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 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...