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Simran Zutshi

Content Strategist|Tech-innovator|National Hackathon Winner | Updated On - Mar 30, 2026

The JEE Main 2023 Physics Question Paper with Solution PDF is available here for download. The exam was successfully conducted by NTA on January 29, 2023, in the first shift.

Students can download JEE Main previous year question papers PDFs with detailed solutions to practice and improve their performance. Solving these papers helps aspirants understand the JEE Main exam pattern, analyze the difficulty level, and prepare according to the latest JEE Main syllabus.

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JEE Main 2023 Physics Question Paper Jan 29 Shift 1 with Solution Pdf

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JEE Main 2023 Question Paper Jan 29 Shift 1 with Solution Pdf

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:

  1. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  3. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  4. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Correct Answer: (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
View Solution

The correct matching pairs are:

  • A-I: Ribosomes are responsible for protein synthesis.
  • B-II: Mitochondria produce energy in the form of ATP through cellular respiration.
  • C-III: Lysosomes are involved in intracellular digestion of waste and cellular debris.
  • D-IV: The Golgi apparatus modifies, sorts, and packages proteins for secretion or use within the cell.

Understanding the functions of cellular organelles is fundamental to the study of cell biology and physiology.


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:

  1. q / 12ε₀
  2. q / 3ε₀
  3. q / 2ε₀
  4. q / 6ε₀
Correct Answer: (4) q / 6ε₀
View Solution

The total flux through the cuboid is given by:

Φtotal = q / ε₀.

As the cuboid has six faces, the flux through each face is:

Φ = q / 6ε₀.


Question 3:

Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is:

  1. 3:1
  2. 1:1
  3. 1:3
  4. 1:27
Correct Answer: (1) 3:1
View Solution

The power dissipated in a resistor is proportional to V2 / R.

For R: Power ∝ 1 / R, and for 3R: Power ∝ 1 / (3R).

Thus, the ratio of powers is:

P1 / P2 = 3:1.


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:

  1. (Not shown)
  2. (Not shown)
  3. (Correct graph)
  4. (Not shown)
Correct Answer: (3)
View Solution

At z = 0 (plane of the loop), By = 0.

By is opposite in sign for +z and −z, as determined by the right-hand rule.


Question 5:

The magnitude of magnetic induction at the mid-point O due to the current arrangement shown in the figure is:

  1. μ₀I / (2πa)
  2. 0
  3. μ₀I / (4πa)
  4. μ₀I / (πa)
Correct Answer: (4) μ₀I / (πa)
View Solution

The magnetic field contributions due to segments BC and ET are outward at point O.

Total magnetic field:

B = μ₀I / (4πr) + μ₀I / (4πr) = μ₀I / (πa).


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:

  1. M = √2μ₀R² / L
  2. M = 2√2μ₀R / L²
  3. M = 2√2μ₀R² / L
  4. M = √2μ₀R / L²
Correct Answer: (3) M = 2√2μ₀R² / L
View Solution

The magnetic flux is given by:

φ = πR² (4μ₀ / 4π) · i / (L√2).

Therefore, mutual inductance:

M = 2√2μ₀R² / L.


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:

  1. M = √2μ₀R² / L
  2. M = 2√2μ₀R / L²
  3. M = 2√2μ₀R² / L
  4. M = √2μ₀R / L²
Correct Answer: (3) M = 2√2μ₀R² / L
View Solution

The magnetic flux is given by:

φ = πR² (4μ₀ / 4π) · i / (L√2).

Therefore, mutual inductance:

M = 2√2μ₀R² / L.


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:

  1. A and C only
  2. B and D only
  3. B and C only
  4. C and D only
Correct Answer: (4) C and D only
View Solution

1. Analysis of Statements:

  • A: Incorrect. The speed of light in vacuum does not depend on the direction of propagation.
  • B: Incorrect. Speed of light in a medium depends on the wavelength (dispersion).
  • C: Correct. The speed of light is constant in vacuum, independent of the source’s motion.
  • D: Correct. The speed of light in a medium is not affected by light intensity.

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:

  1. 0.4 mm
  2. 0.5 mm
  3. 0.2 mm
  4. 0.1 mm
Correct Answer: (3) 0.2 mm
View Solution

1. Condition for Minima: Path difference for the first minimum:

Δx = λ / 2.

2. Slit Separation: From geometry:

a = λD / Δx.

Substituting values:

a = (800 × 10⁻⁹ × 5 × 10⁻²) / (0.5 × 10⁻³) = 0.2 mm.


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:

  1. 1:2
  2. 1:4
  3. 4:1
  4. 4:3
Correct Answer: (4) 4:3
View Solution

1. Kinetic Energy at Projection:

KEprojection = (1/2)mu².

2. Kinetic Energy at the Highest Point: Horizontal velocity remains constant:

KEhighest = (1/2)m(u cos 30°)².

3. Ratio:

KEprojection / KEhighest = u² / (u² cos² 30°) = 1 / cos² 30° = 4 / 3.


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:

  1. (2√3 + 1) / 2
  2. 1 / 2√3
  3. √3 / 2
  4. (2√3 − 1) / 2
Correct Answer: (2) 1 / 2√3
View Solution

1. Force Equation:

mg sin 30° − μmg cos 30° = ma.

2. Substitute Values:

a = g/4, sin 30° = 1/2, cos 30° = √3/2:

(mg / 2) − μ(mg √3 / 2) = mg / 4.

3. Solve for μ:

μ = 1 / 2√3.


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):

  1. 13.4 m/s
  2. 22.4 m/s
  3. 13 m/s
  4. 17 m/s
Correct Answer: (3) 13 m/s
View Solution

1. Maximum Speed on a Curved Road:

The maximum speed of a vehicle on a curved road is given by:

vmax = √(μgr),

where μ is the coefficient of friction, g is the acceleration due to gravity, and r is the radius of curvature.

2. Substitute the Values:

μ = 0.34, g = 10 m/s², r = 50 m:

vmax = √(0.34 × 10 × 50) = √170 ≈ 13 m/s.


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:

  1. √(GM / 2r)
  2. √(4GM / r)
  3. √(GM / r)
  4. √(GM / 4r)
Correct Answer: (4) √(GM / 4r)
View Solution

1. Gravitational Force:

The gravitational force provides the centripetal force:

Gm² / (4r²) = mv² / r.

2. Solve for v:

v² = GM / (4r),

v = √(GM / 4r).


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:

  1. 4.072 × 10⁻³ J
  2. 5.76 × 10⁻³ J
  3. 1.848 × 10⁻³ J
  4. 9.24 × 10⁻³ J
Correct Answer: (3) 1.848 × 10⁻³ J
View Solution

1. Work Done:

Work done is equal to the change in surface energy:

W = Surface tension × ΔA.

2. Surface Area of a Bubble:

Surface area of a sphere: A = 4πR².

Change in surface area for a bubble (two surfaces):

ΔA = 2 × 4π(R₂² − R₁²).

3. Substitute Values:

R₁ = 3.5 cm, R₂ = 7 cm, T = 2 × 10⁻² N/m:

W = 2 × 10⁻² × 2 × 4π [(0.07)² − (0.035)²].

Simplify:

W ≈ 1.848 × 10⁻³ J.


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:

  1. A is correct but R is not correct
  2. A is not correct but R is correct
  3. Both A and R are correct, and R is the correct explanation of A
  4. Both A and R are correct, but R is not the correct explanation of A
Correct Answer: (3) Both A and R are correct, and R is the correct explanation of A
View Solution

1. The first law of thermodynamics states:

ΔQ = ΔU + ΔW.

2. Rearranging gives:

dQ = dU − dW.

3. This law is based on the conservation of energy, explaining that energy can neither be created nor destroyed, only transformed.


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:

  1. 270 kPa
  2. 262 kPa
  3. 278 kPa
  4. 360 kPa
Correct Answer: (3) 278 kPa
View Solution

1. Pressure-Temperature Relation:

Using Gay-Lussac’s law (volume is constant):

P₁ / T₁ = P₂ / T₂.

2. Substitute Values:

P₁ = 270 kPa, T₁ = 300 K, T₂ = 309 K:

P₂ = P₁ × (T₂ / T₁) = 270 × (309 / 300).

3. Calculate P₂:

P₂ ≈ 278 kPa.


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:

  1. 33 Hz
  2. 55 Hz
  3. 80 Hz
  4. 10 Hz
Correct Answer: (2) 55 Hz
View Solution

1. Frequency Observed (Doppler Effect):

The observed frequency for train A (approaching):

fA = f × (v + vo) / v = 300 × (330 / 300) = 330 Hz.

The observed frequency for train B (receding):

fB = f × v / (v + vs) = 300 × (330 / 360) = 275 Hz.

2. Difference in Frequencies:

Δf = fA − fB = 330 − 275 = 55 Hz.


Question 17:

If the height of transmitting and receiving antennas are 80 m each, the maximum line of sight distance will be:

  1. 32 km
  2. 28 km
  3. 36 km
  4. 64 km
Correct Answer: (4) 64 km
View Solution

1. Line of Sight Distance Formula:

The maximum line of sight distance is:

dmax = √2Rht + √2Rhr,

where R = 6.4 × 10⁶ m (Earth’s radius) and ht = hr = 80 m.

2. Substitute Values:

dmax = √(2 × 6.4 × 10⁶ × 80) + √(2 × 6.4 × 10⁶ × 80).

dmax = 2 × √1024000 = 2 × 32 = 64 km.


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:

  1. B and C only
  2. A and D only
  3. C only
  4. C and D only
Correct Answer: (4) C and D only
View Solution

1. Threshold Wavelength:

Photoelectric emission occurs if λ < λthreshold.

Given λthreshold = 5500 Å.

2. Analysis of Radiation:

- Infra-red radiation has λ > 5500 Å (no emission).

- Ultra-violet radiation has λ < 5500 Å (emission occurs).

Correct options: C and D.


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:

  1. 1/8
  2. 1/16
  3. 1/4
  4. 1/2
Correct Answer: (1) 1/8
View Solution

1. Number of Half-Lives:

Time elapsed: t = 90 min, Half-life: T1/2 = 30 min.

Number of half-lives: n = t / T1/2 = 90 / 30 = 3.

2. Remaining Fraction:

Fraction remaining after n half-lives:

N/N₀ = (1/2)n = (1/2)³ = 1/8.


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:

  1. C and D
  2. A
  3. C
  4. B
Correct Answer: (3) C
View Solution

1. Correct Statements:

- A: Correct. LEDs are made from heavily doped p-n junctions.

- B: Correct. LEDs emit light only in forward bias.

- D: Correct. The emitted light energy corresponds to the bandgap.

2. Incorrect Statement:

- C: Incorrect. LEDs do not emit light in reverse bias.


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 ——.

Correct Answer: 87
View Solution

1. Effect of Emission:

  • Emission of an α-particle reduces the mass number by 4 and the atomic number by 2.
  • Emission of an electron (β⁻) increases the atomic number by 1.
  • Emission of a positron (β⁺) decreases the atomic number by 1.

2. Calculate Atomic Number (P):

  • After two α-particles: Atomic number = 92 − 2 × 2 = 88, Mass number = 242 − 2 × 4 = 234.
  • After one electron (β⁻): Atomic number = 88 + 1 = 89.
  • After two positrons (2β⁺): Atomic number = 89 − 2 = 87.

Final Answer: 87


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.

Correct Answer: 120°
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.

Correct Answer: 28°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.

Correct Answer: 40 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)

Correct Answer: 300 J
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)

Correct Answer: 120 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.

Correct Answer: 24 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.

Correct Answer: 2 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.

Correct Answer: 10 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.

Correct Answer: 24 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


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