Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A wire of: Length = 10 cm Diameter = 0.5 mm Temperature = 1727°C Power radiated = 94.2 W Find the emissivity (e) of the wire. 🖼 Question Image ✍️ Short Solution This is a Stefan–Boltzmann Law question. Golden formula: P = e σ A T 4 P = e \sigma A T^4 👉 Convert temperature to Kelvin 👉 Find curved surface area 👉 Substitute and solve for e 🔷 Step 1 — Convert All Units (MOST IMPORTANT 💯) Length: L = 10   cm = 0.1   m L = 10\,\text{cm} = 0.1\,\text{m} Diameter: d = 0.5   mm = 0.0005   m d = 0.5\,\text{mm} = 0.0005\,\text{m} Radius: r = 0.00025   m r = 0.00025\,\text{m} Temperature: T = 1727 + 273 = 2000   K T = 1727 + 273 = 2000\,K 🔷 Step 2 — Surface Area of Wire Wire is cylindrical → Use curved surface area: A = 2 π r L A = 2\pi r L A = 2 π ( 0.00025 ) ( 0.1 ) A = 2\pi (0.00025)(0.1) A = 1.57 × 10 − 4   m 2 A = 1.57 \times 10^{-4} \, m^2 🔷 Step 3 — Apply Stefan’s Law P = e σ A T 4 P = e \sigma A T^4 Given: P = 94.2 P = 94.2 σ = 5.67 ×...