Posts

Hydrogen Spectrum JEE Concept — Which Series Has Bigger λ? ⚡

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❓ Concept 🎬 Largest Wavelength: Lyman vs Balmer — Hydrogen Spectrum in 59 Sec Hydrogen atom mein itni saari spectral lines hoti hain… par exam mein sawal hota hai 👇 👉 Sabse badi wavelength kis series se aati hai? 🌈🤔 Is concept ko samajh liya, toh ratio-based spectrum questions instant ho jaate hain ⚡ 🖼️ Concept Image ✍️ Short Explanation Yeh question ratta nahi , energy-gap logic ka test hota hai. Bas yeh yaad rakho: 👉 Badi wavelength = chhota energy gap . 🔹 Step 1 — Hydrogen Spectral Series Basics Har spectral series ka final energy level fixed hota hai: Lyman series → final level n = 1 n = 1 Balmer series → final level n = 2 n = 2 📌 Transitions higher level → final level ke beech hote hain. 🔹 Step 2 — Energy–Wavelength Relation (FOUNDATION 💯)** Photon energy: E = h c λ​ Iska matlab: λ ∝ 1 Δ E​ 📌 Largest wavelength ⇔ smallest energy difference 🔹 Step 3 — Smallest Energy Gap in Any Series Har spectral series mein: Sabse chhota...

Radius vs Energy in Magnetic Field — JEE Concept in 59 Sec 🔥

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❓ Question A particle of charge q q q , mass m m m and kinetic energy E E E enters a uniform magnetic field perpendicular to its velocity and undergoes a circular arc of radius r r r . Which of the following curves correctly represents the variation of r r r with E E E ? 🖼️ Question Image ✍️ Short Explanation This is a pure relation-based magnetism question . No numbers. No tricks. 👉 Sirf radius–energy dependence samajhni hai. 🔹 Step 1 — Magnetic force causes circular motion For a charged particle moving perpendicular to magnetic field B B B : q v B = m v 2 r So radius of circular path: r = m v q B 📌 Radius speed v v v par depend karta hai. 🔹 Step 2 — Connect speed with kinetic energy Kinetic energy: E = 1 2 m v 2 So: v = 2 E m v=\sqrt{\frac{2E}{m}} ​ Substitute in radius expression: r = m q B 2 E m r=\frac{m}{qB}\sqrt{\frac{2E}{m}} Hence: r ∝ E \boxed{r\propto \sqrt{E}} ​ ​ 🔹 Step 3 — Nature of r vs E graph (MOST IMPORTANT 🔥)** From: r ∝ E r\pro...

Displacement Current Concept — Dimensions Made Easy ⚡

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  ❓ Concept 🎬 EM Waves: Dimension of ε 0 d Φ E d t \varepsilon_0 \dfrac{d\Phi_E}{dt} ​ ​ in 59 Sec Electric flux ka time-rate diya ho… aur uske saath ε 0 \varepsilon_0 ​ multiply ho raha ho… 👉 Toh sawal sirf maths ka nahi hai — physics ka meaning samajhne ka hai ⚡🤔 🖼️ Concept Image ✍️ Short Explanation JEE is concept ko isliye pyaar karta hai kyunki yahin se Maxwell equations aur EM waves ka foundation banta hai. Agar tumhe yeh samajh aa gaya, 👉 dimensions + theory dono clear . 🔹 Step 1 — Electric Flux Meaning (FOUNDATION 💯)** Electric flux: Φ E = ∮ E ⃗ ⋅ d A ⃗ So dimensionally: Electric field E E Area A A [ Φ E ] = [ E ] × [ Area ] 📌 Matlab: Flux measures how much electric field passes through a surface . 🔹 Step 2 — Time Rate of Electric Flux d Φ E d t​ ​ Iska matlab: Electric field time ke saath change ho raha hai Ya surface ke through field flow vary kar raha hai 📌 Yeh quantity batati hai: Electric field kitni tezi se ...

Zener Diode Circuit Trick — Ammeter Reading in 59 Sec! 🔥

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❓ Question In the following circuit, the reading of the ammeter will be ? (Take Zener breakdown voltage = 4 V) 🖼️ Question Image ✍️ Short Solution This is a classic JEE Zener-diode regulator problem . No tricks, no confusion — bas Zener ON/OFF logic + Ohm’s law 🔥 🔹 Step 1 — Identify Zener Condition (MOST IMPORTANT 💯)** Given: Supply voltage = 12 V Zener breakdown voltage = 4 V Since: 12  V > 4  V 👉 Zener diode will be in breakdown (ON) mode . 📌 Golden rule: When Zener is ON, it clamps the voltage across it to Vz . So, V node = 4  V \boxed{V_{\text{node}} = 4\text{ V}} 🔹 Step 2 — Voltage Across Load Branch The 400 Ω resistor + ammeter branch is parallel to the Zener. Hence: V across 400 Ω = 4  V V_{\text{across 400 Ω}} = 4\text{ V} 🔹 Step 3 — Ammeter Reading (Direct Calculation 🔥)** Current through 400 Ω resistor: I = V R = 4 400 = 0.01  A = 10  mA I = \frac{V}{R} = \frac{4}{400} = 0.01\text{ ...

JEE Main Projectile Motion: (45°+α) vs (45°−α) Concept 💡

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❓ Concept 🎬 Projectile at (45° ± α) — Time of Flight Ratio in 59 Sec Dono projectiles same speed se chhode gaye… bas angle ek thoda upar , ek thoda neeche — phir bhi time of flight alag kyun hota hai? ⏳🤔 👉 Reason simple hai: Time of flight sinθ par depend karta hai , na ki range wale sin ⁡ 2 θ \sin 2\theta  par. 🖼️ Concept Image ✍️ Short Explanation Is concept ko pakad liya, toh JEE ke symmetry-based projectile questions seconds mein clear ho jaate hain 😎 Bas yaad rakho: time = vertical motion ka game . 🔹 Step 1 — Time of Flight Formula (FOUNDATION 💯)** Ground-to-ground projectile ke liye: T = 2 u sin ⁡ θ g 📌 Key takeaway: u u  same hai g g  same hai Time of flight sirf sin ⁡ θ \sin\theta  par depend karta hai 🔹 Step 2 — Angles Given θ 1 = 45 ∘ + α , θ 2 = 45 ∘ − α So: T 1 ∝ sin ⁡ ( 45 ∘ + α ) , T 2 ∝ sin ⁡ ( 45 ∘ − α ) 👉 Ratio nikaalne ke liye constants cancel ho jaate hain. 🔹 Step 3 — Golden Trig Identity (VERY...

Equivalent Resistance of Triangular Pyramid — JEE Trick in 60 Sec! 🔥

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  ❓ Question A wire of total resistance R is bent to form a triangular pyramid (tetrahedron) as shown in the figure, with each segment having the same length (and hence same resistance). The equivalent resistance between points A and B is given as R n . Find the value of n n . 🖼️ Question Image ✍️ Short Solution This is a classic JEE symmetry-based resistance problem . No Kirchhoff, no long equations — sirf symmetry + equivalent paths 🔥 🔹 Step 1 — Count number of equal segments A triangular pyramid (tetrahedron) has: 4 vertices 6 edges The wire of total resistance R R  is bent uniformly into these 6 equal segments . So resistance of each edge : r = R 6​ 🔹 Step 2 — Identify symmetry in the network We are asked resistance between A and B . Important symmetry observation 👇 Points C and D are symmetrically placed with respect to A and B So their potentials will be equal 📌 Hence: No current flows in the wire CD We can safely r...

Magnetic Susceptibility Trick: Field Change in Solenoid ⚡

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❓ Concept 🎬 Solenoid + Magnetic Material = New Magnetic Field | Concept in 59 Sec Solenoid ke andar vacuum/air ki jagah agar magnesium ya koi magnetic material bhar diya… 👉 Magnetic field badal jaati hai 🧲 Par kyun , aur kitni ? Yeh concept JEE mein baar-baar aata hai 🔥 🖼️ Concept Image ✍️ Short Explanation Is type ke questions mein calculation kam aur definition zyada important hoti hai. Agar tum μ, μr aur χ ka relation pakad loge, toh answer seconds mein nikal jaata hai 😎 🎯 HOOK (Write & Read First) “Agar solenoid ke andar vacuum ki jagah magnesium bhar diya… toh magnetic field badh jaati hai — par kyun aur kitni?” 🧲🤔 🔹 Step 1 — Magnetic Field of a Solenoid (in Air / Vacuum)** Long solenoid ke liye: B 0 = μ 0 n I Yahaan: n n  = turns per unit length I I  = current μ 0 \mu_0 ​ = permeability of free space 📌 Yeh reference field hota hai (without material). 🔹 Step 2 — When a Magnetic Material is Inserted Material daaln...