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JEE Main 2025: Calculate Percentile from Raw Marks Using NTA Normalisation

A clear breakdown of high-probability Physics questions for board exams, covering photoelectric effect, LCR circuits, De Broglie wavelength, electric fields, and key formulas.
authorImageVarnika Srivastava17 Nov, 2025
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JEE Main 2025

Boards are approaching, and this is the peak moment when every student looks for content that can genuinely boost their score. This session brings together high-probability Physics questions and concepts that are extremely likely to appear in the exam. The explanations here are based strictly on the transcript and preserve every technical detail.

1. Photoelectric Effect: Guaranteed Question Area

A question from the photoelectric effect is almost certain to appear. The common pattern in such questions revolves around:

  • Energy of incident photon: E = hν

  • Maximum kinetic energy: Kₘₐₓ = hν – Φ

  • Work function relations

  • Threshold frequency

  • Photon emission per second

Example Problem: Photon Emission from a 100 W Bulb

Given:

  • Wavelength = 540 nm

  • Power = 100 W

If the bulb emits photons of frequency ν, the number of photons emitted per second is:

nt=Phν\frac{n}{t} = \frac{P}{hν}tn​=hνP​

Where:

  • ν=cλ=3×108540×10−9=5.5×1014 Hzν = \frac{c}{\lambda} = \frac{3×10^8}{540×10^{-9}} = 5.5×10^{14} \text{ Hz}ν=λc​=540×10−93×108​=5.5×1014 Hz

  • Using Planck’s constant h=6×10−34h = 6×10^{-34}h=6×10−34
    Final value (as derived in the transcript):

3.03×1020 photons per second3.03 × 10^{20} \text{ photons per second}3.03×1020 photons per second

This is a high-probability exam question type.

2. Capacitor Fully Charged: Conduction & Displacement Current

When a capacitor becomes fully charged:

  • Conduction current = 0

  • Displacement current = 0

Reason: After full charging, there is no change in electric flux, hence no displacement current exists.

3. Universal Property of Magnetic Substances

Every object exhibits diamagnetism.
Hence, diamagnetism is called the universal magnetic property.

4. Photoelectric Effect: Kinetic Energy Ratio Question

For metals A and B with threshold frequencies ν/2 and ν/3:

Work functions:

  • Φₐ = hν / 2

  • Φᵦ = hν / 3
    Kinetic energies:

Ka=hν−hν/2=hν/2Kₐ = hν - hν/2 = hν/2Ka​=hν−hν/2=hν/2 Kβ=hν−hν/3=2hν/3Kᵦ = hν - hν/3 = 2hν/3Kβ​=hν−hν/3=2hν/3

Ratio:

Ka:Kβ=12:23=3:4Kₐ:Kᵦ = \frac{1}{2} : \frac{2}{3} = 3 : 4Ka​:Kβ​=21​:32​=3:4

Correct answer: 3 : 4

5. Surface Charge Density: Hollow Conducting Sphere

Given:

  • Inner radius = r

  • Outer radius = 2r

  • Charge +q placed at the center

Induced charges:

  • Inner surface: –q

  • Outer surface: +q

Surface charge densities:

  • Inner: −q4πr2\frac{-q}{4πr^2}4πr2−q​

  • Outer: +q4π(2r)2=q16πr2\frac{+q}{4π(2r)^2} = \frac{q}{16πr^2}4π(2r)2+q​=16πr2q​

Ratio:

−q/4πr2q/16πr2=−4:1\frac{-q/4πr^2}{q/16πr^2} = -4 : 1q/16πr2−q/4πr2​=−4:1

Correct answer: –4 : 1

6. Reflection from Rarer to Denser Medium

When a wave reflects from rarer → denser medium:

  • The phase change = 180°

  • Crest reflects as trough

This is a previously asked exam question (PYQ).

7. Electric Field Line Diagram: Identifying Charges

Rules:

  • Field lines originate from positive and terminate at negative.

From the described field figure:

  • A and C are positive (lines emerging)

  • B is negative (lines converging)

  • Charge with most field lines = C (largest magnitude)

8. De Broglie Wavelength vs 1/√V Graph

Graph is plotted between:

  • y-axis: De Broglie wavelength (λ)

  • x-axis: 1 / √V

Using:

λ=h2mqVλ = \frac{h}{\sqrt{2m qV}}λ=2mqV​h​

Slope:

slope∝1m\text{slope} ∝ \frac{1}{\sqrt{m}}slope∝m​1​

Therefore:

  • The line with smaller slope corresponds to larger mass.

  • From the transcript: Particle A has greater mass than B.

9. De Broglie Wavelength of Electrons from X-rays

Given:

  • X-ray wavelength = λ

  • Work function negligible

  • So kinetic energy = hν=hcλhν = \frac{hc}{λ}hν=λhc​

Using:

λd=h2mKλ_d = \frac{h}{\sqrt{2mK}}λd​=2mK​h​

Substituting K:

λd=λ2mcλ_d = \sqrt{\frac{λ}{2mc}}λd​=2mcλ​​

The transcript shows the entire step-by-step algebra leading to this expression.

10. LCR Circuit: Peak Current in Series Circuit

Given:

  • Peak voltage = 300 V

  • Angular frequency = 400 rad/s

  • R, L, C values given in the transcript

Peak current:

I0=V0ZI_0 = \frac{V_0}{Z}I0​=ZV0​​

Impedance:

Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}Z=R2+(XL​−XC​)2​

Where:

  • XL=ωL=8 ΩX_L = ωL = 8 \, ΩXL​=ωL=8Ω

  • XC=1ωCX_C = \frac{1}{ωC}XC​=ωC1​ (value solved later in transcript)

This is a 100% sure-shot topic—LCR is almost guaranteed to appear.

Conclusion

This transcript-based blog covers all key high-probability Physics topics:

  • Photoelectric effect

  • LCR circuits

  • De Broglie wavelength

  • Magnetic properties

  • Electric field line interpretation

  • Electrostatics with conductors

These concepts repeatedly appear in board exams.
Practice every example given above—each one has strong chances of being tested.

 

JEE Main 2025 FAQs

How is percentile calculated in JEE Main?

Percentile is derived using NTA’s normalisation formula based on your relative performance.

Do raw marks directly decide JEE Main rank?

No, ranks are based on percentiles, not raw marks.

Why does JEE Main use normalisation?

To adjust for difficulty variations across different shifts.

Can two students with the same raw marks get different percentiles?

Yes, because percentile depends on shift-wise performance.
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