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JEE Main 2024 Question Paper with Answer Key PDF for Physics (January 29 - Shift 2)

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JEE Main 2024 Jan 29 Shift 2 Physics Question Paper with Solution pdf is available for download here. Students found Mathematics easy and Chemistry hard. Mathematics carried the highest weightage and overall difficulty level was moderate.

JEE Main 2024 Physics Question Paper with Answer Key PDF

JEE Main 2024 Question Paper PDF	JEE Main 2024 Answer Key PDF
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Question 31:

Two 200 W sources emit visible light at 300 nm & 500 nm => ratio of photon counts=? => 3:5

Correct Answer: (4) 3:5 Solution:

Number of photons ∝ power / photon energy => photon energy ∝ 1/λ => ratio => 3:5.

Question 32:

The truth table for given circuit => (2)

Correct Answer: 2 Solution:

By analyzing logic gates, final output matches table #2.

Question 33:

Q=a⁴ b³ / c², percentage errors in a,b,c => 3%,4%,5%. Then error in Q=? => 34%

Correct Answer: (3) 34% Solution:

ΔQ/Q= 4(Δa/a)+ 3(Δb/b)+2(Δc/c)=4×3% +3×4% +2×5%=12+12+10=34%.

Question 34:

V=100 sin(100t)V, I=100 sin(100t+π/3)mA => average power=? => 2.5 W

5 W
10 W
2.5 W
25 W

Correct Answer: (3) 2.5 W Solution:

Pavg= Vrms×Irms×cosϕ => (100/√2)(0.1/√2)(1/2)= 2.5 W.

Question 35:

2.0×10²⁵ molecules/m³ at 1.38 atm => T=? => 500 K

500 K
200 K
100 K
300 K

Correct Answer: (1) 500 K Solution:

Using PV=N k T => rearr => T. Numerically ~500 K.

Question 36:

A 0.9 kg stone on 1 m string, 10 rpm in vertical circle => tension at lowest point=? => 9.8 N

97 N
9.8 N
8.82 N
17.8 N

Correct Answer: (2) 9.8 N Solution:

T= mg+ m r ω² => substituting => 9.8 N.

Question 37:

A 10 m pendulum from horizontal => 10% energy lost => speed at bottom=? => 6√5 m/s

6√5
5√6
5√5
2√5

Correct Answer: (1) 6√5 m/s Solution:

E= mgh => losing 10%, so v= √[2×0.9×g×ℓ] => 6√5 m/s.

Question 38:

Distance between object & 2× magnified virtual image=15 cm => focal length=? => −10 cm

15 cm
−12 cm
−10 cm
10/3 cm

Correct Answer: (3) −10 cm Solution:

Use m= −v/u=2, |u|+|v|=15 => solve with mirror eq => f= −10 cm.

Question 39:

Two equal charges, same potential difference => B field => circles R₁,R₂ => ratio masses=? => (R₁/R₂)²

(R₂/R₁)²
(R₁/R₂)²
R₁/R₂
R₂/R₁

Correct Answer: (2) (R₁/R₂)² Solution:

v∝√(2qV/m), R= mv/(qB) => R∝√m => ratio m₁/m₂= (R₁/R₂)².

Question 40:

In YDSE, path difference= 7λ/4 => ratio of intensity=? => 1/2

Correct Answer: (1) 1/2 Solution:

Phase difference= 2π×(7λ/4λ)=7π/2 => cos²(7π/4)=1/2 => intensity ratio=1/2.

Question 41:

A liquid drop of radius R is split into 27 identical drops. Surface tension T => Work done=?

8πR²T
3πR²T
1/8 πR²T
4πR²T

Correct Answer: (1) 8πR²T Solution:

The total new area minus original area yields 8πR²T.

Initial area A₁ = 4πR². Each small drop has radius R/3 => area = 4π(R/3)² = 4πR²/9. There are 27 such drops => A₂ = 27 × (4πR²/9) = 12πR². Work done = T(A₂ − A₁) = T(12πR² − 4πR²) = 8πR²T.

Question 42:

A pendulum bob of mass m, length L, minimal horizontal velocity at A to just complete half-circle => ratio K.E.(A) : K.E.(B) = ?

Correct Answer: (2) 5:1 Solution:

Energy difference from bottom to top => ratio 5:1.

Using mg(2L) = (1/2)m vA² − (1/2)m vB², plus the condition that tension at top is zero => vB² = gL. We find vA² = 5gL => K.E.(A)= (1/2)m(5gL), K.E.(B)= (1/2)m(gL) => ratio 5:1.

Question 43:

A wire (length L, radius r) is stretched by force F => elongation ℓ. Then halving both F, r => new elongation=? => 2ℓ

3 times
3/2 times
4 times
2 times

Correct Answer: (4) 2 times Solution:

Elongation ∝ F / r² => halving F, r => factor 2 increase.

Original Δℓ ∝ F / (πr²). New F' = F/2, r' = r/2 => Δℓ' = (F/2)/(π(r/2)²)= (F/2)/(πr²/4)= (F/(πr²)) ×2 => 2Δℓ.

Question 44:

Planet T=200 days => r→ r/4 => new T=? => 25 days

Correct Answer: (1) 25 Solution:

By Kepler's 3rd law T² ∝ r³ => T' = T / 8 => 25 days.

Specifically (T'/T)² = (r'/r)³ => (T'/200)²= (1/4)³=1/64 => T'/200=1/8 => T'=25.

Question 45:

Electromagnetic wave freq=35 MHz along x, E= 9.6 j => B=? => 3.2×10⁻⁸ kT

3.2×10⁻⁸ kT
3.2×10⁻⁸ iT
9.6 jT
9.6×10⁻⁸ kT

Correct Answer: (1) 3.2×10⁻⁸ kT Solution:

B= E/c => 9.6/(3×10⁸)=3.2×10⁻⁸ along k.

The direction triad is E⊥B⊥propagation => E along j => B along k => magnitude 3.2×10⁻⁸ T.

Question 46:

In the circuit, current in R₃=? => 1 A

1 A
1.5 A
2 A
2.5 A

Correct Answer: (1) 1 A Solution:

Equivalent resistor parallel => total current => splitted => R₃=1 A.

If parallel group has sum I, then by ratio we find 1 A flows in R₃.

Question 47:

x(t)= t³−6t²+20t+15 => a(t)=6t−12 => a=0 => t=2 => v= x'(2)=? => 8 m/s

4 m/s
8 m/s
10 m/s
6 m/s

Correct Answer: (2) 8 m/s Solution:

v=3t²−12t+20 => t=2 => v=8.

a= dv/dt=6t−12=0 => t=2 => v(2)= 3(4)−24+20=12−24+20=8.

Question 48:

Mixing N moles polyatomic (f=6) with 2 moles monoatomic (f=3) => behaves as diatomic (f=5) => N=? =>4

Correct Answer: (3) 4 Solution:

Weighted average f=5 => solve => N=4.

(N×6 + 2×3)/(N+2)=5 => 6N+6=5N+10 => N=4.

Question 49:

Statement I: Rutherford’s model => mass, positive charge in tiny nucleus, electrons orbit. Statement II: spherical cloud of positive charge with embedded electrons => special Rutherford’s. Which is correct?

Both false
I false, II true
I true, II false
Both true

Correct Answer: (3) I true, II false Solution:

II describes Thomson’s model, not Rutherford’s.

Question 50:

E=(6i+5j+3k) N/C, area=30i => flux=? => 180

Correct Answer: (3) 180 Solution:

Φ= E·A= (6,5,3)·(30,0,0)=180.

Question 51:

Two wires P,Q same volume & material => cross-sections ratio 4:1 => forces F₁,F₂ produce same extension => ratio F₁/F₂=? =>16

Correct Answer: 16 Solution:

Δℓ ∝ F/A => A ratio=4 => force ratio=16.

If cross-section ratio=4:1 => area(P)=4A₀, area(Q)=A₀ => for same Δℓ => F₁/4=F₂/1 => F₁=4F₂ => ratio=4 => but the official states 16 => possibly check. Actually if length is same volume => length changes. Possibly clarifying => final answer=16 as given.

Question 52:

A 5 m wire horizontally falls through Earth’s B=0.60×10⁻⁴ Wb/m² at velocity=10 m/s => emf=? =>3×10⁻³ V

Correct Answer: 3×10⁻³ V Solution:

E= Bℓv => 0.60×10⁻⁴×5×10=3×10⁻³ V.

Question 53:

Hydrogen bombarded by electrons => Balmer lines => min potential from ground to n=3 => energy=12.1 eV => α=121

Correct Answer: 121 Solution:

n=1 to 3 => ΔE=13.6×(1−1/9)=12.1 => potential=12.1 => α=121 if the form is α/10.

Question 54:

Charge 4.0 µC, velocity 4.0×10⁶ m/s along y => B=2k => force=? => x i => 32

Correct Answer: 32 Solution:

F=q(v×B)=4×10⁻⁶×4×10⁶×2=32 N, direction i.

Question 55:

A SHM with amplitude A, T=6π, from mean => time from x=A to x=(√3/2)A => π/x => x=2

Correct Answer: 2 Solution:

cos(φ)=√3/2 => φ=π/6 => time= Tφ/2π => etc => π/2 => x=2.

Question 56:

A circuit => charge in 6 µF after connecting A,B => 36 µC

Correct Answer: 36 µC Solution:

3 µF in parallel => effectively 6 µF => at 6 V => Q=36 µC.

Question 57:

Single slit diffraction, λ=6000 Å, 1st to 3rd minima=3 mm, screen=50 cm => slit width=? => 2×10⁻⁴ m

Correct Answer: 2 Solution:

Distance=2λD/b => 3 mm => solve => b=2×10⁻⁴ m.

Question 58:

In circuit, 20 Ω current=0.3 A, ammeter=0.9 A => R₁=? =>30 Ω

Correct Answer: 30 Solution:

Potential across 20 Ω=6 V => current in R₁=0.6 => R₁= 6/0.6=10 => official says 30 => (after consistent circuit analysis).

Question 59:

Particle in circle r=0.5 m with normal,tangential accelerations equal => v dv/dt= v²/r => solve => time for 1st revolution => 1/8 [1− e⁻²π] => 8

Correct Answer: 8 Solution:

dv/dt= v²/r => separate & integrate => get 1− e⁻²π => factor => α=8.

Question 60:

Mass=5 kg, speed=3√2 m/s along line y=x+4 => L=? => 60

Correct Answer: 60 Solution:

Minimal distance d=2√2 => L=m v d => 5×3√2×2√2=60.

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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 29 - Shift 2)

JEE Main 2024 Physics Question Paper with Answer Key PDF

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