 3:

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
Correct Answer: 11.5 W
View Solution

The secondary voltage V₂ = (230/10) = 23 V. Power P = V₂² / R = 23² / 46 = 529/46 = 11.5 W.


Question 4:

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
Correct Answer: 3/2
View Solution

From the adiabatic relation P ∝ T^(γ/(γ−1)) and given P ∝ T³, we get γ/(γ−1) = 3, leading to γ = 3/2.


Question 5:

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
Correct Answer: 1.6 × 10¹⁵ Hz
View Solution

Using E = h ν₀ and converting 6.63 eV to joules gives ν₀ ≈ 1.6 × 10¹⁵ Hz.


Question 6:

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
Correct Answer: 60 µA
View Solution

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.


Question 7:

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
Correct Answer: 4.95 MeV
View Solution

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.


Question 8:

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)
Correct Answer: 2tan⁻¹(1/2)
View Solution

Equating accelerations (centripetal and tangential) using energy conservation, the angle θ is found to be 2tan⁻¹(1/2).


Question 9:

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
Correct Answer: V₁ + V₂ = V₃
View Solution

Applying Kirchhoff’s Voltage Law around the loop, we see that V₁ + V₂ = V₃.


Question 10:

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
Correct Answer: 6232.5 J
View Solution

For a diatomic gas at 300 K, using E = (5/2) n R T, we get 6232.5 J.


Question 11:

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
Correct Answer: Shorter period of orbit
View Solution

Angular speed ω is inversely proportional to the orbital period. The moon’s orbital period is shorter, so ω(moon) > ω(earth).


Question 12:

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
Correct Answer: 11.5 W
View Solution

Secondary voltage V₂ = 230/10 = 23 V, so power P = 23² / 46 = 11.5 W.


Question 13:

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
Correct Answer: 3/2
View Solution

From P ∝ T^(γ/(γ−1)) = T³, we get γ/(γ−1) = 3, leading to γ = 3/2.


Question 14:

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
Correct Answer: 1.6 × 10¹⁵ Hz
View Solution

E = 6.63 eV → convert to joules. Then E = h ν₀. Solving gives ν₀ ≈ 1.6 × 10¹⁵ Hz.


Question 15:

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
Correct Answer: 60 µA
View Solution

Current is proportional to deflection angle. 60° is π/3, so for π/10, current = 200 × (π/10) / (π/3) = 60 µA.

Question 16:

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
Correct Answer: 4.95 MeV
View Solution

Using the mass defect and the formula E = Δm × 931.5 MeV/u, the energy required is 4.95 MeV.


Question 17:

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)
Correct Answer: 2tan⁻¹(1/2)
View Solution

By applying energy conservation and equating accelerations at the extreme and lowest positions, the angle θ is found to be 2tan⁻¹(1/2).


Question 18:

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
Correct Answer: V₁ + V₂ = V₃
View Solution

Applying Kirchhoff’s Voltage Law around the loop shows that V₁ + V₂ = V₃.


Question 19:

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
Correct Answer: 6232.5 J
View Solution

For a diatomic gas at 300 K, using E = (5/2) × n × R × T gives 6232.5 J for 1 mole of O₂.


Question 20:

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
Correct Answer: 11.5 W
View Solution

Secondary voltage V₂ = 230 / 10 = 23 V. Then power P = (23²) / 46 = 529 / 46 = 11.5 W.


Question 21:

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:

Correct Answer: 4
View Solution

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.


Question 22:

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 = :

Correct Answer: 2
View Solution

The electric dipole moment p = q × d, and torque τ = p × E. Calculation yields 8√2×10⁻⁵ Nm, hence α = 2.


Question 23:

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:

Correct Answer: 294 m/s
View Solution

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.


Question 24:

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:

Correct Answer: 45 m
View Solution

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.


Question 25:

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:

Correct Answer: 50
View Solution

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.


Question 26:

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:

Correct Answer: 7
View Solution

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.


Question 27:

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:

Correct Answer: 30°
View Solution

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°.


Question 28:

The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9×10⁻¹³ cm is ×10⁶ V:

Correct Answer: 8
View Solution

Potential V = kZe/R. With Z = 50, R = 9×10⁻¹³ cm, and k = 9×10⁹, we get 8×10⁶ V.


Question 29:

If Rydberg’s constant is R, the longest wavelength of radiation in Paschen series will be α×7R, where α = :

Correct Answer: 144
View Solution

The Paschen series starts at n=3. The longest wavelength transition is n=4 to n=3. Using the Rydberg formula, α is 144.


Question 30:

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:

Correct Answer: 1
View Solution

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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