Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question In a hydrogen-like ion , the energy difference between the 2nd excited state and the ground state is 108.8 eV . Find the atomic number (Z) of the ion. 🖼️ Question Image ✍️ Short Solution This is a direct Bohr-model energy-level question . No tricks, no approximations — just n-value clarity + Z² scaling 🔥 🔹 Step 1 — Understand ‘2nd Excited State’ (MOST IMPORTANT 💯)** Energy levels in hydrogen-like ions: Ground state → n = 1 n = 1 1st excited state → n = 2 n = 2 2nd excited state → n = 3 n = 3  ✅ 📌 Many students mess up here — 2nd excited ≠ n = 2 🔹 Step 2 — Energy Formula for Hydrogen-like Ion Bohr energy level: E n = − 13.6   Z 2 n 2  eV Where: Z Z  = atomic number n n  = principal quantum number 🔹 Step 3 — Write Energies of Required Levels Ground state ( n = 1 n=1 ): E 1 = − 13.6 Z 2 2nd excited state ( n = 3 n=3 ): E 3 = − 13.6 Z 2 9 🔹 Step 4 — Energy Difference Given Energy difference: Δ E =...

❓ Concept 🎬 Resistance of a Triangular Pyramid — Concept in 59 Sec Wire ko 3D pyramid shape mein bend kar diya… sab segments same length, same material ke… aur puchh liya: 👉 A aur B ke beech equivalent resistance kya hoga? ⚡🤯 Is type ke questions calculation se nahi , symmetry se solve hote hain — bas wahi samajhna hai 🔥 🖼️ Concept Image ✍️ Short Explanation JEE mein jab bhi 3D resistance network dikhe, sabse pehle symmetry dhoondo . 👉 Symmetry mil gayi = problem aadhi solve 😎 🔹 Step 1 — Equal Length ⇒ Equal Resistance (FOUNDATION 💯)** Given: Same material Same length Uniform wire bent into a pyramid 👉 Har edge ka resistance same hoga. Total resistance R R   uniformly distributed hai poore network mein. 📌 Matlab: Har segment = identical resistor 🔹 Step 2 — Use Symmetry First (MOST IMPORTANT 🔥)** Triangular pyramid perfectly symmetric hota hai. Iska matlab: Jo points geometry mein symmetric hain Unka potential bhi same hoga (when current ...

❓ Question Two wires A and B are made of the same material . The ratio of their lengths is L A L B = 1 3 and the ratio of their diameters is d A d B = 2. If both wires are stretched using the same force , what is the ratio of their elongations ? 🖼️ Concept Image ✍️ Short Solution This is a direct Young’s modulus application . No heavy maths — bas elongation dependence samajh lo, answer khud nikal jaata hai 😎 🔹 Step 1 — Elongation Formula (FOUNDATION 💯)** For a wire under tension: Δ L = F L Y A \Delta L = \frac{FL}{YA} Where: F F  = applied force L L  = original length Y Y  = Young’s modulus A A  = cross-sectional area 📌 Same material ⇒ same Y Y Y 📌 Same force ⇒ same F F F So, Δ L ∝ L A \Delta L \propto \frac{L}{A} 🔹 Step 2 — Area in terms of Diameter Cross-sectional area: A = π d 2 4 A=\frac{\pi d^2}{4} So: A ∝ d 2 A \propto d^2 Hence: Δ L ∝ L d 2 \Delta L \propto \frac{L}{d^2} 🧠 Golden relation to remember : El...

❓ 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...

❓ 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...

  ❓ 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 ...

❓ 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{ ...

❓ 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...