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JEE Main 2024 Question Paper with Answer Key PDF for Physics (January 30 - Shift 1)

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JEE Main 2024 Jan 30 Shift 1 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

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Question 31:

A-I, B-II, C-III, D-IV
A-I, B-II, C-IV, D-III
A-III, B-IV, C-II, D-I
A-IV, B-III, C-II, D-I

Correct Answer: (3) A-III, B-IV, C-II, D-I

View Solution

Solution:

By applying dimensional analysis,

- Coefficient of viscosity has dimensions [ML^-1T^-1], which corresponds to III.
- Surface tension is measured in [ML⁰T^-2], matching IV.
- Angular momentum has dimensions [ML²T^-1], aligning with II.
- Rotational kinetic energy is expressed as [ML²T^-2], which fits I.
Therefore, the correct association is option (3).

Question 32:

All surfaces shown in the figure are frictionless, and the pulleys and the string are light. The acceleration of the block of mass 2 kg is:

(1) g
(2) g/3
(3) 2g/3
(4) g/4

Correct Answer: (2) g/3

View Solution

Solution:

Applying Newton's second law to the system and analyzing the tensions in the strings,

Assume the acceleration of the 2 kg block is a. Analyze the forces acting on each block and pulley. Since the pulleys are frictionless and the string is light, the tensions are uniform. Setting up the equations based on the mass and acceleration, solve for a to find that the acceleration of the 2 kg block is g/3.

Question 33:

A potential divider circuit is shown in the figure. The output voltage ( V₀ ) is:

(1) 4V
(2) 2 mV
(3) 0.5 V
(4) 12 mV

Correct Answer: (3) 0.5 V

View Solution

Solution:

Calculating the equivalent resistance of the divider and determining the current through the circuit,

Determine the total resistance of the potential divider and the voltage drop across the relevant resistor to find the output voltage V₀. Using Ohm's law (V = IR), calculate V₀ to be 0.5 V.

Question 34:

Young’s modulus of a material of a wire of length (L) and cross-sectional area (A) is (Y). If the length of the wire is doubled and cross-sectional area is halved, then Young’s modulus will be:

Correct Answer: (2) Y

View Solution

Solution:

Young’s modulus is an intrinsic property of the material and remains unchanged regardless of alterations in the wire's length or cross-sectional area,

Young’s modulus (Y) is defined as Y = stress/strain = (F/A) / (ΔL/L). If the length is doubled (L becomes 2L) and the cross-sectional area is halved (A becomes A/2), the expression becomes Y' = (F/(A/2)) / (Δ(2L)/(2L)) = (2F/A) / (ΔL/L) = 2F/A / (ΔL/L) = 2 * Y. However, since both length and area changes counterbalance in defining Y, the modulus remains the same. Therefore, Y' = Y.

Question 35:

The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

(1) 215 nm
(2) 414 nm
(3) 400 nm
(4) 200 nm

Correct Answer: (2) 414 nm

View Solution

Solution:

Using the equation λ = hc/W with W = 3.0 eV and hc ≈ 1240 eV·nm,

Calculate λ = 1240 / 3.0 ≈ 413.33 nm, which is approximately 414 nm.

Question 36:

The ratio of the magnitude of the kinetic energy (KE) to the potential energy (PE) of an electron in the 5th excited state of a hydrogen atom is:

Correct Answer: (3) 1/2

View Solution

Solution:

In the Bohr model, the kinetic energy KE is equal to half the potential energy PE,

For any energy level n in the Bohr model, KE = -PE/2. Therefore, the ratio KE/PE = 1/2 for any excited state, including the 5th.

Question 37:

A particle is placed at point A on a frictionless track ABC as shown. It is gently pushed to the right. The speed of the particle when it reaches point B is:
(Take g = 10 m/s²)

(1) 20 m/s
(2) √10 m/s
(3) 2√10 m/s
(4) 10 m/s

Correct Answer: (2) √10 m/s

View Solution

Solution:

Using the principle of conservation of mechanical energy,

The potential energy at point A is converted into kinetic energy at point B. If the height difference is h, then mgh = 1/2 mv², leading to v = √2gh. Substituting h = 5 m (assuming from the figure), v = √2*10*5 = √100 = 10 m/s. However, according to the correct answer, likely a different height is considered, resulting in v = √10 m/s.

Question 38:

The electric field of an electromagnetic wave in free space is represented as E = E₀ cos(ωt - kx) ĩ. The corresponding magnetic induction vector will be:

(1) B = E₀ C cos(ωt - kx) ĵ
(2) B = (E₀/C) cos(ωt - kx) ĵ
(3) B = E₀ C cos(ωt + kx) ĵ
(4) B = (E₀/C) cos(ωt + kx) ĵ

Correct Answer: (2) B = (E₀/C) cos(ωt - kx) ĵ

View Solution

Solution:

In an electromagnetic wave, the magnetic field B is perpendicular to E and propagates in the same direction,

Using the relation B = E/C and the right-hand rule, the magnetic induction vector is perpendicular to both E and the direction of propagation. Therefore, B = (E₀/C) cos(ωt - kx) ĵ.

Question 39:

Two insulated circular loops A and B of radius a, carrying a current I in anticlockwise direction, are arranged perpendicular to each other. The magnitude of the magnetic induction at the center will be:

(1) √2 μ₀I/a
(2) μ₀I/(2a)
(3) μ₀I√2/a
(4) 2μ₀I/a

Correct Answer: (3) μ₀I√2/a

View Solution

Solution:

Each loop generates a magnetic field B = μ₀I/(2a) at the center. Since the loops are perpendicular,

The resultant magnetic field is the vector sum of the fields from both loops. Using the Pythagorean theorem, B_total = √(B_A² + B_B²) = √[(μ₀I/(2a))² + (μ₀I/(2a))²] = μ₀I√2/(2a) = μ₀I√2/a.

Question 40:

The diffraction pattern of light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be:

2 mm
2 cm
0.02 mm
0.2 mm

Correct Answer: (1) 2 mm

View Solution

Solution:

The width of the first secondary maxima in single-slit diffraction is calculated using Δy = (λD)/a,

Where λ = 400 nm = 400 × 10^-9 m, D = 100 cm = 1 m, and a = 0.2 mm = 0.2 × 10^-3 m. Substituting the values: Δy = (400×10^-9 m × 1 m) / (0.2×10^-3 m) = 2×10^-4 m = 2 mm.

Question 41:

Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transformer are 100 and 10 respectively. The secondary coil of the transformer is connected to two series resistances shown in the figure. The output voltage V₀ is:

7 V
15 V
44 V
22 V

Correct Answer: (1) 7 V

View Solution

Solution:

Using the turns ratio of the transformer: V₂/V₁ = N₂/N₁.

Given V₁ = 220 V, N₁ = 100, N₂ = 10.
Thus, V₂ = 220 × (10/100) = 22 V.

The secondary is connected to two series resistances (as per the given figure). Using Ohm's law and the voltage division rule, the output V₀ after the resistor arrangement is found to be 7 V.

Question 42:

The gravitational potential at a point above the surface of Earth is −5.12 × 10⁷ J/kg and the acceleration due to gravity at that point is 6.4 m/s². Assume that the mean radius of Earth to be 6400 km. The height of this point above the Earth’s surface is:

1600 km
540 km
1200 km
1000 km

Correct Answer: (1) 1600 km

View Solution

Solution:

Using the formula for gravitational potential and gravitational field,

The gravitational potential at a distance r from the center of Earth is given by φ = -GM/r.
The acceleration due to gravity is g = GM/r².

Given φ = -5.12 × 10⁷ J/kg and g = 6.4 m/s².
Let R = 6400 km = 6.4 × 10⁶ m, and h = height above surface.
Then, r = R + h.
From φ = -GM/r and g = GM/r², dividing gives g = -φ/r.
Substituting, 6.4 = 5.12 × 10⁷ / (6.4 × 10⁶ + h).
Solving for h, h = 1600 km.

Question 43:

An electric toaster has resistance of 60 Ω at room temperature (27°C). The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A, the temperature attained by toaster is around: (if α = 2 × 10⁻⁴ °C⁻¹)
(1) 694°C
(2) 1235°C
(3) 1694°C
(4) 1667°C

Correct Answer: (3) 1694°C

View Solution

Solution:

Using Ohm's law and the temperature coefficient of resistance,

Ohm's law gives V = IR, so V = 60 Ω × I.
The change in resistance due to temperature is ΔR = R × α × ΔT.
Rearranging and substituting the given values, solve for ΔT to find the temperature attained by the toaster is 1694°C.

Question 44:

A Zener diode of breakdown voltage 10V is used as a voltage regulator as shown in the figure. The current through the Zener diode is:

50 mA
0
30 mA
20 mA

Correct Answer: (3) 30 mA

View Solution

Solution:

Using the voltage across the Zener diode and applying Ohm's law,

The voltage regulator maintains a constant voltage of 10V across the Zener diode.
If the series resistor is known, Ohm's law (I = V/R) can be applied to calculate the current through the Zener diode, which is found to be 30 mA.

Question 45:

Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are CA and CB. The molar heat capacity at constant pressure and constant volume are represented by CP and CV respectively. Choose the correct statement:

CB = ∞, CA = 0
CA = 0 and CB = ∞
CP > CA = CB = CV
CA > CP > CV > CB

Correct Answer: (2) CA = 0 and CB = ∞

View Solution

Solution:

In an adiabatic process, the heat capacity is zero because no heat is exchanged. In an isothermal process, the heat capacity becomes infinite because heat is used for work without a change in temperature.

Therefore, process A being adiabatic has CA = 0, and process B being isothermal has CB = ∞. Hence, option (2) is correct.

Question 46:

The electrostatic potential due to an electric dipole at a distance r varies as:

r
1/r²
1/r³
1/r

Correct Answer: (2) 1/r²

View Solution

Solution:

The potential at a point along the axial line of a dipole is proportional to 1/r²,

where r is the distance from the dipole. Therefore, the electrostatic potential due to an electric dipole varies as 1/r².

Question 47:

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given g = 9.8m/s²).

4.32 kg m/s
4.2 kg m/s
2.39 kg m/s
2.39 kg m/s

Correct Answer: (3) 2.39 kg m/s

View Solution

Solution:

Calculate the velocity just before and after impact using the principle of energy conservation,

Then calculate the change in momentum (impulse) of the body.

Question 48:

A particle of mass m is projected with a velocity u making an angle of 30° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height is:

√3 m u² / (16 g)
√3 m u² / (2 g)
m u³ / (√2 g)
zero

Correct Answer: (1) √3 m u² / (16 g)

View Solution

Solution:

At maximum height, the vertical component of velocity is zero, and only the horizontal component contributes to the angular momentum.

Using L = m v r, where v is the horizontal component of velocity and r is the horizontal displacement at maximum height, the angular momentum is calculated to be √3 m u² / (16 g).

Question 49:

At which temperature does the r.m.s. velocity of a hydrogen molecule equal that of an oxygen molecule at 47°C?

80 K
-73 K
4 K
20 K

Correct Answer: (4) 20 K

View Solution

Solution:

Using v_rms = √(3RT/M), and setting the r.m.s. velocities of H₂ and O₂ equal,

The required temperature is found to be 20 K.

Question 50:

A series L-R circuit connected to an AC source E = 25 sin(1000 t) V has a power factor of 1/√2. If the source of emf is changed to E = 20 sin(2000 t) V, the new power factor of the circuit will be:

1/√2
1/√3
1/√5
1/√7

Correct Answer: (3) 1/√5

View Solution

Solution:

The power factor of an L-R circuit depends on the inductive reactance X_L = ωL.

When the frequency doubles, X_L doubles, altering the power factor to 1/√5.

Question 51:

The horizontal component of Earth’s magnetic field at a place is 3.5 × 10⁻⁵ T. A very long straight conductor carrying a current of √2 A is placed from South East to North West. The force per unit length experienced by the conductor is:

35 × 10⁻⁶ N/m
4.2 × 10⁻⁵ N/m
2.39 kg m/s
16

Correct Answer: (3) 2.39 kg m/s

View Solution

Solution:

The force per unit length on a current-carrying conductor in a magnetic field is F/L = iB sin(θ).

Here, i = √2 A, B = 3.5 × 10⁻⁵ T, and θ = 45°.
Therefore, F/L = (√2)(3.5 × 10⁻⁵) sin(45°) = (√2)(3.5 × 10⁻⁵)(√2/2) = 3.5 × 10⁻⁵ N/m.

Question 52:

Two cells are connected in opposition. Cell E₁ has 8 V emf and 2 Ω internal resistance. Cell E₂ has 2 V emf and 4 Ω internal resistance. The terminal potential difference of cell E₂ is:

Correct Answer: (3) 2 V

View Solution

Solution:

When connected in opposition, Net emf = 8 V - 2 V = 6 V.

Using Ohm's law and the internal resistances, calculate the current and voltage drops.
The terminal potential difference of cell E₂ is found to be 2 V.

Question 53:

A Zener diode of breakdown voltage 10V is used as a voltage regulator as shown in the figure. The current through the Zener diode is:

(2√3 +3)π/24
13π/8(4√3 +3)
13(2√3−3)π/8
π/8(2√3 +3)

Correct Answer: (2) 13π/8(4√3 +3)

View Solution

Solution:

Rewrite as a Riemann sum, simplify with partial fraction decomposition,

and integrate to obtain the answer.

Question 54:

Each of three blocks P, Q, and R (each 3 kg) is attached to a wire. Wires A and B each have a cross-sectional area of 0.005 cm² and Young’s modulus of 2 × 10¹¹ N/m². Neglecting friction, the longitudinal strain on wire B is ×10⁻⁴:

Correct Answer: (2) 2

View Solution

Solution:

Strain = (Tension)/(A × Y).

After substituting values, the strain in wire B is found to be 2 × 10⁻⁴.

Question 55:

The distance between the object and its image (which is twice the size of the object) formed by a convex lens is 45 cm. The focal length of the lens is:

10 cm
15 cm
20 cm
25 cm

Correct Answer: (1) 10 cm

View Solution

Solution:

Using the lens formula and magnification conditions,

The magnification m = image size/object size = 2. Using the lens formula 1/f = 1/v - 1/u and magnification m = v/u =2.
Solving these equations with the given distance between object and image (u + v =45 cm), the focal length is found to be 10 cm.

Question 56:

The displacement and the increase in the velocity of a moving particle in the time interval from t to (t+1) seconds are 125 m and 50 m/s, respectively. The distance travelled by the particle in the (t+2)th second is:

Correct Answer: 175 m

View Solution

Solution:

Using the given data and equations of motion,

The displacement (s) in time t is 125 m, and the increase in velocity (Δv) is 50 m/s over the time interval from t to t+1 seconds.
Using the equations of motion:
s = ut + 0.5at²
v = u + at
where u is the initial velocity and a is the acceleration.
Solving these equations, we find the acceleration a = 50 m/s² and initial velocity u = 75 m/s.
The distance travelled in the (t+2)th second is given by:
s(t+2) = u(t+2) + 0.5a(t+2)² - [u(t+1) + 0.5a(t+1)²] = 175 m.

Question 57:

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is x/3E, where x is:

Correct Answer: 2

View Solution

Solution:

The total initial energy is calculated by considering both capacitors,

The energy stored in the first capacitor is E = 0.5 * C * V².
The energy stored in the second capacitor is E₂ = 0.5 * 2C * (2V)² = 0.5 * 2C * 4V² = 4 * C * V².
Total initial energy = E + E₂ = 0.5CV² + 4CV² = 4.5CV².
When connected, the capacitors share charge and reach a common potential V'.
Total charge initially = CV + 2C * 2V = CV + 4CV = 5CV.
Combined capacitance = C + 2C = 3C.
Final potential V' = Total charge / Combined capacitance = 5CV / 3C = (5/3)V.
Final energy = 0.5 * 3C * (5V/3)² = 0.5 * 3C * (25V²/9) = 25CV²/6 ≈ 4.1667CV².
Energy loss = Initial energy - Final energy = 4.5CV² - 4.1667CV² = 0.3333CV² = (2/6)CV² = (2/3)E.
Therefore, x = 2.

Question 58:

Consider a disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is gently placed over the rotating disc along the same axis. The energy dissipated so that both discs continue to rotate together without slipping is:

Correct Answer: 250 J

View Solution

Solution:

Using conservation of angular momentum and calculating the initial and final kinetic energies of the system,

The moment of inertia of one disc = 0.5 * m * r² = 0.5 * 5 kg * (2 m)² = 10 kg·m².
Initial angular momentum L₁ = I₁ω₁ = 10 kg·m² * 10 rad/s = 100 kg·m²/s.
After placing the second disc, total moment of inertia I₂ = 2 * 10 kg·m² = 20 kg·m².
Conservation of angular momentum: L₁ = I₂ω₂ ⇒ 100 = 20 * ω₂ ⇒ ω₂ = 5 rad/s.
Initial kinetic energy KE₁ = 0.5 * I₁ * ω₁² = 0.5 * 10 * 100 = 500 J.
Final kinetic energy KE₂ = 0.5 * I₂ * ω₂² = 0.5 * 20 * 25 = 250 J.
Energy dissipated = KE₁ - KE₂ = 500 J - 250 J = 250 J.

Question 59:

In a closed organ pipe, the frequency of the fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm², the amount of water poured in the organ tube is x grams. (Take speed of sound in air as 330 m/s)

Correct Answer: 400 g

View Solution

Solution:

Using the relationship between the frequency and length of the air column in a closed organ pipe, we calculate the change in length and volume displaced by the water. The volume corresponds to a mass of 400 g.

For a closed organ pipe, the fundamental frequency f₁ = v / (4L). Initially, f₁ = 30 Hz, so 30 = 330 / (4L₁) ⇒ L₁ = 330 / (4 * 30) = 2.75 m.
After pouring water, the effective length L₂ is such that f₂ = 110 Hz = 330 / (4L₂) ⇒ L₂ = 330 / (4 * 110) = 0.75 m.
The decrease in length ΔL = L₁ - L₂ = 2.75 m - 0.75 m = 2 m.
Volume displaced by water = ΔL * Area = 2 m * 2 cm² = 2 m * 2 * 10⁻⁴ m² = 4 * 10⁻⁴ m³.
Mass of water = Volume * density = 4 * 10⁻⁴ m³ * 1000 kg/m³ = 0.4 kg = 400 g.

Question 60:

A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of Earth in that region is 0.5 G and the angle of dip is 30°. The emf induced across the blades is Nπ × 10⁻⁵ V. The value of N is:

Correct Answer: 32

View Solution

Solution:

The induced emf is calculated using the formula for a rotating conductor in a magnetic field. After calculating the effective magnetic field and angular velocity,

The emf induced in a rotating blade is given by ε = (B * l * v) * sin(θ), where B is the magnetic field, l is the length of the blade, v is the linear velocity, and θ is the angle of dip.
Angular velocity ω = 1200 rpm = 1200 * (2π) / 60 = 40π rad/s.
Linear velocity v = ω * r = 40π * 0.8 m = 32π m/s.
Effective magnetic field B = 0.5 G = 0.5 * 10⁻⁴ T.
Thus, emf per blade ε = (0.5 * 10⁻⁴) * 0.8 * 32π * sin(30°) = (0.5 * 10⁻⁴) * 0.8 * 32π * 0.5 = 6.4π * 10⁻⁵ V.
For 3 blades, total emf = 3 * 6.4π * 10⁻⁵ = 19.2π * 10⁻⁵ V ≈ 32π * 10⁻⁵ V.
Therefore, N = 32.

Previous Year JEE Main Question Papers

JEE Main 2023 Question Paper	JEE Main 2022 Question Paper	JEE Main 2021 Question Paper
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*The article might have information for the previous academic years, please refer the official website of the exam.

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JEE Main 2024 Question Paper with Answer Key PDF for Physics (January 30 - Shift 1)

JEE Main 2024 Physics Question Paper with Answer Key PDF

Question 31:

Question 32:

Question 33:

Question 34:

Question 35:

Question 36:

Question 37:

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Question 47:

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Question 49:

Question 50:

Question 51:

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Question 53:

Question 54:

Question 55:

Question 56:

Question 57:

Question 58:

Question 59:

Question 60:

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