
JEE Main 2024 Jan 31 Shift 2 Question Paper with Solution pdf is available for download here. Students found Physics easy and Chemistry hard. Chemistry carried the highest weightage and overall difficulty level was moderate.
| JEE Main 2024 Physics Question Paper with Answer Key 31 Jan Shift 2 | Check Solution |
|---|
A light string passing over a smooth light fixed pulley connects two blocks of masses m1 and m2. If the acceleration of the system is g/8, then the ratio of masses is:
Step 1: Using Newton’s Second Law for each mass:
Step 2: Adding the two equations:
m2g − m1g = (m1 + m2)a
Given a = g/8:
Step 3: Substitute the value of acceleration:
m2g − m1g = (m1 + m2)(g/8)
Step 4: Cancel g (since g ≠ 0):
m2 − m1 = (m1 + m2)/8
Step 5: Multiply both sides by 8 to eliminate the denominator:
8m2 − 8m1 = m1 + m2
Step 6: Rearrange the equation:
8m2 − m2 = 8m1 + m1
7m2 = 9m1
Step 7: Thus, the ratio of masses:
m1/m2 = 7/9 or m2/m1 = 9/7.
Final Answer: The ratio of masses m2/m1 is 9/7.
A uniform magnetic field of 2×10−3 T acts along the positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with a current of 5 A is in the Y-Z plane. The current is in the anticlockwise sense with reference to the negative X-axis. Magnitude and direction of the torque is:
Step 1: Calculate the Magnetic Moment (μ)
Step 2: Calculate the Torque Magnitude (τ)
Step 3: Determine the Direction of Torque
Final Answer: The torque is 2 × 10−4 N-m along the negative Z-direction.
The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is ___%. The value of N is:
The formula for acceleration due to gravity (g) is:
g = (4π2 l) / T2
where l is the length of the pendulum and T is the time period.
Step 1: Calculate the errors
Step 2: Compute relative error in g
Step 3: Convert to percentage
Final Answer: The accuracy in the measurement of acceleration due to gravity is 6%.
Force between two point charges q1 and q2 placed in vacuum at r cm apart is F. Force between them when placed in a medium having dielectric constant K = 5 at r/5 cm apart will be:
Step 1: Coulomb’s law in a medium is:
Fm = F / K
Additionally, force is inversely proportional to the square of the distance:
F ∝ 1/r²
Step 2: When the distance is reduced to r/5:
F' = Fm × (r / (r/5))² = (F / 5) × 25 = 5F
Final Answer: The new force is 5F.
An AC voltage V = 20 sin(200πt) is applied to a series LCR circuit which drives a current I = 10 sin(200πt + π/3). The average power dissipated is:
The average power dissipated in an AC circuit is given by:
P = Vrms × Irms × cos φ
Step 1: Calculate RMS values
Step 2: Find cos φ
Step 3: Compute power
P = Vrms × Irms × cos φ = 14.14 × 7.07 × 0.5 ≈ 50 W
Final Answer: The average power dissipated is 50 W.
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.
*The article might have information for the previous academic years, please refer the official website of the exam.