
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.
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| JEE Main 2023 Physics Question Paper | Check Solution |

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:
The correct matching pairs are:
Understanding the functions of cellular organelles is fundamental to the study of cell biology and physiology.
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:
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.
Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is:
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.
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:
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.
The magnitude of magnetic induction at the mid-point O due to the current arrangement shown in the figure is:
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).
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:
The magnetic flux is given by:
φ = πR² (4μ₀ / 4π) · i / (L√2).
Therefore, mutual inductance:
M = 2√2μ₀R² / L.
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:
The magnetic flux is given by:
φ = πR² (4μ₀ / 4π) · i / (L√2).
Therefore, mutual inductance:
M = 2√2μ₀R² / L.
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. Analysis of Statements:
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. 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.
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. 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.
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. 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.
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. 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.
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. Gravitational Force:
The gravitational force provides the centripetal force:
Gm² / (4r²) = mv² / r.
2. Solve for v:
v² = GM / (4r),
v = √(GM / 4r).
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. 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.
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. 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.
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. 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.
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. 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.
If the height of transmitting and receiving antennas are 80 m each, the maximum line of sight distance will be:
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.
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:
Choose the correct answer from the options given below:
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.
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. 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.
Which of the following statements is not correct in the case of light emitting diodes (LEDs)?
Choose the correct answer from the options given below:
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.
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 ——.
1. Effect of Emission:
2. Calculate Atomic Number (P):
Final Answer: 87
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.
1. Resultant Amplitude Formula:
Aresultant = √(A₁² + A₂² + 2A₁A₂ cos φ).
2. Given Data:
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°.
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.
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.
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.
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
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)
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
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)
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
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.
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
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.
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
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.
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
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.
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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