

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.
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
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.
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.
Every object exhibits diamagnetism.
Hence, diamagnetism is called the universal magnetic property.
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
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
When a wave reflects from rarer → denser medium:
The phase change = 180°
Crest reflects as trough
This is a previously asked exam question (PYQ).
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)
Graph is plotted between:
y-axis: De Broglie wavelength (λ)
x-axis: 1 / √V
Using:
λ=h2mqVλ = \frac{h}{\sqrt{2m qV}}λ=2mqVh
Slope:
slope∝1m\text{slope} ∝ \frac{1}{\sqrt{m}}slope∝m1
Therefore:
The line with smaller slope corresponds to larger mass.
From the transcript: Particle A has greater mass than B.
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=2mKh
Substituting K:
λd=λ2mcλ_d = \sqrt{\frac{λ}{2mc}}λd=2mcλ
The transcript shows the entire step-by-step algebra leading to this expression.
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.
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.