Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 Mechanical Properties of Solids — Wire Elongation Concept in 59 Sec Same material… same force … phir bhi ek wire zyada stretch karegi! 👉 Reason length mein hai ya thickness mein? 🤯 Is concept ko pakad liya, toh ratio-based elasticity questions turant ho jaate hain 🔥 🖼️ Concept Image ✍️ Short Explanation Elasticity ke questions mein formula se zyada dependence matter karta hai. Yahan ek hi idea kaam karta hai: 👉 Elongation kis cheez par depend karti hai? 🔹 Step 1 — Elongation Formula (FOUNDATION 💯)** For a stretched wire: Δ L = F   L A   Y \Delta L=\frac{F\,L}{A\,Y} Where: F F  = applied force L L  = length A A  = cross-sectional area Y Y  = Young’s modulus 📌 Core idea: elongation depends on L and A . 🔹 Step 2 — Same Material ⇒ Same Y Y ** Given: both wires are of the same material . So: Y A = Y B ⇒ Y  cancels in ratio Y_A = Y_B \quad \Rightarrow \quad Y \text{ cancels in ratio} ...

  ❓ Question A uniform rod of total length 5L is bent at a right angle , such that one arm has length 2L and the other arm has length 3L . Find the position of the centre of mass of the system. 🖼️ Question Image ✍️ Short Solution This is a standard JEE Centre of Mass problem . No integration needed — treat each arm separately and use weighted average 🔥 🔹 Step 1 — Split the system into two rods Since the rod is uniform : Mass ∝ length Let: Rod 1: length = 2L , along x-axis Rod 2: length = 3L , along y-axis Assume: Total mass = M Mass per unit length = λ \lambda λ Then: Mass of rod 1 = 2 λ L 2\lambda L Mass of rod 2 = 3 λ L 3\lambda L 🔹 Step 2 — Locate centre of mass of each part For a uniform straight rod , COM lies at its midpoint. COM of rod 1 (along x-axis): ( x 1 , y 1 ) = ( L , 0 ) (x_1, y_1) = (L, 0) COM of rod 2 (along y-axis): ( x 2 , y 2 ) = ( 0 , 3 L 2 ) (x_2, y_2) = (0, \tfrac{3L}{2}) 🔹 Step 3 — Use Centre...

  ❓ Concept 🎬 Zener Diode in Circuit — Ammeter Reading Concept in 59 Sec Circuit mein Zener diode laga ho… aur ammeter ka reading poocha jaaye — toh sabse pehla sawal yeh hai 👇 👉 Zener ON hai ya OFF? ⚡🤔 Is concept ko clear rakhoge, toh numbers khud line mein aa jaate hain 😎 🖼️ Concept Image    ✍️ Short Explanation Zener-based questions calculation se pehle logic maangte hain. Ek baar ON/OFF decision ho gaya, baaki steps straightforward hote hain 🔥 🔹 Step 1 — Zener Diode: Basic Role (FOUNDATION 💯)** Zener diode ka kaam: Voltage regulation Reverse bias mein kaam karta hai Rules: If V applied ≥ V Z V_{\text{applied}} \ge V_Z ​ → Zener ON (breakdown) If V applied < V Z V_{\text{applied}} < V_Z ​ → Zener OFF 📌 First check = breakdown condition 🔹 Step 2 — Voltage Across Parallel Branch (MOST IMPORTANT 🔥)** Agar Zener ON ho jaata hai: Zener node voltage clamp kar deta hai Parallel branch (load + ammeter) ...

  ❓ Question A vehicle starting from rest attains a speed of 72 km/h after covering a distance of 100 m . If the mass of the vehicle is 500 kg , find the force exerted by the vehicle . 🖼️ Concept Image ✍️ Short Solution This is a pure kinematics + Newton’s second law question. No tricks, no time calculation — just v² = u² + 2as and F = ma 💯 🔹 Step 1 — Convert Speed into SI Units (MOST IMPORTANT 💯) Given speed is in km/h , but formulas need m/s . 72  km/h = 20  m/s 📌 NTSE aur school exams mein yahin sabse zyada silly mistakes hoti hain. 🔹 Step 2 — Write Known Values Clearly Initial speed: u = 0 (starting from rest) u = 0 \quad \text{(starting from rest)} Final speed: v = 20  m/s v = 20 \text{ m/s} Distance covered: s = 100  m s = 100 \text{ m} Mass of vehicle: m = 500  kg m = 500 \text{ kg} 🔹 Step 3 — Use Equation of Motion Since u, v, s are given, use: v 2 = u 2 + 2 a s v^2 = u^2 + 2as Substitute values: ( 20 ) ...

  ❓ 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...