JEE Main 2023 Physics Question Paper Jan 29 Shift 1 is available for download. Candidates can download JEE Main 2023 Physics Question Paper PDF with Answer Key for Jan 29 Shift 1 using the link below. JEE Main Physics Question Paper is divided into two sections, Section A with 20 MCQs and Section B with 10 numerical-type questions. Candidates are required to answer all questions from Section A and any 5 questions from section B.
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JEE Main 2023 Physics Question Paper Jan 29 Shift 1- Download PDF
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JEE Main 2023 Jan 29 Shift 1 Physics Question Paper with Solution
Question 1:
Match List-I with List-II:
| List-I (Organelle/Structure) | List-II (Function) |
|---|---|
| (A) Ribosome | (I) Protein synthesis |
| (B) Mitochondria | (II) Energy production |
| (C) Lysosome | (III) Intracellular digestion |
| (D) Golgi apparatus | (IV) Protein modification and packaging |
Choose the correct answer from the options given below:
View Solution
Question 2:
In a cuboid of dimensions 2L × 2L × L, a charge q is placed at the center of the surface S having area 4L2. The flux through the opposite surface to S is:
View Solution
Question 3:
Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is:
View Solution
Question 4:
A single current-carrying loop of wire carrying current I flows in the anticlockwise direction (seen from the +z direction) and lies in the xy plane. The plot of the ĵ component of magnetic field (By) at a distance a (less than the radius of the coil) and on the yz plane vs z coordinate looks like:
View Solution
Question 5:
The magnitude of magnetic induction at the mid-point O due to the current arrangement shown in the figure is:
View Solution
Question 6:
Find the mutual inductance in the arrangement, when a small circular loop of radius R is placed inside a large square loop of side L (L ≫ R). The loops are coplanar and their centers coincide:
View Solution
Question 6:
Find the mutual inductance in the arrangement, when a small circular loop of radius R is placed inside a large square loop of side L (L ≫ R). The loops are coplanar and their centers coincide:
View Solution
Question 7:
Which of the following are true?
A. Speed of light in vacuum depends on the direction of propagation.
B. Speed of light in a medium is independent of the wavelength of light.
C. Speed of light is independent of the motion of the source.
D. Speed of light in a medium is independent of intensity.
Choose the correct answer from the options given below:
View Solution
Question 8:
In a Young’s double slit experiment, two slits are illuminated with light of wavelength 800 nm. The first minimum is detected at P. The value of slit separation a is:
View Solution
Question 9:
A stone is projected at an angle of 30° to the horizontal. The ratio of kinetic energy at projection to its kinetic energy at the highest point is:
View Solution
Question 10:
A block of mass m slides down a plane inclined at an angle of 30° with an acceleration of g/4. The coefficient of kinetic friction is:
View Solution
Question 11:
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of the car will be, if the friction coefficient between tyres and road is 0.34. (Take g = 10 m/s²):
View Solution
Question 12:
Two particles of equal mass m move in a circle of radius r under the action of their mutual gravitational attraction. The speed of each particle will be:
View Solution
Question 13:
The surface tension of a soap bubble is 2 × 10⁻² N/m. Work done to increase the radius of the bubble from 3.5 cm to 7 cm will be:
View Solution
Question 14:
Assertion A: If dQ and dW represent the heat supplied to the system and the work done on the system respectively, then according to the first law of thermodynamics dQ = dU − dW.
Reason R: First law of thermodynamics is based on the law of conservation of energy.
In the light of the above statements, choose the correct answer from the options given below:
View Solution
Question 15:
A bicycle tyre is filled with air at a pressure of 270 kPa at 27°C. The approximate pressure of the air in the tyre when the temperature increases to 36°C is:
View Solution
Question 16:
A person observes two moving trains, A reaching the station and B leaving the station with equal speed of 30 m/s. If both trains emit sounds with frequency 300 Hz, the approximate difference of frequencies heard by the person will be:
View Solution
Question 17:
If the height of transmitting and receiving antennas are 80 m each, the maximum line of sight distance will be:
View Solution
Question 18:
The threshold wavelength for photoelectric emission from a material is 5500 Å. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from:
- A. 75 W infra-red lamp
- B. 10 W infra-red lamp
- C. 75 W ultra-violet lamp
- D. 10 W ultra-violet lamp
Choose the correct answer from the options given below:
View Solution
Question 19:
If a radioactive element with a half-life of 30 min undergoes beta decay, the fraction of the radioactive element that remains undecayed after 90 min is:
View Solution
Question 20:
Which of the following statements is not correct in the case of light emitting diodes (LEDs)?
- A. It is a heavily doped p-n junction.
- B. It emits light only when it is forward biased.
- C. It emits light only when it is reverse biased.
- D. The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used.
Choose the correct answer from the options given below:
View Solution
Question 21:
A radioactive element 24292X emits two α-particles, one electron, and two positrons. The product nucleus is represented by 234PY. The value of P is ——.
View Solution
Question 22:
Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is —– degrees.
View Solution
1. Resultant Amplitude Formula:
Aresultant = √(A₁² + A₂² + 2A₁A₂ cos φ).
2. Given Data:
- A₁ = A₂ = 8 cm, Aresultant = 8 cm.
3. Substitute Values:
8 = √(8² + 8² + 2 × 8 × 8 cos φ).
64 = √(128 + 128 cos φ).
Square both sides:
64 = 128(1 + cos φ).
Solve for cos φ:
cos φ = −1/2.
4. Phase Difference:
φ = 120°.
Question 23:
A body cools from 60°C to 40°C in 6 minutes. If the temperature of the surroundings is 10°C, then after the next 6 minutes, its temperature will be —— °C.
View Solution
1. Newton’s Law of Cooling:
Average rate of cooling:
(T − Ts)/∆t = k(T − Ts),
where Ts is the surrounding temperature.
2. First Interval:
(60 − 40)/6 = k[(60 + 40)/2 − 10],
k = 20 / (6 × 50).
3. Second Interval:
(40 − T)/6 = k[(40 + T)/2 − 10].
Solve:
T = 28°C.
Question 24:
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of the center of mass of the sphere will be —– ms⁻¹.
View Solution
1. Kinetic Energy of a Rolling Sphere:
Total kinetic energy:
KE = (1/2)mv² + (1/2)Iω²,
where I = (2/5)mr² for a sphere.
2. Relation Between v and ω:
For pure rolling: ω = v/r.
3. Substitute Values:
KE = (1/2)mv² + (1/2)(2/5)mr²(v²/r²).
Simplify:
KE = (1/2)mv² + (1/5)mv² = (7/10)mv².
4. Solve for v:
2240 = (7/10) × 2 × v²,
v² = 1600, v = 40 ms⁻¹.
Final Answer: 40 ms⁻¹
Question 25:
A 0.4 kg mass takes 8 seconds to reach the ground when dropped from a certain height P above the surface of Earth. The loss of potential energy in the last second of fall is —— J. (Take g = 10 m/s²)
View Solution
1. Height Traveled in the Last Second:
The distance traveled in the nth second of free fall is:
hₙ = u + (1/2)g(2n − 1).
Here, u = 0, g = 10 m/s², n = 8:
h₈ = (1/2) × 10 × (2 × 8 − 1) = (1/2) × 10 × 15 = 75 m.
2. Loss of Potential Energy:
The loss of potential energy is given by:
ΔPE = mgh₈.
Substituting m = 0.4 kg, g = 10 m/s², and h₈ = 75 m:
ΔPE = 0.4 × 10 × 75 = 300 J.
Final Answer: 300 J
Question 26:
A tennis ball is dropped onto the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. The ball comes in contact with the floor for 0.2 s. The average acceleration during contact is —— m/s². (Given g = 10 m/s²)
View Solution
1. Velocity Just Before Impact:
Using v² = u² + 2gh, where u = 0, h = 9.8 m:
v = √(2 × 10 × 9.8) = √196 = 14 m/s.
2. Velocity Just After Rebound:
Using v² = u² + 2gh, where u = 0, h = 5.0 m:
v = √(2 × 10 × 5.0) = √100 = 10 m/s.
3. Change in Velocity During Contact:
Total change in velocity:
Δv = vbefore impact + vafter rebound = 14 + 10 = 24 m/s.
4. Average Acceleration:
Average acceleration:
a = Δv/Δt = 24/0.2 = 120 m/s².
Final Answer: 120 m/s²
Question 27:
A point charge q₁ = 4q₀ is placed at the origin. Another point charge q₂ = −q₀ is placed at x = 12 cm. The proton is placed on the x-axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is — cm.
View Solution
1. Force Balance Condition: - The electrostatic force on the proton is zero when:
F₁ = F₂.
Using Coulomb’s law:
k · |4q₀| / r² = k · |q₀| / (12 − r)².
2. Simplify: - Cancel k and q₀:
4 / r² = 1 / (12 − r)².
Take the square root:
2 / r = 1 / (12 − r).
Cross-multiply:
2(12 − r) = r ⇒ 24 − 2r = r.
Solve for r:
3r = 24 ⇒ r = 8 cm.
3. Position of Proton: - The proton is 12 + r = 24 cm from the origin.
Final Answer: 24 cm
Question 28:
In a metre bridge experiment, the balance point is obtained if the gaps are closed by 2 Ω and 3 Ω. A shunt of X Ω is added to the 3 Ω resistor to shift the balancing point by 22.5 cm. The value of X is — Ω.
View Solution
1. Initial Balance Point: - The ratio of resistances gives the balance length:
l₁ / l₂ = R₁ / R₂,
where l₁ + l₂ = 100 cm.
2. Initial Condition: - For R₁ = 2 Ω and R₂ = 3 Ω:
l₁ / l₂ = 2 / 3, so l₁ = (2/5) × 100 = 40 cm.
3. After Adding Shunt: - The effective resistance of R₂ with a shunt X:
R'₂ = (R₂X) / (R₂ + X) = (3X) / (3 + X).
- The new balance point shifts by 22.5 cm:
l'₁ = 40 + 22.5 = 62.5 cm.
4. New Condition: - The new ratio is:
l'₁ / l'₂ = R₁ / R'₂.
Substituting l'₁ = 62.5, l'₂ = 37.5, R₁ = 2 Ω, and R'₂ = (3X) / (3 + X):
62.5 / 37.5 = 2 / ((3X) / (3 + X)).
Simplify:
5 / 3 = 2(3 + X) / 3X.
Cross-multiply and solve for X:
15X = 18 + 6X ⇒ 9X = 18 ⇒ X = 2 Ω.
Final Answer: 2 Ω
Question 29:
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released, the radius of the loop starts shrinking at a constant rate of 2 cm/s. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be — mV.
View Solution
1. Magnetic Flux: - The magnetic flux through the loop is:
Φ = B · A = B · πr²,
where B = 0.8 T and r = 10 cm = 0.1 m.
2. Rate of Change of Flux: - The emf induced is:
E = − dΦ/dt.
Differentiate Φ with respect to time:
E = − d/dt(Bπr²) = −B · 2πr (dr/dt).
3. Substitute Values: - B = 0.8 T, r = 0.1 m, dr/dt = −2 cm/s = −0.02 m/s:
E = 0.8 · 2π · 0.1 · 0.02 = 0.010 V.
4. Convert to mV:
E = 10 mV.
Final Answer: 10 mV
Question 30:
Three identical polaroids P₁, P₂, and P₃ are placed one after another. The pass axis of P₂ and P₃ are inclined at angles of 60° and 90° with respect to the axis of P₁. The source S has an intensity of 256 W/m². The intensity of light at point O is —— W/m².
View Solution
1. Intensity After First Polaroid (P₁): - When unpolarized light passes through a polaroid, its intensity is reduced by half:
I₁ = I₀ / 2 = 256 / 2 = 128 W/m².
2. Intensity After Second Polaroid (P₂): - The intensity after P₂ is given by Malus’s Law:
I₂ = I₁ cos²(60°).
- Substituting cos 60° = 1/2:
I₂ = 128 · (1/2)² = 128 · 1/4 = 32 W/m².
3. Intensity After Third Polaroid (P₃): - The intensity after P₃ is again reduced according to Malus’s Law:
I₃ = I₂ cos²(30°),
where the relative angle between P₂ and P₃ is 30° (since 90° − 60° = 30°).
- Substituting cos 30° = √3/2:
I₃ = 32 · (√3/2)² = 32 · 3/4 = 24 W/m².
Final Answer: 24 W/m²
Also Check:
JEE Main 2023 Physics Analysis Jan 29 Shift 1
JEE Main 2023 Paper Analysis for physics paper scheduled on January 29 Shift 1 has also been updated here. According to the student reactions, JEE Main 2023 Physics Question Paper Jan 29 Shift 1 was reported to be moderate in terms of overall difficulty. Candidates will be able to check the topics with the highest weightage, difficulty level and memory-based Physics questions.
JEE Main 2023 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) |
| Exam Duration | 3 hours |
| Sectional Time Limit | None |
| Physics Marks | 100 marks |
| Total Number of Questions Asked | 20 MCQs + 10 Numerical Type Questions |
| Total Number of Questions to be Answered | 20 MCQs + 5 Numerical Type Questions |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Also Check:
JEE Main 2022 Question Paper
JEE Main 2023 aspirants can practice and check their exam prep level by attempting the previous year question papers as well. The table below shows JEE Main 2022 Question Paper PDF for B.E./B.Tech to practice.









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